{"id":1127,"date":"2025-06-30T16:36:39","date_gmt":"2025-06-30T16:36:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1127"},"modified":"2025-09-19T15:35:01","modified_gmt":"2025-09-19T15:35:01","slug":"ratio-and-root-tests-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-apply-it\/","title":{"raw":"Ratio and Root Tests: Apply It","rendered":"Ratio and Root Tests: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the ratio test to check if a series converges absolutely<\/li>\r\n \t<li>Use the root test to check if a series converges absolutely<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Series Converging to [latex]\\pi [\/latex] and [latex]\\frac{1}{\\pi} [\/latex]<\/h2>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Throughout history, mathematicians have been fascinated by the challenge of computing [latex]\\pi [\/latex] with ever-increasing precision. This quest has led to the discovery of numerous infinite series that converge to [latex]\\pi [\/latex], each with its own remarkable properties and convergence characteristics.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Dozens of series exist that converge to [latex]\\pi[\/latex] or an algebraic expression containing [latex]\\pi[\/latex]. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of [latex]\\pi[\/latex] in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">In this exploration, we'll examine three historically significant series for \u03c0, each discovered in different eras and demonstrating increasingly sophisticated mathematical techniques. From the elegant simplicity of the Gregory-Leibniz series to the breathtaking efficiency of Ramanujan's formula, these series showcase the evolution of mathematical understanding and computational power.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">The following series was discovered by Gregory and Leibniz in the late 1600s:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736707603\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi =4\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{2n - 1}=4-\\frac{4}{3}+\\frac{4}{5}-\\frac{4}{7}+\\frac{4}{9}-\\cdots [\/latex]<\/div>\r\nThis result follows from the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex].\u00a0<span data-type=\"newline\">\r\n<\/span>\r\n<ol id=\"fs-id1169739171821\" type=\"a\">\r\n \t<li>Prove that this series converges.<\/li>\r\n \t<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=10,20,50,100[\/latex].<\/li>\r\n \t<li>Use the remainder estimate for alternating series to get a bound on the error [latex]{R}_{n}[\/latex].<\/li>\r\n \t<li>What is the smallest value of [latex]N[\/latex] that guarantees [latex]|{R}_{N}|&lt;0.01\\text{?}[\/latex] Evaluate [latex]{S}_{N}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"15289\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"15289\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">The series can be written as [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{4(-1)^{n+1}}{2n-1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This is an alternating series with [latex]a_n = \\frac{4}{2n-1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">To apply the alternating series test, we need to verify:<\/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]a_n &gt; 0[\/latex] for all [latex]n \\geq 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_{n+1} \\leq a_n[\/latex] for all [latex]n \\geq 1[\/latex] (decreasing)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\lim_{n \\to \\infty} a_n = 0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Verification:<\/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]a_n = \\frac{4}{2n-1} &gt; 0[\/latex] since [latex]2n-1 &gt; 0[\/latex] for all [latex]n \\geq 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">We need [latex]a_{n+1} \\leq a_n[\/latex], which means: [latex]\\frac{4}{2(n+1)-1} \\leq \\frac{4}{2n-1}[\/latex] [latex]\\frac{4}{2n+1} \\leq \\frac{4}{2n-1}[\/latex] Since [latex]2n+1 &gt; 2n-1[\/latex] for [latex]n \\geq 1[\/latex], we have [latex]\\frac{1}{2n+1} &lt; \\frac{1}{2n-1}[\/latex], so the sequence is decreasing.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\lim_{n \\to \\infty} \\frac{4}{2n-1} = 0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Since all conditions are satisfied, the series converges by the alternating series test.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]\r\n\\begin{array}{rcl}\r\nS_{10} &amp; = &amp; 4\\left(1 - \\tfrac{1}{3} + \\tfrac{1}{5} - \\tfrac{1}{7} + \\tfrac{1}{9} - \\tfrac{1}{11} + \\tfrac{1}{13} - \\tfrac{1}{15} + \\tfrac{1}{17} - \\tfrac{1}{19}\\right) \\\\\r\n&amp; = &amp; 4(0.760460) = 3.041840 \\\\[6pt]\r\nS_{20} &amp; = &amp; 4\\sum_{n=1}^{20} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.772840) = 3.091361 \\\\[6pt]\r\nS_{50} &amp; = &amp; 4\\sum_{n=1}^{50} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.780398) = 3.121593 \\\\[6pt]\r\nS_{100} &amp; = &amp; 4\\sum_{n=1}^{100} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.782898) = 3.131593 \\\\\r\n\\end{array}\r\n[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">For an alternating series, the remainder estimate states that [latex]|R_n| \\leq a_{n+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore: [latex]|R_n| \\leq \\frac{4}{2(n+1)-1} = \\frac{4}{2n+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">For specific values:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]|R_{10}| \\leq \\frac{4}{21} \\approx 0.190[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]|R_{20}| \\leq \\frac{4}{41} \\approx 0.098[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]|R_{50}| \\leq \\frac{4}{101} \\approx 0.040[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]|R_{100}| \\leq \\frac{4}{201} \\approx 0.020[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">We need [latex]\\frac{4}{2N+1} &lt; 0.01[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\frac{4}{2N+1} &lt; 0.01[\/latex] [latex]4 &lt; 0.01(2N+1)[\/latex] [latex]4 &lt; 0.02N + 0.01[\/latex] [latex]3.99 &lt; 0.02N[\/latex] [latex]N &gt; 199.5[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]N = 200[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]S_{200} = 4\\sum_{n=1}^{200} \\frac{(-1)^{n+1}}{2n-1} \\approx 3.136593[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">The following series has been attributed to Newton in the late 1600s:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\pi &amp; =6 {\\displaystyle\\sum _{n=0}^{\\infty }} \\frac{\\left(2n\\right)\\text{!}}{{2}^{4n+1}{\\left(n\\text{!}\\right)}^{2}\\left(2n+1\\right)}\\hfill \\\\ &amp; =6\\left(\\frac{1}{2}+\\frac{1}{2\\cdot 3}{\\left(\\frac{1}{2}\\right)}^{3}+\\frac{1\\cdot 3}{2\\cdot 4\\cdot 5}\\cdot {\\left(\\frac{1}{2}\\right)}^{5}+\\frac{1\\cdot 3\\cdot 5}{2\\cdot 4\\cdot 6\\cdot 7}{\\left(\\frac{1}{2}\\right)}^{7}+\\cdots \\right)\\hfill \\end{array}[\/latex]<\/p>\r\nThe proof of this result uses the Maclaurin series for [latex]f\\left(x\\right)={\\sin}^{-1}x[\/latex].\r\n<ol id=\"fs-id1169736851958\" type=\"a\">\r\n \t<li>Prove that the series converges.<\/li>\r\n \t<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=5,10,20[\/latex].<\/li>\r\n \t<li>Compare [latex]{S}_{n}[\/latex] to [latex]\\pi [\/latex] for [latex]n=5,10,20[\/latex] and discuss the number of correct decimal places.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"476703\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"476703\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">Let [latex]a_n = \\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Using the ratio test:\r\n[latex]\r\n\\begin{array}{rcl}\r\n\\dfrac{a_{n+1}}{a_n} &amp; = &amp; \\dfrac{(2n+2)!}{2^{4n+5}((n+1)!)^2(2n+3)} \\cdot \\dfrac{2^{4n+1}(n!)^2(2n+1)}{(2n)!} \\\\\r\n&amp; = &amp; \\dfrac{(2n+2)(2n+1)(2n)!}{2^4(n+1)^2(n!)^2(2n+3)} \\cdot \\dfrac{2^{4n+1}(n!)^2(2n+1)}{2^{4n+1}(2n)!} \\\\\r\n&amp; = &amp; \\dfrac{(2n+2)(2n+1)^2}{16(n+1)^2(2n+3)} \\\\\r\n&amp; = &amp; \\dfrac{2(n+1)(2n+1)^2}{16(n+1)^2(2n+3)} \\\\\r\n&amp; = &amp; \\dfrac{(2n+1)^2}{8(n+1)(2n+3)}\r\n\\end{array}\r\n[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">As [latex]n \\to \\infty[\/latex]: [latex]\\lim_{n \\to \\infty} \\frac{(2n+1)^2}{8(n+1)(2n+3)} = \\lim_{n \\to \\infty} \\frac{4n^2}{16n^2} = \\frac{1}{4} &lt; 1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Since the ratio test limit is less than 1, the series converges.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">[latex]S_5 = 6\\sum_{n=0}^{5} \\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Computing each term:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=0: \\frac{1!}{2^1 \\cdot 1^2 \\cdot 1} = \\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=1: \\frac{2!}{2^5 \\cdot 1^2 \\cdot 3} = \\frac{2}{96} = \\frac{1}{48}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=2: \\frac{4!}{2^9 \\cdot 4 \\cdot 5} = \\frac{24}{10240} = \\frac{3}{1280}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=3: \\frac{6!}{2^{13} \\cdot 36 \\cdot 7} = \\frac{720}{2064384} \\approx 0.000349[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=4: \\frac{8!}{2^{17} \\cdot 576 \\cdot 9} \\approx 0.0000596[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]n=5: \\frac{10!}{2^{21} \\cdot 14400 \\cdot 11} \\approx 0.0000109[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">[latex]S_5 = 6(0.5 + 0.02083 + 0.00234 + 0.000349 + 0.0000596 + 0.0000109) \\approx 3.14159[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]S_{10} \\approx 3.141592653[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]S_{20} \\approx 3.141592653589793[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">[latex]\\pi = 3.141592653589793...[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_5 \\approx 3.14159[\/latex]: Correct to 5 decimal places<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_{10} \\approx 3.141592653[\/latex]: Correct to 9 decimal places<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_{20} \\approx 3.141592653589793[\/latex]: Correct to 15 decimal places<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Newton's series converges much faster than the Gregory-Leibniz series, gaining approximately 6 additional correct decimal places with each 10 additional terms.<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">The following series was discovered by Ramanujan in the early 1900s :\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\pi }=\\frac{\\sqrt{8}}{9801}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\left(4n\\right)\\text{!}\\left(1103+26390n\\right)}{{\\left(n\\text{!}\\right)}^{4}{396}^{4n}}[\/latex]<\/p>\r\nWilliam Gosper, Jr., used this series to calculate [latex]\\pi [\/latex] to an accuracy of more than [latex]17[\/latex] million digits in the mid 1980s. At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for [latex]\\pi [\/latex] and [latex]\\frac{1}{\\pi} [\/latex].\r\n<ol id=\"fs-id1169739252686\" type=\"a\">\r\n \t<li>Prove that this series converges.<\/li>\r\n \t<li>Evaluate the first term in this series. Compare this number with the value of [latex]\\pi [\/latex] from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"703420\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"703420\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">Let [latex]a_n = \\frac{(4n)!(1103+26390n)}{(n!)^4 \\cdot 396^{4n}}[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Using the ratio test:<\/p>\r\n[latex]\r\n\\begin{array}{rcl}\r\n\\dfrac{a_{n+1}}{a_n}\r\n&amp; = &amp; \\dfrac{(4n+4)!\\,(1103+26390(n+1))}{((n+1)!)^4 \\cdot 396^{4(n+1)}}\r\n\\cdot \\dfrac{(n!)^4 \\cdot 396^{4n}}{(4n)!\\,(1103+26390n)} \\\\\r\n&amp; = &amp; \\dfrac{(4n+4)(4n+3)(4n+2)(4n+1)(1103+26390n+26390)}{(n+1)^4 \\cdot 396^4}\r\n\\cdot \\dfrac{1}{1103+26390n}\r\n\\end{array}\r\n[\/latex]\r\n<p class=\"whitespace-pre-wrap break-words\">As [latex]n \\to \\infty[\/latex], the dominant terms give:<\/p>\r\n[latex]\r\n\\begin{array}{rcl}\r\n\\lim_{n \\to \\infty} \\dfrac{a_{n+1}}{a_n}\r\n&amp; = &amp; \\lim_{n \\to \\infty}\r\n\\dfrac{256n^4 \\cdot 26390n}{n^4 \\cdot 396^4 \\cdot 26390n} \\\\\r\n&amp; = &amp; \\dfrac{256}{396^4} \\\\\r\n&amp; = &amp; \\dfrac{256}{24596196096} &lt; 1 \\\\\\end{array}\r\n[\/latex]\r\n<p class=\"whitespace-normal break-words\">Since the ratio is less than 1, the series converges by the ratio test.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\"><strong>First term ([latex]n=0[\/latex]):\r\n<\/strong> [latex]a_0 = \\frac{0! \\cdot 1103}{(0!)^4 \\cdot 396^0} = 1103[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\pi} \\approx \\frac{\\sqrt{8}}{9801} \\cdot 1103 = \\frac{2\\sqrt{2} \\cdot 1103}{9801} \\approx 0.318309886183791[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore: [latex]\\pi \\approx \\frac{1}{0.318309886183791} \\approx 3.141592653589793[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Comparing with [latex]\\pi = 3.141592653589793...[\/latex], the first term alone gives [latex]\\pi[\/latex] correct to <strong>15 decimal places<\/strong>!<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Adding the second term ([latex]n=1[\/latex]):<\/strong>\r\n[latex]a_1 = \\frac{4! \\cdot (1103+26390)}{(1!)^4 \\cdot 396^4} = \\frac{24 \\cdot 27493}{396^4} \\approx 2.68 \\times 10^{-8}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\pi} \\approx \\frac{\\sqrt{8}}{9801}(1103 + 2.68 \\times 10^{-8})[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives [latex]\\pi[\/latex] accurate to approximately <strong>23 decimal places<\/strong>.<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the ratio test to check if a series converges absolutely<\/li>\n<li>Use the root test to check if a series converges absolutely<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Series Converging to [latex]\\pi[\/latex] and [latex]\\frac{1}{\\pi}[\/latex]<\/h2>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Throughout history, mathematicians have been fascinated by the challenge of computing [latex]\\pi[\/latex] with ever-increasing precision. This quest has led to the discovery of numerous infinite series that converge to [latex]\\pi[\/latex], each with its own remarkable properties and convergence characteristics.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Dozens of series exist that converge to [latex]\\pi[\/latex] or an algebraic expression containing [latex]\\pi[\/latex]. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of [latex]\\pi[\/latex] in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">In this exploration, we&#8217;ll examine three historically significant series for \u03c0, each discovered in different eras and demonstrating increasingly sophisticated mathematical techniques. From the elegant simplicity of the Gregory-Leibniz series to the breathtaking efficiency of Ramanujan&#8217;s formula, these series showcase the evolution of mathematical understanding and computational power.<\/p>\n<\/div>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">The following series was discovered by Gregory and Leibniz in the late 1600s:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736707603\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\pi =4\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{2n - 1}=4-\\frac{4}{3}+\\frac{4}{5}-\\frac{4}{7}+\\frac{4}{9}-\\cdots[\/latex]<\/div>\n<p>This result follows from the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex].\u00a0<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1169739171821\" type=\"a\">\n<li>Prove that this series converges.<\/li>\n<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=10,20,50,100[\/latex].<\/li>\n<li>Use the remainder estimate for alternating series to get a bound on the error [latex]{R}_{n}[\/latex].<\/li>\n<li>What is the smallest value of [latex]N[\/latex] that guarantees [latex]|{R}_{N}|<0.01\\text{?}[\/latex] Evaluate [latex]{S}_{N}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q15289\">Show Answer<\/button><\/p>\n<div id=\"q15289\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">The series can be written as [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{4(-1)^{n+1}}{2n-1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This is an alternating series with [latex]a_n = \\frac{4}{2n-1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To apply the alternating series test, we need to verify:<\/p>\n<ol class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]a_n > 0[\/latex] for all [latex]n \\geq 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_{n+1} \\leq a_n[\/latex] for all [latex]n \\geq 1[\/latex] (decreasing)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\lim_{n \\to \\infty} a_n = 0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Verification:<\/p>\n<ol class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]a_n = \\frac{4}{2n-1} > 0[\/latex] since [latex]2n-1 > 0[\/latex] for all [latex]n \\geq 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">We need [latex]a_{n+1} \\leq a_n[\/latex], which means: [latex]\\frac{4}{2(n+1)-1} \\leq \\frac{4}{2n-1}[\/latex] [latex]\\frac{4}{2n+1} \\leq \\frac{4}{2n-1}[\/latex] Since [latex]2n+1 > 2n-1[\/latex] for [latex]n \\geq 1[\/latex], we have [latex]\\frac{1}{2n+1} < \\frac{1}{2n-1}[\/latex], so the sequence is decreasing.<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\lim_{n \\to \\infty} \\frac{4}{2n-1} = 0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Since all conditions are satisfied, the series converges by the alternating series test.<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{array}{rcl}  S_{10} & = & 4\\left(1 - \\tfrac{1}{3} + \\tfrac{1}{5} - \\tfrac{1}{7} + \\tfrac{1}{9} - \\tfrac{1}{11} + \\tfrac{1}{13} - \\tfrac{1}{15} + \\tfrac{1}{17} - \\tfrac{1}{19}\\right) \\\\  & = & 4(0.760460) = 3.041840 \\\\[6pt]  S_{20} & = & 4\\sum_{n=1}^{20} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.772840) = 3.091361 \\\\[6pt]  S_{50} & = & 4\\sum_{n=1}^{50} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.780398) = 3.121593 \\\\[6pt]  S_{100} & = & 4\\sum_{n=1}^{100} \\frac{(-1)^{\\,n+1}}{2n-1} = 4(0.782898) = 3.131593 \\\\  \\end{array}[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">For an alternating series, the remainder estimate states that [latex]|R_n| \\leq a_{n+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore: [latex]|R_n| \\leq \\frac{4}{2(n+1)-1} = \\frac{4}{2n+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">For specific values:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]|R_{10}| \\leq \\frac{4}{21} \\approx 0.190[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]|R_{20}| \\leq \\frac{4}{41} \\approx 0.098[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]|R_{50}| \\leq \\frac{4}{101} \\approx 0.040[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]|R_{100}| \\leq \\frac{4}{201} \\approx 0.020[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">We need [latex]\\frac{4}{2N+1} < 0.01[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\frac{4}{2N+1} < 0.01[\/latex] [latex]4 < 0.01(2N+1)[\/latex] [latex]4 < 0.02N + 0.01[\/latex] [latex]3.99 < 0.02N[\/latex] [latex]N > 199.5[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]N = 200[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">[latex]S_{200} = 4\\sum_{n=1}^{200} \\frac{(-1)^{n+1}}{2n-1} \\approx 3.136593[\/latex]<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The following series has been attributed to Newton in the late 1600s:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\pi & =6 {\\displaystyle\\sum _{n=0}^{\\infty }} \\frac{\\left(2n\\right)\\text{!}}{{2}^{4n+1}{\\left(n\\text{!}\\right)}^{2}\\left(2n+1\\right)}\\hfill \\\\ & =6\\left(\\frac{1}{2}+\\frac{1}{2\\cdot 3}{\\left(\\frac{1}{2}\\right)}^{3}+\\frac{1\\cdot 3}{2\\cdot 4\\cdot 5}\\cdot {\\left(\\frac{1}{2}\\right)}^{5}+\\frac{1\\cdot 3\\cdot 5}{2\\cdot 4\\cdot 6\\cdot 7}{\\left(\\frac{1}{2}\\right)}^{7}+\\cdots \\right)\\hfill \\end{array}[\/latex]<\/p>\n<p>The proof of this result uses the Maclaurin series for [latex]f\\left(x\\right)={\\sin}^{-1}x[\/latex].<\/p>\n<ol id=\"fs-id1169736851958\" type=\"a\">\n<li>Prove that the series converges.<\/li>\n<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=5,10,20[\/latex].<\/li>\n<li>Compare [latex]{S}_{n}[\/latex] to [latex]\\pi[\/latex] for [latex]n=5,10,20[\/latex] and discuss the number of correct decimal places.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q476703\">Show Answer<\/button><\/p>\n<div id=\"q476703\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">Let [latex]a_n = \\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Using the ratio test:<br \/>\n[latex]\\begin{array}{rcl}  \\dfrac{a_{n+1}}{a_n} & = & \\dfrac{(2n+2)!}{2^{4n+5}((n+1)!)^2(2n+3)} \\cdot \\dfrac{2^{4n+1}(n!)^2(2n+1)}{(2n)!} \\\\  & = & \\dfrac{(2n+2)(2n+1)(2n)!}{2^4(n+1)^2(n!)^2(2n+3)} \\cdot \\dfrac{2^{4n+1}(n!)^2(2n+1)}{2^{4n+1}(2n)!} \\\\  & = & \\dfrac{(2n+2)(2n+1)^2}{16(n+1)^2(2n+3)} \\\\  & = & \\dfrac{2(n+1)(2n+1)^2}{16(n+1)^2(2n+3)} \\\\  & = & \\dfrac{(2n+1)^2}{8(n+1)(2n+3)}  \\end{array}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">As [latex]n \\to \\infty[\/latex]: [latex]\\lim_{n \\to \\infty} \\frac{(2n+1)^2}{8(n+1)(2n+3)} = \\lim_{n \\to \\infty} \\frac{4n^2}{16n^2} = \\frac{1}{4} < 1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Since the ratio test limit is less than 1, the series converges.<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">[latex]S_5 = 6\\sum_{n=0}^{5} \\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Computing each term:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]n=0: \\frac{1!}{2^1 \\cdot 1^2 \\cdot 1} = \\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]n=1: \\frac{2!}{2^5 \\cdot 1^2 \\cdot 3} = \\frac{2}{96} = \\frac{1}{48}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]n=2: \\frac{4!}{2^9 \\cdot 4 \\cdot 5} = \\frac{24}{10240} = \\frac{3}{1280}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]n=3: \\frac{6!}{2^{13} \\cdot 36 \\cdot 7} = \\frac{720}{2064384} \\approx 0.000349[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]n=4: \\frac{8!}{2^{17} \\cdot 576 \\cdot 9} \\approx 0.0000596[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]n=5: \\frac{10!}{2^{21} \\cdot 14400 \\cdot 11} \\approx 0.0000109[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">[latex]S_5 = 6(0.5 + 0.02083 + 0.00234 + 0.000349 + 0.0000596 + 0.0000109) \\approx 3.14159[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]S_{10} \\approx 3.141592653[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]S_{20} \\approx 3.141592653589793[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">[latex]\\pi = 3.141592653589793...[\/latex]<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]S_5 \\approx 3.14159[\/latex]: Correct to 5 decimal places<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_{10} \\approx 3.141592653[\/latex]: Correct to 9 decimal places<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_{20} \\approx 3.141592653589793[\/latex]: Correct to 15 decimal places<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Newton&#8217;s series converges much faster than the Gregory-Leibniz series, gaining approximately 6 additional correct decimal places with each 10 additional terms.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The following series was discovered by Ramanujan in the early 1900s :<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\pi }=\\frac{\\sqrt{8}}{9801}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\left(4n\\right)\\text{!}\\left(1103+26390n\\right)}{{\\left(n\\text{!}\\right)}^{4}{396}^{4n}}[\/latex]<\/p>\n<p>William Gosper, Jr., used this series to calculate [latex]\\pi[\/latex] to an accuracy of more than [latex]17[\/latex] million digits in the mid 1980s. At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for [latex]\\pi[\/latex] and [latex]\\frac{1}{\\pi}[\/latex].<\/p>\n<ol id=\"fs-id1169739252686\" type=\"a\">\n<li>Prove that this series converges.<\/li>\n<li>Evaluate the first term in this series. Compare this number with the value of [latex]\\pi[\/latex] from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q703420\">Show Answer<\/button><\/p>\n<div id=\"q703420\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">Let [latex]a_n = \\frac{(4n)!(1103+26390n)}{(n!)^4 \\cdot 396^{4n}}[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Using the ratio test:<\/p>\n<p>[latex]\\begin{array}{rcl}  \\dfrac{a_{n+1}}{a_n}  & = & \\dfrac{(4n+4)!\\,(1103+26390(n+1))}{((n+1)!)^4 \\cdot 396^{4(n+1)}}  \\cdot \\dfrac{(n!)^4 \\cdot 396^{4n}}{(4n)!\\,(1103+26390n)} \\\\  & = & \\dfrac{(4n+4)(4n+3)(4n+2)(4n+1)(1103+26390n+26390)}{(n+1)^4 \\cdot 396^4}  \\cdot \\dfrac{1}{1103+26390n}  \\end{array}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">As [latex]n \\to \\infty[\/latex], the dominant terms give:<\/p>\n<p>[latex]\\begin{array}{rcl}  \\lim_{n \\to \\infty} \\dfrac{a_{n+1}}{a_n}  & = & \\lim_{n \\to \\infty}  \\dfrac{256n^4 \\cdot 26390n}{n^4 \\cdot 396^4 \\cdot 26390n} \\\\  & = & \\dfrac{256}{396^4} \\\\  & = & \\dfrac{256}{24596196096} < 1 \\\\\\end{array}[\/latex]\n\n\n<p class=\"whitespace-normal break-words\">Since the ratio is less than 1, the series converges by the ratio test.<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\"><strong>First term ([latex]n=0[\/latex]):<br \/>\n<\/strong> [latex]a_0 = \\frac{0! \\cdot 1103}{(0!)^4 \\cdot 396^0} = 1103[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\pi} \\approx \\frac{\\sqrt{8}}{9801} \\cdot 1103 = \\frac{2\\sqrt{2} \\cdot 1103}{9801} \\approx 0.318309886183791[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore: [latex]\\pi \\approx \\frac{1}{0.318309886183791} \\approx 3.141592653589793[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Comparing with [latex]\\pi = 3.141592653589793...[\/latex], the first term alone gives [latex]\\pi[\/latex] correct to <strong>15 decimal places<\/strong>!<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Adding the second term ([latex]n=1[\/latex]):<\/strong><br \/>\n[latex]a_1 = \\frac{4! \\cdot (1103+26390)}{(1!)^4 \\cdot 396^4} = \\frac{24 \\cdot 27493}{396^4} \\approx 2.68 \\times 10^{-8}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]\\frac{1}{\\pi} \\approx \\frac{\\sqrt{8}}{9801}(1103 + 2.68 \\times 10^{-8})[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives [latex]\\pi[\/latex] accurate to approximately <strong>23 decimal places<\/strong>.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":36,"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\/1127"}],"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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1127\/revisions"}],"predecessor-version":[{"id":2390,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1127\/revisions\/2390"}],"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\/1127\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1127"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1127"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1127"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1127"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}