{"id":1135,"date":"2025-06-30T16:37:09","date_gmt":"2025-06-30T16:37:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1135"},"modified":"2025-08-14T15:56:46","modified_gmt":"2025-08-14T15:56:46","slug":"sequences-and-series-foundations-get-stronger","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-series-foundations-get-stronger\/","title":{"raw":"Sequences and Series Foundations: Get Stronger","rendered":"Sequences and Series Foundations: Get Stronger"},"content":{"raw":"<h2><span data-sheets-root=\"1\">Sequences and Their Properties<\/span><\/h2>\r\n<p dir=\"ltr\" data-pm-slice=\"1 3 []\"><strong>Find the first six terms of each of the following sequences (1-2), starting with<\/strong> [latex]n=1[\/latex].<\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" data-tight=\"true\">\r\n \t<li>[latex]a_n = 1 + (-1)^n[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\r\n \t<li>[latex]a_1 = 1[\/latex] and [latex]a_n = a_{n-1} + n[\/latex] for [latex]n \\ge 2[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Find an explicit formula for<\/strong> [latex]a_n[\/latex] <strong>for each of the following sequences (3-6).<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"3\" data-tight=\"true\">\r\n \t<li>[latex]a_1 = 1[\/latex] and [latex]a_n = a_{n-1} + n[\/latex] for [latex]n \\ge 2[\/latex].<\/li>\r\n \t<li>Find a formula [latex]a_n[\/latex] for the [latex]n[\/latex]th term of the arithmetic sequence whose first term is [latex]a_1 = -3[\/latex] such that [latex]a_{n-1} - a_n = 4[\/latex] for [latex]n \\ge 1[\/latex].<\/li>\r\n \t<li>Find a formula [latex]a_n[\/latex] for the [latex]n[\/latex]th term of the geometric sequence whose first term is [latex]a_1 = 3[\/latex] such that [latex]\\dfrac{a_{n+1}}{a_n} = \\dfrac{1}{10}[\/latex] for [latex]n \\ge 1[\/latex].<\/li>\r\n \t<li>Find an explicit formula for the [latex]n[\/latex]th term of the sequence satisfying [latex]a_1 = 0[\/latex] and [latex]a_n = 2 a_{n-1} + 1[\/latex] for [latex]n \\ge 2[\/latex].<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Find a formula for the general term<\/strong> [latex]a_n[\/latex] <strong>of each of the following sequence.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"7\" data-tight=\"true\">\r\n \t<li>[latex]{ 1, -\\dfrac{1}{3}, \\dfrac{1}{5}, -\\dfrac{1}{7}, \\ldots }[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Find a function<\/strong> [latex]f(n)[\/latex] <strong>that identifies the<\/strong> [latex]n[\/latex]th <strong>term<\/strong> [latex]a_n[\/latex] <strong>of the following recursively defined sequences (8-9), as<\/strong> [latex]a_n = f(n)[\/latex].<\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"8\" data-tight=\"true\">\r\n \t<li>[latex]a_1 = 2[\/latex] and [latex]a_{n+1} = 2 a_n[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\r\n \t<li>[latex]a_1 = 2[\/latex] and [latex]a_{n+1} = \\dfrac{(n+1) a_n}{2}[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Plot the first<\/strong> [latex]N[\/latex] <strong>terms of each sequence (10-12). State whether the graphical evidence suggests that the sequence converges or diverges.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"10\" data-tight=\"true\">\r\n \t<li>[latex]a_1 = 1[\/latex], [latex]a_2 = 2[\/latex], and for [latex]n \\ge 2[\/latex], [latex]a_n = \\dfrac{1}{2} (a_{n-1} + a_{n-2})[\/latex]; [latex]N = 30[\/latex]<\/li>\r\n \t<li>[latex]a_1 = 1[\/latex], [latex]a_2 = 2[\/latex], and for [latex]n \\ge 3[\/latex], [latex]a_n = \\sqrt{a_{n-1} a_{n-2}}[\/latex]; [latex]N = 30[\/latex]<\/li>\r\n \t<li>latex]a_n = \\cos{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Suppose that<\/strong> [latex]\\lim_{n \\to \\infty} a_n = 1[\/latex], [latex]\\lim_{n \\to \\infty} b_n = -1[\/latex], <strong>and<\/strong> [latex]0 &lt; -b_n &lt; a_n[\/latex] <strong>for all<\/strong> [latex]n[\/latex]. <strong>Evaluate each of the following limits (13-14), or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"13\" data-tight=\"true\">\r\n \t<li>[latex]\\lim_{n \\to \\infty} (3 a_n - 4 b_n)[\/latex]<\/li>\r\n \t<li>[latex]\\lim_{n \\to \\infty} \\dfrac{a_n + b_n}{a_n - b_n}[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Find the limit of each of the following sequences (15-16), using L\u2019H\u00f4pital\u2019s rule when appropriate.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"15\" data-tight=\"true\">\r\n \t<li>[latex]\\dfrac{n^2}{2^n}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>For the following sequences (17-20), whose<\/strong> [latex]n[\/latex]th <strong>terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"17\" data-tight=\"true\">\r\n \t<li>[latex]\\dfrac{n}{2^n}[\/latex], [latex]n \\ge 2[\/latex]<\/li>\r\n \t<li>[latex]\\sin{n}[\/latex]<\/li>\r\n \t<li>[latex]n^{\\dfrac{1}{n}}[\/latex], [latex]n \\ge 3[\/latex]<\/li>\r\n \t<li>[latex]\\tan{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"21\" data-tight=\"true\">\r\n \t<li>[latex]a_1 = 3[\/latex], [latex]a_n = \\sqrt{2 a_{n-1}}[\/latex], [latex]n = 2, 3, \\ldots[\/latex].<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>Use the Squeeze Theorem to find the limit of each of the following sequences (22-23).<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"22\" data-tight=\"true\">\r\n \t<li>[latex]\\dfrac{\\cos\\left(\\dfrac{1}{n}\\right) - 1}{\\dfrac{1}{n}}[\/latex]<\/li>\r\n \t<li>[latex]a_n = \\sin{n} \\sin\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p dir=\"ltr\"><strong>For the following exercises (24-28), determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.<\/strong><\/p>\r\n\r\n<ol class=\"tight\" dir=\"ltr\" start=\"24\" data-tight=\"true\">\r\n \t<li>[latex]a_n = (2n)^{\\dfrac{1}{n}} - n^{\\dfrac{1}{n}}[\/latex]<\/li>\r\n \t<li>[latex]a_n = \\left(1 - \\dfrac{2}{n}\\right)^n[\/latex]<\/li>\r\n \t<li>[latex]a_n = \\dfrac{2^n + 3^n}{4^n}[\/latex]<\/li>\r\n \t<li>[latex]a_n = \\dfrac{(1000)^n}{n!}[\/latex]<\/li>\r\n \t<li>[latex]a_n = \\dfrac{(n!)^2}{(2n)!}[\/latex]<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (29-30), solve each problem.<\/strong>\r\n<ol class=\"tight\" dir=\"ltr\" start=\"29\" data-tight=\"true\">\r\n \t<li>A lake initially contains [latex]2000[\/latex] fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by [latex]6%[\/latex] each month. However, factoring in all causes, [latex]150[\/latex] fish are lost each month.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Explain why the fish population after [latex]n[\/latex] months is modeled by [latex]P_n = 1.06 P_{n-1} - 150[\/latex] with [latex]P_0 = 2000[\/latex].<\/li>\r\n \t<li>How many fish will be in the pond after one year?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A student takes out a college loan of [latex]$10,000[\/latex] at an annual percentage rate of [latex]6 \\%[\/latex], compounded monthly.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>If the student makes payments of [latex]$100[\/latex] per month, how much does the student owe after [latex]12[\/latex] months?<\/li>\r\n \t<li>After how many months will the loan be paid off?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Introduction to Series<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>Using sigma notation, write the following expressions (1-2) as infinite series.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li>[latex]1+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\cdots[\/latex]<\/li>\r\n \t<li>[latex]1-\\dfrac{1}{2}+\\dfrac{1}{3}-\\dfrac{1}{4}+\\ldots[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (3-4), compute the first four partial sums [latex]{S}_{1},\\ldots,{S}_{4}[\/latex] for the series having [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] starting with [latex]n=1[\/latex] as follows.<\/strong><\/p>\r\n\r\n<ol start=\"3\">\r\n \t<li>[latex]{a}_{n}=n[\/latex]<\/li>\r\n \t<li>[latex]{a}_{n}=\\sin\\left(\\dfrac{n\\pi}{2}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (5-6), compute the general term [latex]{a}_{n}[\/latex] of the series with the given partial sum [latex]{S}_{n}[\/latex]. If the sequence of partial sums converges, find its limit [latex]S[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"5\">\r\n \t<li>[latex]{S}_{n}=1-\\dfrac{1}{n}[\/latex], [latex]n\\ge 2[\/latex]<\/li>\r\n \t<li>[latex]{S}_{n}=\\sqrt{n},n\\ge 2[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For each of the following series (7-8), use the sequence of partial sums to determine whether the series converges or diverges.<\/strong><\/p>\r\n\r\n<ol start=\"7\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{n+2}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\left(n+1\\right)\\left(n+2\\right)}[\/latex] (<em>Hint<\/em>: Use a partial fraction decomposition like that for [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n\\left(n+1\\right)}.[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }a_{n}=1[\/latex], that [latex]\\displaystyle\\sum _{n=1}^{\\infty }b_{n}=-1[\/latex], that [latex]a_{1}=2[\/latex], and [latex]b_{1}=-3[\/latex]. Find the sum of the indicated series (9-10).<\/strong><\/p>\r\n\r\n<ol start=\"9\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(a_{n}+b_{n}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\left(a_{n}-b_{n}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (11-13), state whether the given series converges and explain why.<\/strong><\/p>\r\n\r\n<ol start=\"11\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n+1000}[\/latex] (Hint: Rewrite using a change of index.)<\/li>\r\n \t<li>[latex]1+\\dfrac{1}{10}+\\dfrac{1}{100}+\\dfrac{1}{1000}+\\cdots[\/latex]<\/li>\r\n \t<li>[latex]1+\\dfrac{\\pi }{\\text{e}^{2}}+\\dfrac{{\\pi }^{2}}{{e}^{4}}+\\dfrac{{\\pi }^{3}}{{e}^{6}}+\\dfrac{{\\pi }^{4}}{{e}^{8}}+\\cdots[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For [latex]{a}_{n}[\/latex] as follows, write the sum as a geometric series of the form [latex]\\displaystyle\\sum {n=1}^{\\infty }a{r}^{n}[\/latex]. State whether the series converges and if it does, find the value of [latex]\\displaystyle\\sum {a}{n}[\/latex].<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"14\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}{1}=-1[\/latex] and [latex]\\dfrac{{a}{n}}{{a}_{n+1}}=-5[\/latex] for [latex]n\\ge 1[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}{1}=10[\/latex] and [latex]\\dfrac{{a}{n}}{{a}_{n+1}}=10[\/latex] for [latex]n\\ge 1[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Use the identity [latex]\\dfrac{1}{1-y}=\\displaystyle\\sum _{n=0}^{\\infty }{y}^{n}[\/latex] to express the following functions (16-17) as a geometric series in the indicated term.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"16\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{x}{1+x}[\/latex] in [latex]x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{1+{\\sin}^{2}x}[\/latex] in [latex]\\sin{x}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Evaluate the following (18-18) telescoping series or state whether the series diverges.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"18\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\dfrac{1}{n}}-{2}^{\\dfrac{1}{\\left(n+1\\right)}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\sqrt{n}-\\sqrt{n+1}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Express the following (<\/strong><strong>20-21) series as a telescoping sum and evaluate its [latex]n[\/latex]th partial sum.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"20\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\text{ln}\\left(\\dfrac{n}{n+1}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{\\text{ln}\\left(1{+}\\dfrac{1}{n}\\right)}{\\text{ln}n\\text{ln}\\left(n+1\\right)}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (22-25), solve each problem. A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"22\">\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]{a}_{n}=f\\left(n\\right)-2f\\left(n+1\\right)+f\\left(n+2\\right)[\/latex], in which [latex]f\\left(n\\right)\\to 0[\/latex] as [latex]n\\to \\infty [\/latex]. Find [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}_{n}={c}_{0}f\\left(n\\right)+{c}_{1}f\\left(n+1\\right)+{c}_{2}f\\left(n+2\\right)+{c}_{3}f\\left(n+3\\right)+{c}_{4}f\\left(n+4\\right)[\/latex], where [latex]f\\left(n\\right)\\to 0[\/latex] as [latex]n\\to \\infty [\/latex]. Find a condition on the coefficients [latex]{c}_{0}\\text{...},{c}_{4}[\/latex] that make this a general telescoping series.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{2}{{n}^{3}-n}[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Define a sequence [latex]{t}_{k}=\\displaystyle\\sum _{n=1}^{k - 1}\\left(\\dfrac{1}{k}\\right)-\\text{ln}k[\/latex]. Use the graph of [latex]\\dfrac{1}{x}[\/latex] to verify that [latex]{t}_{k}[\/latex] is increasing. Plot [latex]{t}_{k}[\/latex] for [latex]k=1\\text{...}100[\/latex] and state whether it appears that the sequence converges.<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Each of the following (26-27) infinite series converges to the given multiple of [latex]\\pi [\/latex] or [latex]\\dfrac{1}{\\pi}[\/latex]. In each case, find the minimum value of [latex]N[\/latex] such that the [latex]N\\text{th}[\/latex] partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to [latex]15[\/latex] decimals place, [latex]\\pi =3.141592653589793...[\/latex].<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"26\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\pi =-3+\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n{2}^{n}n{\\text{!}}^{2}}{\\left(2n\\right)\\text{!}}[\/latex], error [latex]&lt;0.0001[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{9801}{2\\pi }=\\dfrac{4}{9801}\\displaystyle\\sum _{k=0}^{\\infty }\\dfrac{\\left(4k\\right)\\text{!}\\left(1103+26390k\\right)}{{\\left(k\\text{!}\\right)}^{4}{396}^{4k}}[\/latex], error [latex]&lt;{10}^{-12}[\/latex]<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (28-30), solve each problem.<\/strong>\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"28\">\r\n \t<li class=\"whitespace-normal break-words\">A fair coin is one that has probability [latex]\\dfrac{1}{2}[\/latex] of coming up heads when flipped.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li class=\"whitespace-normal break-words\">What is the probability that a fair coin will come up tails [latex]n[\/latex] times in a row?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find the probability that a fair coin will come up heads for the second time after an even number of flips.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that the amount of a drug in a patient's system diminishes by a multiplicative factor [latex]r&lt;1[\/latex] each hour. Suppose that a new dose is administered every [latex]N[\/latex] hours. Find an expression that gives the amount [latex]A\\left(n\\right)[\/latex] in the patient's system after [latex]n[\/latex] hours for each [latex]n[\/latex] in terms of the dosage [latex]d[\/latex] and the ratio [latex]r[\/latex]. (Hint: Write [latex]n=mN+k[\/latex], where [latex]0\\le k&lt;N[\/latex], and sum over values from the different doses administered.)<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">The Divergence and Integral Tests<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>For each of the following sequences (1-7), if the divergence test applies, either state that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}{n}[\/latex] does not exist or find [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}{n}[\/latex]. If the divergence test does not apply, state why.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{n}{5{n}^{2}-3}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{\\left(2n+1\\right)\\left(n - 1\\right)}{{\\left(n+1\\right)}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{2}^{n}}{{3}^{\\frac{n}{2}}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={e}^{\\frac{-2}{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\tan{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(1-\\dfrac{1}{n}\\right)}^{2n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{\\left(\\text{ln}n\\right)}^{2}}{\\sqrt{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (8-10), state whether the given [latex]p[\/latex] -series converges.<\/strong><\/p>\r\n\r\n<ol start=\"8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n\\sqrt{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\sqrt[3]{{n}^{4}}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{\\pi }}{{n}^{2e}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (11-13), use the integral test to determine whether the following sums converge.<\/strong><\/p>\r\n\r\n<ol start=\"11\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\sqrt[3]{n+5}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{1+{n}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{2n}{1+{n}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Express the following (14-15) sums as [latex]p[\/latex] -series and determine whether each converges.<\/strong><\/p>\r\n\r\n<ol start=\"14\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}n}[\/latex] (Hint: [latex]{2}^{\\text{-}\\text{ln}n}=\\dfrac{1}{{n}^{\\text{ln}2}}[\/latex] .)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{2}^{-2\\text{ln}n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (16-17), use the estimate [latex]{R}{N}\\le {\\displaystyle\\int }{N}^{\\infty }f\\left(t\\right)dt[\/latex] to find a bound for the remainder [latex]{R}{N}=\\displaystyle\\sum {n=1}^{\\infty }{a}{n}-\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] where [latex]{a}{n}=f\\left(n\\right)[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"16\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{{n}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{1+{n}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (18-20), find the minimum value of [latex]N[\/latex] such that the remainder estimate [latex]{\\displaystyle\\int }{N+1}^{\\infty }f&lt;{R}{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }f[\/latex] guarantees that [latex]\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] estimates [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex], accurate to within the given error.<\/strong><\/p>\r\n\r\n<ol start=\"18\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{2}}[\/latex], error [latex]&lt;{10}^{-4}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{1.01}}[\/latex], error [latex]&lt;{10}^{-4}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{1+{n}^{2}}[\/latex], error [latex]&lt;{10}^{-3}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (21-23), find a value of [latex]N[\/latex] such that [latex]{R}_{N}[\/latex] is smaller than the desired error. Compute the corresponding sum [latex]\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] and compare it to the given estimate of the infinite series.<\/strong><\/p>\r\n\r\n<ol start=\"21\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{e}^{n}}[\/latex], error [latex]&lt;{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{e}^{n}}=\\dfrac{1}{e - 1}=0.581976\\text{...}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{4}}[\/latex], error [latex]&lt;{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{4}}=\\dfrac{{\\pi }^{4}}{90}=1.08232...[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find the limit as [latex]n\\to \\infty [\/latex] of [latex]\\dfrac{1}{n}+\\dfrac{1}{n+1}+\\text{...}+\\dfrac{1}{2n}[\/latex]. (Hint: Compare to [latex]{\\displaystyle\\int }_{n}^{2n}\\dfrac{1}{t}dt.\\text{)}[\/latex]<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1169737168069\" class=\"whitespace-normal break-words\"><strong>The next few exercises (24-28) are intended to give a sense of applications in which partial sums of the harmonic series arise.<\/strong><\/p>\r\n\r\n<ol start=\"24\">\r\n \t<li class=\"whitespace-normal break-words\">In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number [latex]{H}_{k}=\\left(1+\\dfrac{1}{2}+\\dfrac{1}{3}+\\text{...}+\\dfrac{1}{k}\\right)[\/latex]. Recall that [latex]{T}_{k}={H}_{k}-\\text{ln}k[\/latex] is decreasing. Compute [latex]T=\\underset{k\\to \\infty }{\\text{lim}}{T}_{k}[\/latex] to four decimal places. (Hint: [latex]\\dfrac{1}{k+1}&lt;{\\displaystyle\\int }_{k}^{k+1}\\dfrac{1}{x}dx[\/latex] .)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has [latex]n[\/latex] cards, then the probability that the insertion will be below the card initially at the bottom (call this card [latex]B[\/latex]) is [latex]\\dfrac{1}{n}[\/latex]. Thus the expected number of top random insertions before [latex]B[\/latex] is no longer at the bottom is n. Once one card is below [latex]B[\/latex], there are two places below [latex]B[\/latex] and the probability that a randomly inserted card will fall below [latex]B[\/latex] is [latex]\\dfrac{2}{n}[\/latex]. The expected number of top random insertions before this happens is [latex]\\dfrac{n}{2}[\/latex]. The two cards below [latex]B[\/latex] are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Show that for the remainder estimate to apply on [latex]\\left[N,\\infty \\right)[\/latex] it is sufficient that [latex]f\\left(x\\right)[\/latex] be decreasing on [latex]\\left[N,\\infty \\right)[\/latex], but [latex]f[\/latex] need not be decreasing on [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{1}{n{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">A fast computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=2}^{N}\\dfrac{1}{n\\text{ln}n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Comparison Tests<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>Use the comparison test to determine whether the following (1-6) series converge.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] where [latex]{a}_{n}=\\dfrac{1}{n\\left(n+\\dfrac{1}{2}\\right)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{2n - 1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{\\left(n+2\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sin\\left(\\dfrac{1}{n}\\right)}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sin\\left(\\dfrac{1}{n}\\right)}{\\sqrt{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sqrt{n+1}-\\sqrt{n}}{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Use the limit comparison test to determine whether each of the following (7-14) series converges or diverges.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{\\text{ln}n}{n}\\right)}^{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\text{ln}\\left(1+\\dfrac{1}{n}\\right)}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{4}^{n}-{3}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{e}^{\\left(1.1\\right)n}-{3}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{1+\\dfrac{1}{n}}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\dfrac{1}{n}-\\sin\\left(\\dfrac{1}{n}\\right)\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n}\\left(\\dfrac{\\pi }{2}-{\\tan}^{-1}n\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(1-{e}^{-\\dfrac{1}{n}}\\right)[\/latex] (Hint: [latex]\\dfrac{1}{e}\\approx {\\left(1 - \\dfrac{1}{n}\\right)}^{n}[\/latex], so [latex]1-{e}^{-\\dfrac{1}{n}}\\approx \\dfrac{1}{n}.[\/latex])<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (15-26), solve each problem.<\/strong>\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"15\">\r\n \t<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{\\left(\\text{ln}n\\right)}{n}\\right)}^{p}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For which [latex]p&gt;0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{p}}{{2}^{n}}[\/latex] converge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For which [latex]r&gt;0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{2}^{n}}{{r}^{{n}^{2}}}[\/latex] converge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\sin}^{2}\\left(\\dfrac{nr}{2}\\right)}{n}[\/latex] converge or diverge? Explain.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}\\ge 0[\/latex] and [latex]{b}{n}\\ge 0[\/latex] and that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{b}^{2}{}{n}[\/latex] converge. Prove that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}[\/latex] converges and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}\\le \\dfrac{1}{2}\\left(\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{2}+\\displaystyle\\sum {n=1}^{\\infty }{b}{n}^{2}\\right)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}[\/latex] converge? (Hint: Use [latex]n={e}^{\\text{ln}\\left(n\\right)}[\/latex] to compare to a [latex]p-\\text{series}\\text{.}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, then [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges. If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges, does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] necessarily converge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}&gt;0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] diverges. Suppose that [latex]{b}{n}[\/latex] is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}_{n}[\/latex] necessarily diverge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges, then [latex]\\displaystyle\\sum {n=1}^{\\infty }{\\sin}^{2}\\left({a}{n}\\right)[\/latex] converges.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]{b}_{n}[\/latex] be an infinite sequence of zeros and ones. What is the largest possible value of [latex]x=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\text{?}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Explain why, if [latex]x&gt;\\dfrac{1}{2}[\/latex], then [latex]x[\/latex] cannot be written [latex]x=\\displaystyle\\sum {n=2}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\left({b}{n}=0\\text{ or }1,{b}{1}=0\\right)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of [latex]1\\text{-kg}[\/latex] weights, and nine each of [latex]0.1\\text{-kg,}[\/latex] [latex]0.01\\text{-kg},0.001\\text{-kg,}[\/latex] and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Alternating Series<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>State whether each of the following (1-15) series converges absolutely, conditionally, or not at all.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{\\sqrt{n}+1}{\\sqrt{n}+3}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{\\sqrt{n+3}}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{3}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\dfrac{n+1}{n}\\right)}^{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(1+\\dfrac{1}{n}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{n}^{e}}{1+{n}^{\\pi }}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{n}^{\\dfrac{1}{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left(1-\\cos\\left(\\dfrac{1}{n}\\right)\\right)[\/latex] (Hint: [latex]\\cos\\left(\\dfrac{1}{n}\\right)\\approx 1 - \\dfrac{1}{{n}^{2}}[\/latex] for large [latex]n.[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\dfrac{1}{\\sqrt{n}}-\\dfrac{1}{\\sqrt{n+1}}\\right)[\/latex] (Hint: Find common denominator then rationalize numerator.)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left({\\tan}^{-1}\\left(n+1\\right)-{\\tan}^{-1}n\\right)[\/latex] (Hint: Use Mean Value Theorem.)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+1}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\cos\\left(n\\pi \\right)}{{n}^{\\dfrac{1}{n}}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\sin\\left(\\dfrac{n\\pi}{2}\\right)\\sin\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In each of the following problems (16-18), use the estimate [latex]|{R}{N}|\\le {b}{N+1}[\/latex] to find a value of [latex]N[\/latex] that guarantees that the sum of the first [latex]N[\/latex] terms of the alternating series [latex]\\displaystyle\\sum {n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}{n}[\/latex] differs from the infinite sum by at most the given error. Calculate the partial sum [latex]{S}_{N}[\/latex] for this [latex]N[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"16\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{ln}\\left(n\\right)}[\/latex], [latex]n\\ge 2[\/latex], error [latex]&lt;{10}^{-1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{2}^{n}}[\/latex], error [latex]&lt;{10}^{-6}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{n}^{2}}[\/latex], error [latex]&lt;{10}^{-6}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (19-22), indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.<\/strong><\/p>\r\n\r\n<ol start=\"19\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]{b}{n}\\ge 0[\/latex] is decreasing, then [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left({b}{2n - 1}-{b}{2n}\\right)[\/latex] converges absolutely.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]{b}{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left({b}{3n - 2}+{b}{3n - 1}-{b}_{3n}\\right)[\/latex] converges then [latex]\\displaystyle\\sum {n=1}^{\\infty }{b}{3n - 2}[\/latex] converges.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]{a}{n}^{+}={a}{n}[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{a}{n}^{-}=\\text{-}{a}{n}[\/latex] if [latex]{a}{n}&lt;0[\/latex]. (Also, [latex]{a}{n}^{+}=0\\text{ if }{a}{n}&lt;0[\/latex] and [latex]{a}{n}^{-}=0\\text{ if }{a}{n}\\ge 0.[\/latex]) If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges conditionally but not absolutely, then neither [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{+}[\/latex] nor [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{-}[\/latex] converge.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}[\/latex] is a sequence such that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}[\/latex] converges for every possible sequence [latex]{b}_{n}[\/latex] of zeros and ones. Does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converge absolutely?<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>The following series (23-24) do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.<\/strong><\/p>\r\n\r\n<ol start=\"23\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{\\cos}^{2}n}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]1+\\dfrac{1}{2}-\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}-\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}-\\dfrac{1}{9}+\\text{...}[\/latex]<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (25-28), solve each problem.<\/strong>\r\n<ol start=\"25\">\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]\\displaystyle\\sum {a}{n}[\/latex] converges absolutely. Show that the series consisting of the positive terms [latex]{a}{n}[\/latex] also converges.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The formula [latex]\\cos\\theta =1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}-\\dfrac{{\\theta }^{6}}{6\\text{!}}+\\text{...}[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\cos\\theta [\/latex] by the fifth partial sum [latex]1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}\\dfrac{\\text{-}{\\theta }^{6}}{6\\text{!}}+\\dfrac{{\\theta }^{8}}{8\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\dfrac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi [\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">How many terms in [latex]\\cos\\theta =1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}-\\dfrac{{\\theta }^{6}}{6\\text{!}}+\\text{...}[\/latex] are needed to approximate [latex]\\cos1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Sometimes the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges to a certain fraction of an absolutely convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] at a faster rate. Given that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{2}}=\\dfrac{{\\pi }^{2}}{6}[\/latex], find [latex]12=1-\\dfrac{1}{{2}^{2}}+\\dfrac{1}{{3}^{2}}-\\dfrac{1}{{4}^{2}}+\\text{...}[\/latex]. Which of the series [latex]6\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{2}}[\/latex] and [latex]S\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}[\/latex] gives a better estimation of [latex]{\\pi }^{2}[\/latex] using [latex]1000[\/latex] terms?<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Ratio and Root Tests<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>Use the ratio test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, where [latex]{a}_{n}[\/latex] is given in the following problems (1-6). State if the ratio test is inconclusive.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{n\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{n}^{2}}{{2}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\left(2n\\right)\\text{!}}{{n}^{2n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{{\\left(\\dfrac{n}{e}\\right)}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left({2}^{n}n\\text{!}\\right)}^{2}}{{\\left(2n\\right)}^{2n}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (7-11), use the root test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, where [latex]{a}_{n}[\/latex] is as follows.<\/strong><\/p>\r\n\r\n<ol start=\"7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}={\\left(\\dfrac{2{k}^{2}-1}{{k}^{2}+3}\\right)}^{k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{n}{{2}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{{k}^{e}}{{e}^{k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(\\dfrac{1}{e}+\\dfrac{1}{n}\\right)}^{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{\\left(\\text{ln}\\left(1+\\text{ln}n\\right)\\right)}^{n}}{{\\left(\\text{ln}n\\right)}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (12-14), use either the ratio test or the root test as appropriate to determine whether the series [latex]\\displaystyle\\sum {k=1}^{\\infty }{a}{k}[\/latex] with given terms [latex]{a}_{k}[\/latex] converges, or state if the test is inconclusive.<\/strong><\/p>\r\n\r\n<ol start=\"12\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{2\\cdot 4\\cdot 6\\text{...}2k}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(1-\\dfrac{1}{n}\\right)}^{{n}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}={\\left(\\dfrac{1}{k+1}+\\dfrac{1}{k+2}+\\text{...}+\\dfrac{1}{3k}\\right)}^{k}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the ratio test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, or state if the ratio test is inconclusive.<\/strong><\/p>\r\n\r\n<ol start=\"15\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the root and limit comparison tests to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges.<\/strong><\/p>\r\n\r\n<ol start=\"16\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}{n}=\\dfrac{1}{{x}{n}^{n}}[\/latex] where [latex]{x}{n+1}=\\dfrac{1}{2}{x}{n}+\\dfrac{1}{{x}{n}}[\/latex], [latex]{x}{1}=1[\/latex] (Hint: Find limit of [latex]{{x}_{n}}.[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-22), use an appropriate test to determine whether the series converges.<\/strong><\/p>\r\n\r\n<ol start=\"17\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n+1}\\left(n+1\\right)}{{n}^{3}+3{n}^{2}+3n+1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n - 1\\right)}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{1}{{2}^{{\\sin}^{2}k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{\\left(\\begin{array}{c}n+2\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\ k\\end{array}\\right)=\\dfrac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{{2}^{k}}{\\left(\\begin{array}{l}3k\\ k\\end{array}\\right)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}{k}={\\left(\\dfrac{k}{k+\\text{ln}k}\\right)}^{2k}[\/latex] (Hint: [latex]{a}{k}={\\left(1+\\dfrac{\\text{ln}k}{k}\\right)}^{\\text{-}\\left(\\dfrac{k}{\\text{ln}k}\\right)\\text{ln}{k}^{2}}.[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>The following (23-24) series converge by the ratio test. Use summation by parts, [latex]\\displaystyle\\sum {k=1}^{n}{a}{k}\\left({b}{k+1}-{b}{k}\\right)=\\left[{a}{n+1}{b}{n+1}-{a}{1}{b}{1}\\right]-\\displaystyle\\sum {k=1}^{n}{b}{k+1}\\left({a}{k+1}-{a}{k}\\right)[\/latex], to find the sum of the given series.<\/strong><\/p>\r\n\r\n<ol start=\"23\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {k=1}^{\\infty }\\dfrac{k}{{c}^{k}}[\/latex], where [latex]c&gt;1[\/latex] (Hint: Take [latex]{a}{k}=k[\/latex] and [latex]{b}_{k}=\\dfrac{{c}^{1-k}}{\\left(c - 1\\right)}.[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n+1\\right)}^{2}}{{2}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>The kth term of each of the following (25-26) series has a factor [latex]{x}^{k}[\/latex]. Find the range of [latex]x[\/latex] for which the ratio test implies that the series converges.<\/strong><\/p>\r\n\r\n<ol start=\"25\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{2k}}{{k}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{k}}{k\\text{!}}[\/latex]<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (27-30), solve each problem.<\/strong>\r\n<ol start=\"27\">\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]0 &lt; r &lt; 1[\/latex]. For which real numbers [latex]p[\/latex] does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{n}^{p}{r}^{n}[\/latex] converge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]\\underset{n\\to \\infty }{\\text{lim}}|\\dfrac{{a}{n+1}}{{a}{n}}|=p[\/latex]. For which values of [latex]r&gt;0[\/latex] is [latex]\\displaystyle\\sum {n=1}^{\\infty }{r}^{n}{a}{n}[\/latex] guaranteed to converge?<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For which values of [latex]r&gt;0[\/latex], if any, does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{r}^{\\sqrt{n}}[\/latex] converge? (Hint: [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}=\\displaystyle\\sum _{k=1}^{\\infty }\\displaystyle\\sum {n={k}^{2}}^{{\\left(k+1\\right)}^{2}-1}{a}{n}.[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Let [latex]{a}_{n}={2}^{\\text{-}\\left[\\dfrac{n}{2}\\right]}[\/latex] where [latex]\\left[x\\right][\/latex] is the greatest integer less than or equal to [latex]x[\/latex]. Determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges and justify your answer.<\/li>\r\n<\/ol>","rendered":"<h2><span data-sheets-root=\"1\">Sequences and Their Properties<\/span><\/h2>\n<p dir=\"ltr\" data-pm-slice=\"1 3 []\"><strong>Find the first six terms of each of the following sequences (1-2), starting with<\/strong> [latex]n=1[\/latex].<\/p>\n<ol class=\"tight\" dir=\"ltr\" data-tight=\"true\">\n<li>[latex]a_n = 1 + (-1)^n[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\n<li>[latex]a_1 = 1[\/latex] and [latex]a_n = a_{n-1} + n[\/latex] for [latex]n \\ge 2[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Find an explicit formula for<\/strong> [latex]a_n[\/latex] <strong>for each of the following sequences (3-6).<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"3\" data-tight=\"true\">\n<li>[latex]a_1 = 1[\/latex] and [latex]a_n = a_{n-1} + n[\/latex] for [latex]n \\ge 2[\/latex].<\/li>\n<li>Find a formula [latex]a_n[\/latex] for the [latex]n[\/latex]th term of the arithmetic sequence whose first term is [latex]a_1 = -3[\/latex] such that [latex]a_{n-1} - a_n = 4[\/latex] for [latex]n \\ge 1[\/latex].<\/li>\n<li>Find a formula [latex]a_n[\/latex] for the [latex]n[\/latex]th term of the geometric sequence whose first term is [latex]a_1 = 3[\/latex] such that [latex]\\dfrac{a_{n+1}}{a_n} = \\dfrac{1}{10}[\/latex] for [latex]n \\ge 1[\/latex].<\/li>\n<li>Find an explicit formula for the [latex]n[\/latex]th term of the sequence satisfying [latex]a_1 = 0[\/latex] and [latex]a_n = 2 a_{n-1} + 1[\/latex] for [latex]n \\ge 2[\/latex].<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Find a formula for the general term<\/strong> [latex]a_n[\/latex] <strong>of each of the following sequence.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"7\" data-tight=\"true\">\n<li>[latex]{ 1, -\\dfrac{1}{3}, \\dfrac{1}{5}, -\\dfrac{1}{7}, \\ldots }[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Find a function<\/strong> [latex]f(n)[\/latex] <strong>that identifies the<\/strong> [latex]n[\/latex]th <strong>term<\/strong> [latex]a_n[\/latex] <strong>of the following recursively defined sequences (8-9), as<\/strong> [latex]a_n = f(n)[\/latex].<\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"8\" data-tight=\"true\">\n<li>[latex]a_1 = 2[\/latex] and [latex]a_{n+1} = 2 a_n[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\n<li>[latex]a_1 = 2[\/latex] and [latex]a_{n+1} = \\dfrac{(n+1) a_n}{2}[\/latex] for [latex]n \\ge 1[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Plot the first<\/strong> [latex]N[\/latex] <strong>terms of each sequence (10-12). State whether the graphical evidence suggests that the sequence converges or diverges.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"10\" data-tight=\"true\">\n<li>[latex]a_1 = 1[\/latex], [latex]a_2 = 2[\/latex], and for [latex]n \\ge 2[\/latex], [latex]a_n = \\dfrac{1}{2} (a_{n-1} + a_{n-2})[\/latex]; [latex]N = 30[\/latex]<\/li>\n<li>[latex]a_1 = 1[\/latex], [latex]a_2 = 2[\/latex], and for [latex]n \\ge 3[\/latex], [latex]a_n = \\sqrt{a_{n-1} a_{n-2}}[\/latex]; [latex]N = 30[\/latex]<\/li>\n<li>latex]a_n = \\cos{n}[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Suppose that<\/strong> [latex]\\lim_{n \\to \\infty} a_n = 1[\/latex], [latex]\\lim_{n \\to \\infty} b_n = -1[\/latex], <strong>and<\/strong> [latex]0 < -b_n < a_n[\/latex] <strong>for all<\/strong> [latex]n[\/latex]. <strong>Evaluate each of the following limits (13-14), or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"13\" data-tight=\"true\">\n<li>[latex]\\lim_{n \\to \\infty} (3 a_n - 4 b_n)[\/latex]<\/li>\n<li>[latex]\\lim_{n \\to \\infty} \\dfrac{a_n + b_n}{a_n - b_n}[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Find the limit of each of the following sequences (15-16), using L\u2019H\u00f4pital\u2019s rule when appropriate.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"15\" data-tight=\"true\">\n<li>[latex]\\dfrac{n^2}{2^n}[\/latex]<\/li>\n<li>[latex]\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>For the following sequences (17-20), whose<\/strong> [latex]n[\/latex]th <strong>terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"17\" data-tight=\"true\">\n<li>[latex]\\dfrac{n}{2^n}[\/latex], [latex]n \\ge 2[\/latex]<\/li>\n<li>[latex]\\sin{n}[\/latex]<\/li>\n<li>[latex]n^{\\dfrac{1}{n}}[\/latex], [latex]n \\ge 3[\/latex]<\/li>\n<li>[latex]\\tan{n}[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"21\" data-tight=\"true\">\n<li>[latex]a_1 = 3[\/latex], [latex]a_n = \\sqrt{2 a_{n-1}}[\/latex], [latex]n = 2, 3, \\ldots[\/latex].<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>Use the Squeeze Theorem to find the limit of each of the following sequences (22-23).<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"22\" data-tight=\"true\">\n<li>[latex]\\dfrac{\\cos\\left(\\dfrac{1}{n}\\right) - 1}{\\dfrac{1}{n}}[\/latex]<\/li>\n<li>[latex]a_n = \\sin{n} \\sin\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\n<\/ol>\n<p dir=\"ltr\"><strong>For the following exercises (24-28), determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"24\" data-tight=\"true\">\n<li>[latex]a_n = (2n)^{\\dfrac{1}{n}} - n^{\\dfrac{1}{n}}[\/latex]<\/li>\n<li>[latex]a_n = \\left(1 - \\dfrac{2}{n}\\right)^n[\/latex]<\/li>\n<li>[latex]a_n = \\dfrac{2^n + 3^n}{4^n}[\/latex]<\/li>\n<li>[latex]a_n = \\dfrac{(1000)^n}{n!}[\/latex]<\/li>\n<li>[latex]a_n = \\dfrac{(n!)^2}{(2n)!}[\/latex]<\/li>\n<\/ol>\n<p><strong>For the following exercises (29-30), solve each problem.<\/strong><\/p>\n<ol class=\"tight\" dir=\"ltr\" start=\"29\" data-tight=\"true\">\n<li>A lake initially contains [latex]2000[\/latex] fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by [latex]6%[\/latex] each month. However, factoring in all causes, [latex]150[\/latex] fish are lost each month.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Explain why the fish population after [latex]n[\/latex] months is modeled by [latex]P_n = 1.06 P_{n-1} - 150[\/latex] with [latex]P_0 = 2000[\/latex].<\/li>\n<li>How many fish will be in the pond after one year?<\/li>\n<\/ol>\n<\/li>\n<li>A student takes out a college loan of [latex]$10,000[\/latex] at an annual percentage rate of [latex]6 \\%[\/latex], compounded monthly.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>If the student makes payments of [latex]$100[\/latex] per month, how much does the student owe after [latex]12[\/latex] months?<\/li>\n<li>After how many months will the loan be paid off?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Introduction to Series<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>Using sigma notation, write the following expressions (1-2) as infinite series.<\/strong><\/p>\n<ol>\n<li>[latex]1+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\cdots[\/latex]<\/li>\n<li>[latex]1-\\dfrac{1}{2}+\\dfrac{1}{3}-\\dfrac{1}{4}+\\ldots[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (3-4), compute the first four partial sums [latex]{S}_{1},\\ldots,{S}_{4}[\/latex] for the series having [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] starting with [latex]n=1[\/latex] as follows.<\/strong><\/p>\n<ol start=\"3\">\n<li>[latex]{a}_{n}=n[\/latex]<\/li>\n<li>[latex]{a}_{n}=\\sin\\left(\\dfrac{n\\pi}{2}\\right)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (5-6), compute the general term [latex]{a}_{n}[\/latex] of the series with the given partial sum [latex]{S}_{n}[\/latex]. If the sequence of partial sums converges, find its limit [latex]S[\/latex].<\/strong><\/p>\n<ol start=\"5\">\n<li>[latex]{S}_{n}=1-\\dfrac{1}{n}[\/latex], [latex]n\\ge 2[\/latex]<\/li>\n<li>[latex]{S}_{n}=\\sqrt{n},n\\ge 2[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For each of the following series (7-8), use the sequence of partial sums to determine whether the series converges or diverges.<\/strong><\/p>\n<ol start=\"7\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{n+2}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\left(n+1\\right)\\left(n+2\\right)}[\/latex] (<em>Hint<\/em>: Use a partial fraction decomposition like that for [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n\\left(n+1\\right)}.[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }a_{n}=1[\/latex], that [latex]\\displaystyle\\sum _{n=1}^{\\infty }b_{n}=-1[\/latex], that [latex]a_{1}=2[\/latex], and [latex]b_{1}=-3[\/latex]. Find the sum of the indicated series (9-10).<\/strong><\/p>\n<ol start=\"9\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(a_{n}+b_{n}\\right)[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\left(a_{n}-b_{n}\\right)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (11-13), state whether the given series converges and explain why.<\/strong><\/p>\n<ol start=\"11\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n+1000}[\/latex] (Hint: Rewrite using a change of index.)<\/li>\n<li>[latex]1+\\dfrac{1}{10}+\\dfrac{1}{100}+\\dfrac{1}{1000}+\\cdots[\/latex]<\/li>\n<li>[latex]1+\\dfrac{\\pi }{\\text{e}^{2}}+\\dfrac{{\\pi }^{2}}{{e}^{4}}+\\dfrac{{\\pi }^{3}}{{e}^{6}}+\\dfrac{{\\pi }^{4}}{{e}^{8}}+\\cdots[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For [latex]{a}_{n}[\/latex] as follows, write the sum as a geometric series of the form [latex]\\displaystyle\\sum {n=1}^{\\infty }a{r}^{n}[\/latex]. State whether the series converges and if it does, find the value of [latex]\\displaystyle\\sum {a}{n}[\/latex].<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"14\">\n<li class=\"whitespace-normal break-words\">[latex]{a}{1}=-1[\/latex] and [latex]\\dfrac{{a}{n}}{{a}_{n+1}}=-5[\/latex] for [latex]n\\ge 1[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}{1}=10[\/latex] and [latex]\\dfrac{{a}{n}}{{a}_{n+1}}=10[\/latex] for [latex]n\\ge 1[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Use the identity [latex]\\dfrac{1}{1-y}=\\displaystyle\\sum _{n=0}^{\\infty }{y}^{n}[\/latex] to express the following functions (16-17) as a geometric series in the indicated term.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"16\">\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{x}{1+x}[\/latex] in [latex]x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{1+{\\sin}^{2}x}[\/latex] in [latex]\\sin{x}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Evaluate the following (18-18) telescoping series or state whether the series diverges.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"18\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\dfrac{1}{n}}-{2}^{\\dfrac{1}{\\left(n+1\\right)}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\sqrt{n}-\\sqrt{n+1}\\right)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Express the following (<\/strong><strong>20-21) series as a telescoping sum and evaluate its [latex]n[\/latex]th partial sum.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"20\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\text{ln}\\left(\\dfrac{n}{n+1}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{\\text{ln}\\left(1{+}\\dfrac{1}{n}\\right)}{\\text{ln}n\\text{ln}\\left(n+1\\right)}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (22-25), solve each problem. A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"22\">\n<li class=\"whitespace-normal break-words\">Let [latex]{a}_{n}=f\\left(n\\right)-2f\\left(n+1\\right)+f\\left(n+2\\right)[\/latex], in which [latex]f\\left(n\\right)\\to 0[\/latex] as [latex]n\\to \\infty[\/latex]. Find [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}_{n}={c}_{0}f\\left(n\\right)+{c}_{1}f\\left(n+1\\right)+{c}_{2}f\\left(n+2\\right)+{c}_{3}f\\left(n+3\\right)+{c}_{4}f\\left(n+4\\right)[\/latex], where [latex]f\\left(n\\right)\\to 0[\/latex] as [latex]n\\to \\infty[\/latex]. Find a condition on the coefficients [latex]{c}_{0}\\text{...},{c}_{4}[\/latex] that make this a general telescoping series.<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{2}{{n}^{3}-n}[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Define a sequence [latex]{t}_{k}=\\displaystyle\\sum _{n=1}^{k - 1}\\left(\\dfrac{1}{k}\\right)-\\text{ln}k[\/latex]. Use the graph of [latex]\\dfrac{1}{x}[\/latex] to verify that [latex]{t}_{k}[\/latex] is increasing. Plot [latex]{t}_{k}[\/latex] for [latex]k=1\\text{...}100[\/latex] and state whether it appears that the sequence converges.<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Each of the following (26-27) infinite series converges to the given multiple of [latex]\\pi[\/latex] or [latex]\\dfrac{1}{\\pi}[\/latex]. In each case, find the minimum value of [latex]N[\/latex] such that the [latex]N\\text{th}[\/latex] partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to [latex]15[\/latex] decimals place, [latex]\\pi =3.141592653589793...[\/latex].<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"26\">\n<li class=\"whitespace-normal break-words\">[latex]\\pi =-3+\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n{2}^{n}n{\\text{!}}^{2}}{\\left(2n\\right)\\text{!}}[\/latex], error [latex]<0.0001[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{9801}{2\\pi }=\\dfrac{4}{9801}\\displaystyle\\sum _{k=0}^{\\infty }\\dfrac{\\left(4k\\right)\\text{!}\\left(1103+26390k\\right)}{{\\left(k\\text{!}\\right)}^{4}{396}^{4k}}[\/latex], error [latex]<{10}^{-12}[\/latex]<\/li>\n<\/ol>\n<p><strong>For the following exercises (28-30), solve each problem.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"28\">\n<li class=\"whitespace-normal break-words\">A fair coin is one that has probability [latex]\\dfrac{1}{2}[\/latex] of coming up heads when flipped.\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-normal break-words\">What is the probability that a fair coin will come up tails [latex]n[\/latex] times in a row?<\/li>\n<li class=\"whitespace-normal break-words\">Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.<\/li>\n<\/ol>\n<\/li>\n<li class=\"whitespace-normal break-words\">Find the probability that a fair coin will come up heads for the second time after an even number of flips.<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that the amount of a drug in a patient&#8217;s system diminishes by a multiplicative factor [latex]r<1[\/latex] each hour. Suppose that a new dose is administered every [latex]N[\/latex] hours. Find an expression that gives the amount [latex]A\\left(n\\right)[\/latex] in the patient&#8217;s system after [latex]n[\/latex] hours for each [latex]n[\/latex] in terms of the dosage [latex]d[\/latex] and the ratio [latex]r[\/latex]. (Hint: Write [latex]n=mN+k[\/latex], where [latex]0\\le k<N[\/latex], and sum over values from the different doses administered.)<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">The Divergence and Integral Tests<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>For each of the following sequences (1-7), if the divergence test applies, either state that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}{n}[\/latex] does not exist or find [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}{n}[\/latex]. If the divergence test does not apply, state why.<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{n}{5{n}^{2}-3}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{\\left(2n+1\\right)\\left(n - 1\\right)}{{\\left(n+1\\right)}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{2}^{n}}{{3}^{\\frac{n}{2}}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={e}^{\\frac{-2}{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\tan{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(1-\\dfrac{1}{n}\\right)}^{2n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{\\left(\\text{ln}n\\right)}^{2}}{\\sqrt{n}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (8-10), state whether the given [latex]p[\/latex] -series converges.<\/strong><\/p>\n<ol start=\"8\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n\\sqrt{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\sqrt[3]{{n}^{4}}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{\\pi }}{{n}^{2e}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (11-13), use the integral test to determine whether the following sums converge.<\/strong><\/p>\n<ol start=\"11\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{\\sqrt[3]{n+5}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{1+{n}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{2n}{1+{n}^{4}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Express the following (14-15) sums as [latex]p[\/latex] -series and determine whether each converges.<\/strong><\/p>\n<ol start=\"14\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}n}[\/latex] (Hint: [latex]{2}^{\\text{-}\\text{ln}n}=\\dfrac{1}{{n}^{\\text{ln}2}}[\/latex] .)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{2}^{-2\\text{ln}n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (16-17), use the estimate [latex]{R}{N}\\le {\\displaystyle\\int }{N}^{\\infty }f\\left(t\\right)dt[\/latex] to find a bound for the remainder [latex]{R}{N}=\\displaystyle\\sum {n=1}^{\\infty }{a}{n}-\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] where [latex]{a}{n}=f\\left(n\\right)[\/latex].<\/strong><\/p>\n<ol start=\"16\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{{n}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{1+{n}^{2}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (18-20), find the minimum value of [latex]N[\/latex] such that the remainder estimate [latex]{\\displaystyle\\int }{N+1}^{\\infty }f<{R}{N}<{\\displaystyle\\int }_{N}^{\\infty }f[\/latex] guarantees that [latex]\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] estimates [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex], accurate to within the given error.<\/strong><\/p>\n<ol start=\"18\">\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{1.01}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{1+{n}^{2}}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (21-23), find a value of [latex]N[\/latex] such that [latex]{R}_{N}[\/latex] is smaller than the desired error. Compute the corresponding sum [latex]\\displaystyle\\sum {n=1}^{N}{a}{n}[\/latex] and compare it to the given estimate of the infinite series.<\/strong><\/p>\n<ol start=\"21\">\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{e}^{n}}[\/latex], error [latex]<{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{e}^{n}}=\\dfrac{1}{e - 1}=0.581976\\text{...}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{{n}^{4}}[\/latex], error [latex]<{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{4}}=\\dfrac{{\\pi }^{4}}{90}=1.08232...[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Find the limit as [latex]n\\to \\infty[\/latex] of [latex]\\dfrac{1}{n}+\\dfrac{1}{n+1}+\\text{...}+\\dfrac{1}{2n}[\/latex]. (Hint: Compare to [latex]{\\displaystyle\\int }_{n}^{2n}\\dfrac{1}{t}dt.\\text{)}[\/latex]<\/li>\n<\/ol>\n<p id=\"fs-id1169737168069\" class=\"whitespace-normal break-words\"><strong>The next few exercises (24-28) are intended to give a sense of applications in which partial sums of the harmonic series arise.<\/strong><\/p>\n<ol start=\"24\">\n<li class=\"whitespace-normal break-words\">In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number [latex]{H}_{k}=\\left(1+\\dfrac{1}{2}+\\dfrac{1}{3}+\\text{...}+\\dfrac{1}{k}\\right)[\/latex]. Recall that [latex]{T}_{k}={H}_{k}-\\text{ln}k[\/latex] is decreasing. Compute [latex]T=\\underset{k\\to \\infty }{\\text{lim}}{T}_{k}[\/latex] to four decimal places. (Hint: [latex]\\dfrac{1}{k+1}<{\\displaystyle\\int }_{k}^{k+1}\\dfrac{1}{x}dx[\/latex] .)<\/li>\n<li class=\"whitespace-normal break-words\">The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has [latex]n[\/latex] cards, then the probability that the insertion will be below the card initially at the bottom (call this card [latex]B[\/latex]) is [latex]\\dfrac{1}{n}[\/latex]. Thus the expected number of top random insertions before [latex]B[\/latex] is no longer at the bottom is n. Once one card is below [latex]B[\/latex], there are two places below [latex]B[\/latex] and the probability that a randomly inserted card will fall below [latex]B[\/latex] is [latex]\\dfrac{2}{n}[\/latex]. The expected number of top random insertions before this happens is [latex]\\dfrac{n}{2}[\/latex]. The two cards below [latex]B[\/latex] are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.<\/li>\n<li class=\"whitespace-normal break-words\">Show that for the remainder estimate to apply on [latex]\\left[N,\\infty \\right)[\/latex] it is sufficient that [latex]f\\left(x\\right)[\/latex] be decreasing on [latex]\\left[N,\\infty \\right)[\/latex], but [latex]f[\/latex] need not be decreasing on [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{1}{n{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">A fast computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=2}^{N}\\dfrac{1}{n\\text{ln}n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Comparison Tests<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>Use the comparison test to determine whether the following (1-6) series converge.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] where [latex]{a}_{n}=\\dfrac{1}{n\\left(n+\\dfrac{1}{2}\\right)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{2n - 1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{\\left(n+2\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sin\\left(\\dfrac{1}{n}\\right)}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sin\\left(\\dfrac{1}{n}\\right)}{\\sqrt{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\sqrt{n+1}-\\sqrt{n}}{n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Use the limit comparison test to determine whether each of the following (7-14) series converges or diverges.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"7\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{\\text{ln}n}{n}\\right)}^{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\text{ln}\\left(1+\\dfrac{1}{n}\\right)}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{4}^{n}-{3}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{e}^{\\left(1.1\\right)n}-{3}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{1+\\dfrac{1}{n}}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\dfrac{1}{n}-\\sin\\left(\\dfrac{1}{n}\\right)\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n}\\left(\\dfrac{\\pi }{2}-{\\tan}^{-1}n\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(1-{e}^{-\\dfrac{1}{n}}\\right)[\/latex] (Hint: [latex]\\dfrac{1}{e}\\approx {\\left(1 - \\dfrac{1}{n}\\right)}^{n}[\/latex], so [latex]1-{e}^{-\\dfrac{1}{n}}\\approx \\dfrac{1}{n}.[\/latex])<\/li>\n<\/ol>\n<p><strong>For the following exercises (15-26), solve each problem.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"15\">\n<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{\\left(\\text{ln}n\\right)}{n}\\right)}^{p}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">For which [latex]p>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{p}}{{2}^{n}}[\/latex] converge?<\/li>\n<li class=\"whitespace-normal break-words\">For which [latex]r>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{2}^{n}}{{r}^{{n}^{2}}}[\/latex] converge?<\/li>\n<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\sin}^{2}\\left(\\dfrac{nr}{2}\\right)}{n}[\/latex] converge or diverge? Explain.<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}\\ge 0[\/latex] and [latex]{b}{n}\\ge 0[\/latex] and that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{b}^{2}{}{n}[\/latex] converge. Prove that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}[\/latex] converges and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}\\le \\dfrac{1}{2}\\left(\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{2}+\\displaystyle\\sum {n=1}^{\\infty }{b}{n}^{2}\\right)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}[\/latex] converge? (Hint: Use [latex]n={e}^{\\text{ln}\\left(n\\right)}[\/latex] to compare to a [latex]p-\\text{series}\\text{.}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, then [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges. If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges, does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] necessarily converge?<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}>0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] diverges. Suppose that [latex]{b}{n}[\/latex] is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}_{n}[\/latex] necessarily diverge?<\/li>\n<li class=\"whitespace-normal break-words\">Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges, then [latex]\\displaystyle\\sum {n=1}^{\\infty }{\\sin}^{2}\\left({a}{n}\\right)[\/latex] converges.<\/li>\n<li class=\"whitespace-normal break-words\">Let [latex]{b}_{n}[\/latex] be an infinite sequence of zeros and ones. What is the largest possible value of [latex]x=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\text{?}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Explain why, if [latex]x>\\dfrac{1}{2}[\/latex], then [latex]x[\/latex] cannot be written [latex]x=\\displaystyle\\sum {n=2}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\left({b}{n}=0\\text{ or }1,{b}{1}=0\\right)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of [latex]1\\text{-kg}[\/latex] weights, and nine each of [latex]0.1\\text{-kg,}[\/latex] [latex]0.01\\text{-kg},0.001\\text{-kg,}[\/latex] and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Alternating Series<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>State whether each of the following (1-15) series converges absolutely, conditionally, or not at all.<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{\\sqrt{n}+1}{\\sqrt{n}+3}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{\\sqrt{n+3}}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{3}^{n}}{n\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\dfrac{n+1}{n}\\right)}^{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(1+\\dfrac{1}{n}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{n}^{e}}{1+{n}^{\\pi }}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{n}^{\\dfrac{1}{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left(1-\\cos\\left(\\dfrac{1}{n}\\right)\\right)[\/latex] (Hint: [latex]\\cos\\left(\\dfrac{1}{n}\\right)\\approx 1 - \\dfrac{1}{{n}^{2}}[\/latex] for large [latex]n.[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\dfrac{1}{\\sqrt{n}}-\\dfrac{1}{\\sqrt{n+1}}\\right)[\/latex] (Hint: Find common denominator then rationalize numerator.)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left({\\tan}^{-1}\\left(n+1\\right)-{\\tan}^{-1}n\\right)[\/latex] (Hint: Use Mean Value Theorem.)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+1}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\cos\\left(n\\pi \\right)}{{n}^{\\dfrac{1}{n}}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\sin\\left(\\dfrac{n\\pi}{2}\\right)\\sin\\left(\\dfrac{1}{n}\\right)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In each of the following problems (16-18), use the estimate [latex]|{R}{N}|\\le {b}{N+1}[\/latex] to find a value of [latex]N[\/latex] that guarantees that the sum of the first [latex]N[\/latex] terms of the alternating series [latex]\\displaystyle\\sum {n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}{n}[\/latex] differs from the infinite sum by at most the given error. Calculate the partial sum [latex]{S}_{N}[\/latex] for this [latex]N[\/latex].<\/strong><\/p>\n<ol start=\"16\">\n<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{ln}\\left(n\\right)}[\/latex], [latex]n\\ge 2[\/latex], error [latex]<{10}^{-1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{2}^{n}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{b}_{n}=\\dfrac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>For the following exercises (19-22), indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.<\/strong><\/p>\n<ol start=\"19\">\n<li class=\"whitespace-normal break-words\">If [latex]{b}{n}\\ge 0[\/latex] is decreasing, then [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left({b}{2n - 1}-{b}{2n}\\right)[\/latex] converges absolutely.<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]{b}{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left({b}{3n - 2}+{b}{3n - 1}-{b}_{3n}\\right)[\/latex] converges then [latex]\\displaystyle\\sum {n=1}^{\\infty }{b}{3n - 2}[\/latex] converges.<\/li>\n<li class=\"whitespace-normal break-words\">Let [latex]{a}{n}^{+}={a}{n}[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{a}{n}^{-}=\\text{-}{a}{n}[\/latex] if [latex]{a}{n}<0[\/latex]. (Also, [latex]{a}{n}^{+}=0\\text{ if }{a}{n}<0[\/latex] and [latex]{a}{n}^{-}=0\\text{ if }{a}{n}\\ge 0.[\/latex]) If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges conditionally but not absolutely, then neither [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{+}[\/latex] nor [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}^{-}[\/latex] converge.<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]{a}{n}[\/latex] is a sequence such that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}[\/latex] converges for every possible sequence [latex]{b}_{n}[\/latex] of zeros and ones. Does [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converge absolutely?<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>The following series (23-24) do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.<\/strong><\/p>\n<ol start=\"23\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{\\cos}^{2}n}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]1+\\dfrac{1}{2}-\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}-\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}-\\dfrac{1}{9}+\\text{...}[\/latex]<\/li>\n<\/ol>\n<p><strong>For the following exercises (25-28), solve each problem.<\/strong><\/p>\n<ol start=\"25\">\n<li class=\"whitespace-normal break-words\">Suppose that [latex]\\displaystyle\\sum {a}{n}[\/latex] converges absolutely. Show that the series consisting of the positive terms [latex]{a}{n}[\/latex] also converges.<\/li>\n<li class=\"whitespace-normal break-words\">The formula [latex]\\cos\\theta =1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}-\\dfrac{{\\theta }^{6}}{6\\text{!}}+\\text{...}[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\cos\\theta[\/latex] by the fifth partial sum [latex]1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}\\dfrac{\\text{-}{\\theta }^{6}}{6\\text{!}}+\\dfrac{{\\theta }^{8}}{8\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\dfrac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">How many terms in [latex]\\cos\\theta =1-\\dfrac{{\\theta }^{2}}{2\\text{!}}+\\dfrac{{\\theta }^{4}}{4\\text{!}}-\\dfrac{{\\theta }^{6}}{6\\text{!}}+\\text{...}[\/latex] are needed to approximate [latex]\\cos1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Sometimes the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges to a certain fraction of an absolutely convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] at a faster rate. Given that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{2}}=\\dfrac{{\\pi }^{2}}{6}[\/latex], find [latex]12=1-\\dfrac{1}{{2}^{2}}+\\dfrac{1}{{3}^{2}}-\\dfrac{1}{{4}^{2}}+\\text{...}[\/latex]. Which of the series [latex]6\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{n}^{2}}[\/latex] and [latex]S\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}[\/latex] gives a better estimation of [latex]{\\pi }^{2}[\/latex] using [latex]1000[\/latex] terms?<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Ratio and Root Tests<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>Use the ratio test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, where [latex]{a}_{n}[\/latex] is given in the following problems (1-6). State if the ratio test is inconclusive.<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{n\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{n}^{2}}{{2}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{\\left(2n\\right)\\text{!}}{{n}^{2n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{{\\left(\\dfrac{n}{e}\\right)}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left({2}^{n}n\\text{!}\\right)}^{2}}{{\\left(2n\\right)}^{2n}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (7-11), use the root test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, where [latex]{a}_{n}[\/latex] is as follows.<\/strong><\/p>\n<ol start=\"7\">\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}={\\left(\\dfrac{2{k}^{2}-1}{{k}^{2}+3}\\right)}^{k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{n}{{2}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{{k}^{e}}{{e}^{k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(\\dfrac{1}{e}+\\dfrac{1}{n}\\right)}^{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{{\\left(\\text{ln}\\left(1+\\text{ln}n\\right)\\right)}^{n}}{{\\left(\\text{ln}n\\right)}^{n}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (12-14), use either the ratio test or the root test as appropriate to determine whether the series [latex]\\displaystyle\\sum {k=1}^{\\infty }{a}{k}[\/latex] with given terms [latex]{a}_{k}[\/latex] converges, or state if the test is inconclusive.<\/strong><\/p>\n<ol start=\"12\">\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{2\\cdot 4\\cdot 6\\text{...}2k}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}={\\left(1-\\dfrac{1}{n}\\right)}^{{n}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}={\\left(\\dfrac{1}{k+1}+\\dfrac{1}{k+2}+\\text{...}+\\dfrac{1}{3k}\\right)}^{k}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the ratio test to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges, or state if the ratio test is inconclusive.<\/strong><\/p>\n<ol start=\"15\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the root and limit comparison tests to determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges.<\/strong><\/p>\n<ol start=\"16\">\n<li class=\"whitespace-normal break-words\">[latex]{a}{n}=\\dfrac{1}{{x}{n}^{n}}[\/latex] where [latex]{x}{n+1}=\\dfrac{1}{2}{x}{n}+\\dfrac{1}{{x}{n}}[\/latex], [latex]{x}{1}=1[\/latex] (Hint: Find limit of [latex]{{x}_{n}}.[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-22), use an appropriate test to determine whether the series converges.<\/strong><\/p>\n<ol start=\"17\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n+1}\\left(n+1\\right)}{{n}^{3}+3{n}^{2}+3n+1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n - 1\\right)}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{1}{{2}^{{\\sin}^{2}k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{n}=\\dfrac{1}{\\left(\\begin{array}{c}n+2\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\ k\\end{array}\\right)=\\dfrac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}_{k}=\\dfrac{{2}^{k}}{\\left(\\begin{array}{l}3k\\ k\\end{array}\\right)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}{k}={\\left(\\dfrac{k}{k+\\text{ln}k}\\right)}^{2k}[\/latex] (Hint: [latex]{a}{k}={\\left(1+\\dfrac{\\text{ln}k}{k}\\right)}^{\\text{-}\\left(\\dfrac{k}{\\text{ln}k}\\right)\\text{ln}{k}^{2}}.[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>The following (23-24) series converge by the ratio test. Use summation by parts, [latex]\\displaystyle\\sum {k=1}^{n}{a}{k}\\left({b}{k+1}-{b}{k}\\right)=\\left[{a}{n+1}{b}{n+1}-{a}{1}{b}{1}\\right]-\\displaystyle\\sum {k=1}^{n}{b}{k+1}\\left({a}{k+1}-{a}{k}\\right)[\/latex], to find the sum of the given series.<\/strong><\/p>\n<ol start=\"23\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {k=1}^{\\infty }\\dfrac{k}{{c}^{k}}[\/latex], where [latex]c>1[\/latex] (Hint: Take [latex]{a}{k}=k[\/latex] and [latex]{b}_{k}=\\dfrac{{c}^{1-k}}{\\left(c - 1\\right)}.[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n+1\\right)}^{2}}{{2}^{n}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>The kth term of each of the following (25-26) series has a factor [latex]{x}^{k}[\/latex]. Find the range of [latex]x[\/latex] for which the ratio test implies that the series converges.<\/strong><\/p>\n<ol start=\"25\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{2k}}{{k}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{k}}{k\\text{!}}[\/latex]<\/li>\n<\/ol>\n<p><strong>For the following exercises (27-30), solve each problem.<\/strong><\/p>\n<ol start=\"27\">\n<li class=\"whitespace-normal break-words\">Let [latex]0 < r < 1[\/latex]. For which real numbers [latex]p[\/latex] does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{n}^{p}{r}^{n}[\/latex] converge?<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]\\underset{n\\to \\infty }{\\text{lim}}|\\dfrac{{a}{n+1}}{{a}{n}}|=p[\/latex]. For which values of [latex]r>0[\/latex] is [latex]\\displaystyle\\sum {n=1}^{\\infty }{r}^{n}{a}{n}[\/latex] guaranteed to converge?<\/li>\n<li class=\"whitespace-normal break-words\">For which values of [latex]r>0[\/latex], if any, does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{r}^{\\sqrt{n}}[\/latex] converge? (Hint: [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}=\\displaystyle\\sum _{k=1}^{\\infty }\\displaystyle\\sum {n={k}^{2}}^{{\\left(k+1\\right)}^{2}-1}{a}{n}.[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Let [latex]{a}_{n}={2}^{\\text{-}\\left[\\dfrac{n}{2}\\right]}[\/latex] where [latex]\\left[x\\right][\/latex] is the greatest integer less than or equal to [latex]x[\/latex]. Determine whether [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges and justify your answer.<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":38,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1135"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":26,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1135\/revisions"}],"predecessor-version":[{"id":1800,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1135\/revisions\/1800"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1135\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1135"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1135"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1135"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1135"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}