{"id":183,"date":"2026-01-12T16:00:54","date_gmt":"2026-01-12T16:00:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=183"},"modified":"2026-01-12T16:00:54","modified_gmt":"2026-01-12T16:00:54","slug":"sequences-and-series-foundations-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/sequences-and-series-foundations-get-stronger-answer-key\/","title":{"raw":"Sequences and Series Foundations: Get Stronger Answer Key","rendered":"Sequences and Series Foundations: Get Stronger Answer Key"},"content":{"raw":"

Sequences and Their Properties<\/span><\/h2>\r\n
    \r\n \t
  1. [latex]a_n = 0[\/latex] if [latex]n[\/latex] is odd and [latex]a_n = 2[\/latex] if [latex]n[\/latex] is even<\/li>\r\n \t
  2. [latex]{ a_n } = { 1, 3, 6, 10, 15, 21, \\ldots }[\/latex]<\/li>\r\n \t
  3. [latex]a_n = \\dfrac{n (n+1)}{2}[\/latex]<\/li>\r\n \t
  4. [latex]a_n = 4n - 7[\/latex]<\/li>\r\n \t
  5. [latex]a_n = 3 \\cdot 10^{1-n} = 30 \\cdot 10^{-n}[\/latex]<\/li>\r\n \t
  6. [latex]a_n = 2^n - 1[\/latex]<\/li>\r\n \t
  7. [latex]a_n = \\dfrac{(-1)^{n-1}}{2n-1}[\/latex]<\/li>\r\n \t
  8. [latex]f(n) = 2^n[\/latex]<\/li>\r\n \t
  9. [latex]f(n) = \\dfrac{n!}{2^{n-2}}[\/latex]<\/li>\r\n \t
  10. Terms oscillate above and below [latex]\\dfrac{5}{3}[\/latex] and appear to converge to [latex]\\dfrac{5}{3}[\/latex].<\/li>\r\n \t
  11. Terms oscillate above and below [latex]y \\approx 1.57 \\ldots[\/latex] and appear to converge to a limit.<\/li>\r\n \t
  12. Graph oscillates and suggests no limit.<\/li>\r\n \t
  13. [latex]7[\/latex]<\/li>\r\n \t
  14. [latex]0[\/latex]<\/li>\r\n \t
  15. [latex]0[\/latex]<\/li>\r\n \t
  16. [latex]1[\/latex]<\/li>\r\n \t
  17. bounded, decreasing for [latex]n \\ge 1[\/latex]<\/li>\r\n \t
  18. bounded, not monotone<\/li>\r\n \t
  19. bounded, decreasing<\/li>\r\n \t
  20. not monotone, not bounded<\/li>\r\n \t
  21. [latex]a_n[\/latex] is decreasing and bounded below by [latex]2[\/latex]. The limit [latex]a[\/latex] must satisfy [latex]a = \\sqrt{2a}[\/latex] so [latex]a = 2[\/latex], independent of the initial value.<\/li>\r\n \t
  22. [latex]0[\/latex]<\/li>\r\n \t
  23. [latex]0: |\\sin{x}| \\le |x|[\/latex] and [latex]|\\sin{x}| \\le 1[\/latex] so [latex]-\\dfrac{1}{n} \\le a_n \\le \\dfrac{1}{n}[\/latex]<\/li>\r\n \t
  24. [latex]n^{\\dfrac{1}{n}} \\to 1[\/latex] and [latex]2^{\\dfrac{1}{n}} \\to 1[\/latex], so [latex]a_n \\to 0[\/latex]<\/li>\r\n \t
  25. Since [latex]\\left(1 + \\dfrac{1}{n}\\right)^n \\to e[\/latex], one has [latex]\\left(1 - \\dfrac{2}{n}\\right)^n \\approx \\left(1 + k\\right)^{-2k} \\to e^{-2}[\/latex] as [latex]k \\to \\infty[\/latex].<\/li>\r\n \t
  26. [latex]2^n + 3^n \\le 2 \\cdot 3^n[\/latex] and [latex]\\dfrac{3^n}{4^n} \\to 0[\/latex] as [latex]n \\to \\infty[\/latex], so [latex]a_n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\r\n \t
  27. [latex]0[\/latex]<\/li>\r\n \t
  28. [latex]\\dfrac{a_{n+1}}{a_n} = \\dfrac{n!}{(n+1)(n+2) \\ldots (2n)} = \\dfrac{1 \\cdot 2 \\cdot 3 \\ldots n}{(n+1)(n+2) \\ldots (2n)} < \\dfrac{1}{2^n}[\/latex]. In particular, [latex]\\dfrac{a_{n+1}}{a_n} \\le \\dfrac{1}{2}[\/latex], so [latex]a_n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\r\n \t
  29. a. Without losses, the population would obey [latex]P_n = 1.06 P_{n-1}[\/latex]. The subtraction of [latex]150[\/latex] accounts for fish losses.\r\nb. After [latex]12[\/latex] months, we have [latex]P_{12} \\approx 1494[\/latex].<\/li>\r\n \t
  30. a. The student owes [latex]$9383[\/latex] after [latex]12[\/latex] months.\r\nb. The loan will be paid in full after [latex]139[\/latex] months or eleven and a half years.<\/li>\r\n<\/ol>\r\n

    Introduction to Series<\/span><\/h2>\r\n
      \r\n \t
    1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n}[\/latex]<\/li>\r\n \t
    2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}[\/latex]<\/li>\r\n \t
    3. [latex]1,3,6,10[\/latex]<\/li>\r\n \t
    4. [latex]1,1,0,0[\/latex]<\/li>\r\n \t
    5. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\\dfrac{1}{n - 1}-\\dfrac{1}{n}[\/latex]. Series converges to [latex]S=1[\/latex].<\/li>\r\n \t
    6. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\\sqrt{n}-\\sqrt{n - 1}=\\dfrac{1}{\\sqrt{n - 1}+\\sqrt{n}}[\/latex]. Series diverges because partial sums are unbounded.<\/li>\r\n \t
    7. [latex]{S}{1}=\\dfrac{1}{3}[\/latex], [latex]{S}{2}=\\dfrac{1}{3}+\\dfrac{2}{4}>\\dfrac{1}{3}+\\dfrac{1}{3}=\\dfrac{2}{3}[\/latex], [latex]{S}{3}=\\dfrac{1}{3}+\\dfrac{2}{4}+\\dfrac{3}{5}>3\\cdot \\left(\\dfrac{1}{3}\\right)=1[\/latex]. In general [latex]{S}{k}>\\dfrac{k}{3}[\/latex]. Series diverges.<\/li>\r\n \t
    8. [latex]\\begin{array}{l}{S}{1}=\\dfrac{1}{\\left(2.3\\right)}=\\dfrac{1}{6}=\\dfrac{2}{3} - \\dfrac{1}{2},\\hfill \\\\ {S}{2}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}=\\dfrac{2}{12}+\\dfrac{1}{12}=\\dfrac{1}{4}=\\dfrac{3}{4} - \\dfrac{1}{2},\\hfill \\\\ {S}{3}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}+\\dfrac{1}{\\left(4.5\\right)}=\\dfrac{10}{60}+\\dfrac{5}{60}+\\dfrac{3}{60}=\\dfrac{3}{10}=\\dfrac{4}{5} - \\dfrac{1}{2},\\hfill \\\\ {S}{4}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}+\\dfrac{1}{\\left(4.5\\right)}+\\dfrac{1}{\\left(5.6\\right)}=\\dfrac{10}{60}+\\dfrac{5}{60}+\\dfrac{3}{60}+\\dfrac{2}{60}=\\dfrac{1}{3}=\\dfrac{5}{6} - \\dfrac{1}{2}.\\hfill \\end{array}[\/latex]The pattern is [latex]{S}_{k}=\\dfrac{\\left(k+1\\right)}{\\left(k+2\\right)}-\\dfrac{1}{2}[\/latex] and the series converges to [latex]\\dfrac{1}{2}[\/latex].<\/li>\r\n \t
    9. [latex]0[\/latex]<\/li>\r\n \t
    10. [latex]-3[\/latex]<\/li>\r\n \t
    11. diverges, [latex]\\displaystyle\\sum _{n=1001}^{\\infty }\\dfrac{1}{n}[\/latex]<\/li>\r\n \t
    12. convergent geometric series, [latex]r=\\dfrac{1}{10}<1[\/latex]<\/li>\r\n \t
    13. convergent geometric series, [latex]r=\\dfrac{\\pi}{{e}^{2}}<1[\/latex]<\/li>\r\n \t
    14. [latex]\\displaystyle\\sum _{n=1}^{\\infty }5\\cdot {\\left(-\\dfrac{1}{5}\\right)}^{n}[\/latex], converges to [latex]-\\dfrac{5}{6}[\/latex]<\/li>\r\n \t
    15. [latex]\\displaystyle\\sum _{n=1}^{\\infty }100\\cdot {\\left(\\dfrac{1}{10}\\right)}^{n}[\/latex], converges to [latex]\\dfrac{100}{9}[\/latex]<\/li>\r\n \t
    16. [latex]x\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\text{-}x\\right)}^{n}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{x}^{n}[\/latex]<\/li>\r\n \t
    17. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\sin}^{2n}\\left(x\\right)[\/latex]<\/li>\r\n \t
    18. [latex]{S}_{k}=2-{2}^{\\dfrac{1}{\\left(k+1\\right)}}\\to 1[\/latex] as [latex]k\\to \\infty [\/latex].<\/li>\r\n \t
    19. [latex]{S}_{k}=1-\\sqrt{k+1}[\/latex] diverges<\/li>\r\n \t
    20. [latex]\\displaystyle\\sum {n=1}^{\\infty }\\text{ln}n-\\text{ln}\\left(n+1\\right),{S}{k}=\\text{-}\\text{ln}\\left(k+1\\right)[\/latex]<\/li>\r\n \t
    21. [latex]{a}{n}=\\dfrac{1}{\\text{ln}n}-\\dfrac{1}{\\text{ln}\\left(n+1\\right)}[\/latex] and [latex]{S}{k}=\\dfrac{1}{\\text{ln}\\left(2\\right)}-\\dfrac{1}{\\text{ln}\\left(k+1\\right)}\\to \\dfrac{1}{\\text{ln}\\left(2\\right)}[\/latex]<\/li>\r\n \t
    22. [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}=f\\left(1\\right)-f\\left(2\\right)[\/latex]<\/li>\r\n \t
    23. [latex]{c}{0}+{c}{1}+{c}{2}+{c}{3}+{c}_{4}=0[\/latex]<\/li>\r\n \t
    24. [latex]\\dfrac{2}{{n}^{3}-1}=\\dfrac{1}{n - 1}-\\dfrac{2}{n}+\\dfrac{1}{n+1}[\/latex], [latex]{S}_{n}=\\left(1 - 1+\\dfrac{1}{3}\\right)+\\left(\\dfrac{1}{2} - \\dfrac{2}{3}+\\dfrac{1}{4}\\right)[\/latex] [latex]+\\left(\\dfrac{1}{3} - \\dfrac{2}{4}+\\dfrac{1}{5}\\right)+\\left(\\dfrac{1}{4} - \\dfrac{2}{5}+\\dfrac{1}{6}\\right)+\\text{\u22ef }=\\dfrac{1}{2}[\/latex]<\/li>\r\n \t
    25. [latex]{t}_{k}[\/latex] converges to [latex]0.57721\\text{\u22ef }{t}_{k}[\/latex] is a sum of rectangles of height [latex]\\dfrac{1}{k}[\/latex] over the interval [latex]\\left[k,k+1\\right][\/latex] which lie above the graph of [latex]\\dfrac{1}{x}[\/latex].<\/li>\r\n \t
    26. [latex]N=22[\/latex], [latex]{S}_{N}=6.1415[\/latex]<\/li>\r\n \t
    27. [latex]N=3[\/latex], [latex]{S}_{N}=1.559877597243667..[\/latex].<\/li>\r\n \t
    28. a. The probability of any given ordered sequence of outcomes for [latex]n[\/latex] coin flips is [latex]\\dfrac{1}{{2}^{n}}[\/latex]. b. The probability of coming up heads for the first time on the [latex]n[\/latex] th flip is the probability of the sequence [latex]TT\\text{$\\ldots$\u00a0}TH[\/latex] which is [latex]\\dfrac{1}{{2}^{n}}[\/latex]. The probability of coming up heads for the first time on an even flip is [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{2}^{2n}}[\/latex] or [latex]\\dfrac{1}{3}[\/latex].<\/li>\r\n \t
    29. [latex]\\dfrac{5}{9}[\/latex]<\/li>\r\n \t
    30. The part of the first dose after [latex]n[\/latex] hours is [latex]d{r}^{n}[\/latex], the part of the second dose is [latex]d{r}^{n-N}[\/latex], and, in general, the part remaining of the [latex]m\\text{th}[\/latex] dose is [latex]d{r}^{n-mN}[\/latex], so [latex]A\\left(n\\right)=\\displaystyle\\sum _{l=0}^{m}d{r}^{n-lN}=\\displaystyle\\sum _{l=0}^{m}d{r}^{k+\\left(m-l\\right)N}=\\displaystyle\\sum _{q=0}^{m}d{r}^{k+qN}=d{r}^{k}\\displaystyle\\sum _{q=0}^{m}{r}^{Nq}=d{r}^{k}\\dfrac{1-{r}^{\\left(m+1\\right)N}}{1-{r}^{N}},n=k+mN[\/latex].<\/li>\r\n<\/ol>\r\n

      The Divergence and Integral Tests<\/span><\/h2>\r\n
        \r\n \t
      1. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/li>\r\n \t
      2. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=2[\/latex]. Series diverges.<\/li>\r\n \t
      3. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\infty [\/latex] (does not exist). Series diverges.<\/li>\r\n \t
      4. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex]. Series diverges.<\/li>\r\n \t
      5. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist. Series diverges.<\/li>\r\n \t
      6. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\dfrac{1}{{e}^{2}}[\/latex]. Series diverges.<\/li>\r\n \t
      7. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/li>\r\n \t
      8. Series converges, [latex]p>1[\/latex].<\/li>\r\n \t
      9. Series converges, [latex]p=\\dfrac{4}{3}>1[\/latex].<\/li>\r\n \t
      10. Series converges, [latex]p=2e-\\pi >1[\/latex].<\/li>\r\n \t
      11. Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{dx}{{\\left(x+5\\right)}^{\\frac{1}{3}}}[\/latex].<\/li>\r\n \t
      12. Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{x}{1+{x}^{2}}dx[\/latex].<\/li>\r\n \t
      13. Series converges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{2x}{1+{x}^{4}}dx[\/latex].<\/li>\r\n \t
      14. [latex]{2}^{\\text{-}\\text{ln}n}=\\dfrac{1}{{n}^{\\text{ln}2}}[\/latex]. Since [latex]\\text{ln}2<1[\/latex], diverges by [latex]p[\/latex] -series.<\/li>\r\n \t
      15. [latex]{2}^{-2\\text{ln}n}=\\dfrac{1}{{n}^{2\\text{ln}2}}[\/latex]. Since [latex]2\\text{ln}2 - 1<1[\/latex], diverges by [latex]p[\/latex] -series.<\/li>\r\n \t
      16. [latex]{R}{1000}\\le {\\displaystyle\\int }{1000}^{\\infty }\\dfrac{dt}{{t}^{2}}=-\\dfrac{1}{t}{|}_{1000}^{\\infty }=0.001[\/latex]<\/li>\r\n \t
      17. [latex]{R}{1000}\\le {\\displaystyle\\int }{1000}^{\\infty }\\dfrac{dt}{1+{t}^{2}}={\\tan}^{-1}\\infty -{\\tan}^{-1}\\left(1000\\right)=\\dfrac{\\pi}{2}-{\\tan}^{-1}\\left(1000\\right)\\approx 0.000999[\/latex]<\/li>\r\n \t
      18. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{x}^{2}}=\\dfrac{1}{N},N>{10}^{4}[\/latex]<\/li>\r\n \t
      19. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{x}^{1.01}}=100{N}^{-0.01},N>{10}^{600}[\/latex]<\/li>\r\n \t
      20. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{1+{x}^{2}}=\\dfrac{\\pi}{2}-{\\tan}^{-1}\\left(N\\right),N>\\tan\\left(\\dfrac{\\pi}{2}-{10}^{-3}\\right)\\approx 1000[\/latex]<\/li>\r\n \t
      21. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{e}^{x}}={e}^{\\text{-}N},N>5\\text{ln}\\left(10\\right)[\/latex], okay if [latex]N=12;\\displaystyle\\sum _{n=1}^{12}{e}^{\\text{-}n}=0.581973...[\/latex]. Estimate agrees with [latex]\\dfrac{1}{\\left(e - 1\\right)}[\/latex] to five decimal places.<\/li>\r\n \t
      22. [latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\dfrac{dx}{{x}^{4}}=\\dfrac{4}{{N}^{3}},N>{\\left({4.10}^{4}\\right)}^{\\frac{1}{3}}[\/latex], okay if [latex]N=35[\/latex];[latex]\\displaystyle\\sum _{n=1}^{35}\\dfrac{1}{{n}^{4}}=1.08231\\text{...}[\/latex]. Estimate agrees with the sum to four decimal places.<\/li>\r\n \t
      23. [latex]\\text{ln}\\left(2\\right)[\/latex]<\/li>\r\n \t
      24. [latex]T=0.5772..[\/latex].<\/li>\r\n \t
      25. The expected number of random insertions to get [latex]B[\/latex] to the top is [latex]n+\\dfrac{n}{2}+\\dfrac{n}{3}+\\text{...}+\\dfrac{n}{\\left(n - 1\\right)}[\/latex]. Then one more insertion puts [latex]B[\/latex] back in at random. Thus, the expected number of shuffles to randomize the deck is [latex]n\\left(1+\\dfrac{1}{2}+\\text{...}+\\dfrac{1}{n}\\right)[\/latex].<\/li>\r\n \t
      26. Set [latex]{b}{n}={a}{n+N}[\/latex] and [latex]g\\left(t\\right)=f\\left(t+N\\right)[\/latex] such that [latex]f[\/latex] is decreasing on [latex]\\left[t,\\infty \\right)[\/latex].<\/li>\r\n \t
      27. The series converges for [latex]p>1[\/latex] by integral test using change of variable.<\/li>\r\n \t
      28. [latex]N={e}^{{e}^{100}}\\approx {e}^{{10}^{43}}[\/latex] terms are needed.<\/li>\r\n<\/ol>\r\n

        Comparison Tests<\/span><\/h2>\r\n
          \r\n \t
        1. Converges by comparison with [latex]\\dfrac{1}{{n}^{2}}[\/latex].<\/li>\r\n \t
        2. Diverges by comparison with harmonic series, since [latex]2n - 1\\ge n[\/latex].<\/li>\r\n \t
        3. [latex]{a}_{n}=\\dfrac{1}{\\left(n+1\\right)\\left(n+2\\right)}<\\dfrac{1}{{n}^{2}}[\/latex]. Converges by comparison with p-series, [latex]p=2[\/latex].<\/li>\r\n \t
        4. [latex]\\sin\\left(\\dfrac{1}{n}\\right)\\le \\dfrac{1}{n}[\/latex], so converges by comparison with p-series, [latex]p=2[\/latex].<\/li>\r\n \t
        5. [latex]\\sin\\left(\\dfrac{1}{n}\\right)\\le 1[\/latex], so converges by comparison with p-series, [latex]p=\\dfrac{3}{2}[\/latex].<\/li>\r\n \t
        6. Since [latex]\\sqrt{n+1}-\\sqrt{n}=\\dfrac{1}{\\left(\\sqrt{n+1}+\\sqrt{n}\\right)}\\le \\dfrac{2}{\\sqrt{n}}[\/latex], series converges by comparison with p-series for [latex]p=1.5[\/latex].<\/li>\r\n \t
        7. Converges by limit comparison with p-series for [latex]p>1[\/latex].<\/li>\r\n \t
        8. Converges by limit comparison with p-series, [latex]p=2[\/latex].<\/li>\r\n \t
        9. Converges by limit comparison with [latex]{4}^{\\text{-}n}[\/latex].<\/li>\r\n \t
        10. Converges by limit comparison with [latex]\\dfrac{1}{{e}^{1.1n}}[\/latex].<\/li>\r\n \t
        11. Diverges by limit comparison with harmonic series.<\/li>\r\n \t
        12. Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/li>\r\n \t
        13. Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/li>\r\n \t
        14. Diverges by limit comparison with [latex]\\dfrac{1}{n}[\/latex].<\/li>\r\n \t
        15. Converges for [latex]p>1[\/latex] by comparison with a [latex]p[\/latex] series for slightly smaller [latex]p[\/latex].<\/li>\r\n \t
        16. Converges for all [latex]p>0[\/latex].<\/li>\r\n \t
        17. Converges for all [latex]r>1[\/latex]. If [latex]r>1[\/latex] then [latex]{r}^{n}>4[\/latex], say, once [latex]n>\\dfrac{\\text{ln}\\left(2\\right)}{\\text{ln}\\left(r\\right)}[\/latex] and then the series converges by limit comparison with a geometric series with ratio [latex]\\dfrac{1}{2}[\/latex].<\/li>\r\n \t
        18. The numerator is equal to [latex]1[\/latex] when [latex]n[\/latex] is odd and [latex]0[\/latex] when [latex]n[\/latex] is even, so the series can be rewritten [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{2n+1}[\/latex], which diverges by limit comparison with the harmonic series.<\/li>\r\n \t
        19. [latex]{\\left(a-b\\right)}^{2}={a}^{2}-2ab+{b}^{2}[\/latex] or [latex]{a}^{2}+{b}^{2}\\ge 2ab[\/latex], so convergence follows from comparison of [latex]2{a}{n}{b}{n}[\/latex] with [latex]{a}^{2}{}{n}+{b}^{2}{}{n}[\/latex]. Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.<\/li>\r\n \t
        20. [latex]{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}={e}^{\\text{-}\\text{ln}\\left(n\\right)\\text{ln}\\text{ln}\\left(n\\right)}[\/latex]. If [latex]n[\/latex] is sufficiently large, then [latex]\\text{ln}\\text{ln}n>2[\/latex], so [latex]{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}<\\dfrac{1}{{n}^{2}}[\/latex], and the series converges by comparison to a [latex]p-\\text{series}\\text{.}[\/latex]<\/li>\r\n \t
        21. [latex]{a}{n}\\to 0[\/latex], so [latex]{a}^{2}{}{n}\\le |{a}_{n}|[\/latex] for large [latex]n[\/latex]. Convergence follows from limit comparison. [latex]\\displaystyle\\sum \\dfrac{1}{{n}^{2}}[\/latex] converges, but [latex]\\displaystyle\\sum \\dfrac{1}{n}[\/latex] does not, so the fact that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges does not imply that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges.<\/li>\r\n \t
        22. No. [latex]\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{1}{n}[\/latex] diverges. Let [latex]{b}{k}=0[\/latex] unless [latex]k={n}^{2}[\/latex] for some [latex]n[\/latex]. Then [latex]\\displaystyle\\sum {k}\\dfrac{{b}{k}}{k}=\\displaystyle\\sum \\dfrac{1}{{k}^{2}}[\/latex] converges.<\/li>\r\n \t
        23. [latex]|\\sin{t}|\\le |t|[\/latex], so the result follows from the comparison test.<\/li>\r\n \t
        24. By the comparison test, [latex]x=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\le \\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{2}^{n}}=1[\/latex].<\/li>\r\n \t
        25. If [latex]{b}_{1}=0[\/latex], then, by comparison, [latex]x\\le \\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{1}{{2}^{n}}=\\dfrac{1}{2}[\/latex].<\/li>\r\n \t
        26. Yes. Keep adding [latex]1\\text{-kg}[\/latex] weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the [latex]1\\text{-kg}[\/latex] weights, and add [latex]0.1\\text{-kg}[\/latex] weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last [latex]0.1\\text{-kg}[\/latex] weight. Start adding [latex]0.01\\text{-kg}[\/latex] weights. If it balances, stop. If it tips to the side with the weights, remove the last [latex]0.01\\text{-kg}[\/latex] weight that was added. Continue in this way for the [latex]0.001\\text{-kg}[\/latex] weights, and so on. After a finite number of steps, one has a finite series of the form [latex]A+\\displaystyle\\sum {n=1}^{N}\\dfrac{{s}{n}}{{10}^{n}}[\/latex] where [latex]A[\/latex] is the number of full kg weights and [latex]{d}_{n}[\/latex] is the number of [latex]\\dfrac{1}{{10}^{n}}\\text{-kg}[\/latex] weights that were added. If at some state this series is Robert's exact weight, the process will stop. Otherwise it represents the [latex]N\\text{th}[\/latex] partial sum of an infinite series that gives Robert's exact weight, and the error of this sum is at most [latex]\\dfrac{1}{{10}^{N}}[\/latex].<\/li>\r\n<\/ol>\r\n

          Alternating Series<\/span><\/h2>\r\n
            \r\n \t
          1. Does not converge by divergence test. Terms do not tend to zero.<\/li>\r\n \t
          2. Converges conditionally by alternating series test, since [latex]\\dfrac{\\sqrt{n+3}}{n}[\/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\\dfrac{1}{2}[\/latex].<\/li>\r\n \t
          3. Converges absolutely by limit comparison to [latex]\\dfrac{{3}^{n}}{{4}^{n}}[\/latex], for example.<\/li>\r\n \t
          4. Diverges by divergence test since [latex]\\underset{n\\to \\infty }{\\text{lim}}|{a}_{n}|=e[\/latex].<\/li>\r\n \t
          5. Does not converge. Terms do not tend to zero.<\/li>\r\n \t
          6. [latex]\\underset{n\\to \\infty }{\\text{lim}}{\\cos}^{2}\\left(\\dfrac{1}{n}\\right)=1[\/latex]. Diverges by divergence test.<\/li>\r\n \t
          7. Converges by alternating series test.<\/li>\r\n \t
          8. Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\\pi -e[\/latex]<\/li>\r\n \t
          9. Diverges; terms do not tend to zero.<\/li>\r\n \t
          10. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/li>\r\n \t
          11. Converges absolutely by limit comparison with p-series, [latex]p=\\dfrac{3}{2}[\/latex], after applying the hint.<\/li>\r\n \t
          12. Converges by alternating series test since [latex]n\\left({\\tan}^{-1}\\left(n+1\\right)\\text{-}{\\tan}^{-1}n\\right)[\/latex] is decreasing to zero for large [latex]n[\/latex]. Does not converge absolutely by limit comparison with harmonic series after applying hint.<\/li>\r\n \t
          13. Converges absolutely, since [latex]{a}_{n}=\\dfrac{1}{n}-\\dfrac{1}{n+1}[\/latex] are terms of a telescoping series.<\/li>\r\n \t
          14. Terms do not tend to zero. Series diverges by divergence test.<\/li>\r\n \t
          15. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/li>\r\n \t
          16. [latex]\\text{ln}\\left(N+1\\right)>10[\/latex], [latex]N+1>{e}^{10}[\/latex], [latex]N\\ge 22026[\/latex]; [latex]{S}_{22026}=0.0257\\text{...}[\/latex]<\/li>\r\n \t
          17. [latex]{2}^{N+1}>{10}^{6}[\/latex] or [latex]N+1>\\dfrac{6\\text{ln}\\left(10\\right)}{\\text{ln}\\left(2\\right)}=19.93[\/latex]. or [latex]N\\ge 19[\/latex]; [latex]{S}_{19}=0.333333969\\text{...}[\/latex]<\/li>\r\n \t
          18. [latex]{\\left(N+1\\right)}^{2}>{10}^{6}[\/latex] or [latex]N>999[\/latex]; [latex]{S}_{1000}\\approx 0.822466[\/latex].<\/li>\r\n \t
          19. True. [latex]{b}{n}[\/latex] need not tend to zero since if [latex]{c}{n}={b}{n}-\\text{lim}{b}{n}[\/latex], then [latex]{c}{2n - 1}-{c}{2n}={b}{2n - 1}-{b}{2n}[\/latex].<\/li>\r\n \t
          20. True. [latex]{b}{3n - 1}-{b}{3n}\\ge 0[\/latex], so convergence of [latex]\\displaystyle\\sum {b}_{3n - 2}[\/latex] follows from the comparison test.<\/li>\r\n \t
          21. True. If one converges, then so must the other, implying absolute convergence.<\/li>\r\n \t
          22. Yes. Take [latex]{b}{n}=1[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{b}{n}=0[\/latex] if [latex]{a}{n}<0[\/latex]. Then [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}=\\displaystyle\\sum {n:{a}{n}\\ge 0}{a}{n}[\/latex] converges. Similarly, one can show [latex]\\displaystyle\\sum {n:{a}{n}<0}{a}_{n}[\/latex] converges. Since both series converge, the series must converge absolutely.<\/li>\r\n \t
          23. Not decreasing. Does not converge absolutely.<\/li>\r\n \t
          24. Not alternating. Can be expressed as [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\dfrac{1}{3n - 2}+\\dfrac{1}{3n - 1}-\\dfrac{1}{3n}\\right)[\/latex], which diverges by comparison with [latex]\\displaystyle\\sum \\dfrac{1}{3n - 2}[\/latex].<\/li>\r\n \t
          25. Let [latex]{a}^{+}{}{n}={a}{n}[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{a}^{+}{}{n}=0[\/latex] if [latex]{a}{n}<0[\/latex]. Then [latex]{a}^{+}{}{n}\\le |{a}{n}|[\/latex] for all [latex]n[\/latex] so the sequence of partial sums of [latex]{a}^{+}{}{n}[\/latex] is increasing and bounded above by the sequence of partial sums of [latex]|{a}_{n}|[\/latex], which converges; hence, [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{+}{}{n}[\/latex] converges.<\/li>\r\n \t
          26. For [latex]N=5[\/latex] one has [latex]|{R}{N}|{b}{6}=\\dfrac{{\\theta }^{10}}{10\\text{!}}[\/latex]. When [latex]\\theta =1[\/latex], [latex]{R}{5}\\le \\dfrac{1}{10\\text{!}}\\approx 2.75\\times {10}^{-7}[\/latex]. When [latex]\\theta =\\dfrac{\\pi}{6}[\/latex], [latex]{R}{5}\\le {\\left(\\dfrac{\\pi}{6}\\right)}^\\dfrac{{10}}{10\\text{!}}\\approx 4.26\\times {10}^{-10}[\/latex]. When [latex]\\theta =\\pi [\/latex], [latex]{R}_{5}\\le \\dfrac{{\\pi }^{10}}{10\\text{!}}=0.0258[\/latex].<\/li>\r\n \t
          27. Let [latex]{b}_{n}=\\dfrac{1}{\\left(2n - 2\\right)}\\text{!}[\/latex]. Then [latex]{R}_{N}\\le \\dfrac{1}{\\left(2N\\right)\\text{!}}<0.00001[\/latex] when [latex]\\left(2N\\right)\\text{!}>{10}^{5}[\/latex] or [latex]N=5[\/latex] and [latex]1-\\dfrac{1}{2\\text{!}}+\\dfrac{1}{4\\text{!}}-\\dfrac{1}{6\\text{!}}+\\dfrac{1}{8\\text{!}}=0.540325\\text{...}[\/latex], whereas [latex]\\cos1=0.5403023\\text{...}[\/latex]<\/li>\r\n \t
          28. Let [latex]T=\\displaystyle\\sum \\dfrac{1}{{n}^{2}}[\/latex]. Then [latex]T-S=\\dfrac{1}{2}T[\/latex], so [latex]S=\\dfrac{T}{2}[\/latex]. [latex]\\sqrt{6\\times \\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{{n}^{2}}}=3.140638\\text{...}[\/latex]; [latex]\\sqrt{12\\times \\displaystyle\\sum _{n=1}^{1000}\\dfrac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}}=3.141591\\text{...}[\/latex]; [latex]\\pi =3.141592\\text{...}[\/latex]. The alternating series is more accurate for [latex]1000[\/latex] terms.<\/li>\r\n<\/ol>\r\n

            Ratio and Root Tests<\/span><\/h2>\r\n
              \r\n \t
            1. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to 0[\/latex]. Converges.<\/li>\r\n \t
            2. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=\\dfrac{1}{2}{\\left(\\dfrac{n+1}{n}\\right)}^{2}\\to \\dfrac{1}{2}<1[\/latex]. Converges.<\/li>\r\n \t
            3. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to \\dfrac{1}{27}<1[\/latex]. Converges.<\/li>\r\n \t
            4. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to \\dfrac{4}{{e}^{2}}<1[\/latex]. Converges.<\/li>\r\n \t
            5. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to 1[\/latex]. Ratio test is inconclusive.<\/li>\r\n \t
            6. [latex]\\dfrac{{a}{n}}{{a}{n+1}}\\to \\dfrac{1}{{e}^{2}}[\/latex]. Converges.<\/li>\r\n \t
            7. [latex]{\\left({a}_{k}\\right)}^{\\dfrac{1}{k}}\\to 2>1[\/latex]. Diverges.<\/li>\r\n \t
            8. [latex]{\\left({a}_{n}\\right)}^{\\dfrac{1}{n}}\\to \\dfrac{1}{2}<1[\/latex]. Converges.<\/li>\r\n \t
            9. [latex]{\\left({a}_{k}\\right)}^{\\dfrac{1}{k}}\\to \\dfrac{1}{e}<1[\/latex]. Converges.<\/li>\r\n \t
            10. [latex]{a}_{n}^{\\dfrac{1}{n}}=\\dfrac{1}{e}+\\dfrac{1}{n}\\to \\dfrac{1}{e}<1[\/latex]. Converges.<\/li>\r\n \t
            11. [latex]{a}_{n}^{\\dfrac{1}{n}}=\\dfrac{\\left(\\text{ln}\\left(1+\\text{ln}n\\right)\\right)}{\\left(\\text{ln}n\\right)}\\to 0[\/latex] by L'H\u00f4pital's rule. Converges.<\/li>\r\n \t
            12. [latex]\\dfrac{{a}{k+1}}{{a}{k}}=\\dfrac{1}{2k+1}\\to 0[\/latex]. Converges by ratio test.<\/li>\r\n \t
            13. [latex]{\\left({a}_{n}\\right)}^{\\dfrac{1}{n}}\\to \\dfrac{1}{e}[\/latex]. Converges by root test.<\/li>\r\n \t
            14. [latex]{a}_{k}^{\\dfrac{1}{k}}\\to \\text{ln}\\left(3\\right)>1[\/latex]. Diverges by root test.<\/li>\r\n \t
            15. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=[\/latex] [latex]\\dfrac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\\to 0[\/latex]. Converge.<\/li>\r\n \t
            16. Converges by root test and limit comparison test since [latex]{x}_{n}\\to \\sqrt{2}[\/latex].<\/li>\r\n \t
            17. Converges absolutely by limit comparison with [latex]p-\\text{series,}[\/latex] [latex]p=2[\/latex].<\/li>\r\n \t
            18. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\dfrac{1}{{e}^{2}}\\ne 0[\/latex]. Series diverges.<\/li>\r\n \t
            19. Terms do not tend to zero: [latex]{a}_{k}\\ge \\dfrac{1}{2}[\/latex], since [latex]{\\sin}^{2}x\\le 1[\/latex].<\/li>\r\n \t
            20. [latex]{a}_{n}=\\dfrac{2}{\\left(n+1\\right)\\left(n+2\\right)}[\/latex], which converges by comparison with [latex]p-\\text{series}[\/latex] for [latex]p=2[\/latex].<\/li>\r\n \t
            21. [latex]{a}_{k}=\\dfrac{{2}^{k}1\\cdot 2\\text{...}k}{\\left(2k+1\\right)\\left(2k+2\\right)\\text{...}3k}\\le {\\left(\\dfrac{2}{3}\\right)}^{k}[\/latex] converges by comparison with geometric series.<\/li>\r\n \t
            22. [latex]{a}_{k}\\approx {e}^{\\text{-}\\text{ln}{k}^{2}}=\\dfrac{1}{{k}^{2}}[\/latex]. Series converges by limit comparison with [latex]p-\\text{series,}[\/latex] [latex]p=2[\/latex].<\/li>\r\n \t
            23. If [latex]{b}{k}=\\dfrac{{c}^{1-k}}{\\left(c - 1\\right)}[\/latex] and [latex]{a}{k}=k[\/latex], then [latex]{b}{k+1}-{b}{k}=\\text{-}{c}^{\\text{-}k}[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{k}{{c}^{k}}={a}{1}{b}_{1}+\\dfrac{1}{c - 1}\\displaystyle\\sum _{k=1}^{\\infty }{c}^{\\text{-}k}=\\dfrac{c}{{\\left(c - 1\\right)}^{2}}[\/latex].<\/li>\r\n \t
            24. [latex]6+4+1=11[\/latex]<\/li>\r\n \t
            25. [latex]|x|\\le 1[\/latex]<\/li>\r\n \t
            26. [latex]|x|<\\infty [\/latex]<\/li>\r\n \t
            27. All real numbers [latex]p[\/latex] by the ratio test.<\/li>\r\n \t
            28. [latex]r<\\dfrac{1}{p}[\/latex]<\/li>\r\n \t
            29. [latex]0 < r < 1[\/latex]. Note that the ratio and root tests are inconclusive. Using the hint, there are [latex]2k[\/latex] terms [latex]{r}^{\\sqrt{n}}[\/latex] for [latex]{k}^{2}\\le n<{\\left(k+1\\right)}^{2}[\/latex], and for [latex]r<1[\/latex] each term is at least [latex]{r}^{k}[\/latex]. Thus, [latex]\\displaystyle\\sum _{n=1}^{\\infty }{r}^{\\sqrt{n}}=\\displaystyle\\sum _{k=1}^{\\infty }\\displaystyle\\sum _{n={k}^{2}}^{{\\left(k+1\\right)}^{2}-1}{r}^{\\sqrt{n}}[\/latex] [latex]\\ge \\displaystyle\\sum _{k=1}^{\\infty }2k{r}^{k}[\/latex], which converges by the ratio test for [latex]r<1[\/latex]. For [latex]r\\ge 1[\/latex] the series diverges by the divergence test.<\/li>\r\n \t
            30. One has [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}={a}_{3}=\\dfrac{1}{2}\\text{...}{a}_{2n}={a}_{2n+1}=\\dfrac{1}{{2}^{n}}[\/latex]. The ratio test does not apply because [latex]\\dfrac{{a}_{n+1}}{{a}_{n}}=1[\/latex] if [latex]n[\/latex] is even. However, [latex]\\dfrac{{a}_{n+2}}{{a}_{n}}=\\dfrac{1}{2}[\/latex], so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.<\/li>\r\n<\/ol>","rendered":"

              Sequences and Their Properties<\/span><\/h2>\n
                \n
              1. [latex]a_n = 0[\/latex] if [latex]n[\/latex] is odd and [latex]a_n = 2[\/latex] if [latex]n[\/latex] is even<\/li>\n
              2. [latex]{ a_n } = { 1, 3, 6, 10, 15, 21, \\ldots }[\/latex]<\/li>\n
              3. [latex]a_n = \\dfrac{n (n+1)}{2}[\/latex]<\/li>\n
              4. [latex]a_n = 4n - 7[\/latex]<\/li>\n
              5. [latex]a_n = 3 \\cdot 10^{1-n} = 30 \\cdot 10^{-n}[\/latex]<\/li>\n
              6. [latex]a_n = 2^n - 1[\/latex]<\/li>\n
              7. [latex]a_n = \\dfrac{(-1)^{n-1}}{2n-1}[\/latex]<\/li>\n
              8. [latex]f(n) = 2^n[\/latex]<\/li>\n
              9. [latex]f(n) = \\dfrac{n!}{2^{n-2}}[\/latex]<\/li>\n
              10. Terms oscillate above and below [latex]\\dfrac{5}{3}[\/latex] and appear to converge to [latex]\\dfrac{5}{3}[\/latex].<\/li>\n
              11. Terms oscillate above and below [latex]y \\approx 1.57 \\ldots[\/latex] and appear to converge to a limit.<\/li>\n
              12. Graph oscillates and suggests no limit.<\/li>\n
              13. [latex]7[\/latex]<\/li>\n
              14. [latex]0[\/latex]<\/li>\n
              15. [latex]0[\/latex]<\/li>\n
              16. [latex]1[\/latex]<\/li>\n
              17. bounded, decreasing for [latex]n \\ge 1[\/latex]<\/li>\n
              18. bounded, not monotone<\/li>\n
              19. bounded, decreasing<\/li>\n
              20. not monotone, not bounded<\/li>\n
              21. [latex]a_n[\/latex] is decreasing and bounded below by [latex]2[\/latex]. The limit [latex]a[\/latex] must satisfy [latex]a = \\sqrt{2a}[\/latex] so [latex]a = 2[\/latex], independent of the initial value.<\/li>\n
              22. [latex]0[\/latex]<\/li>\n
              23. [latex]0: |\\sin{x}| \\le |x|[\/latex] and [latex]|\\sin{x}| \\le 1[\/latex] so [latex]-\\dfrac{1}{n} \\le a_n \\le \\dfrac{1}{n}[\/latex]<\/li>\n
              24. [latex]n^{\\dfrac{1}{n}} \\to 1[\/latex] and [latex]2^{\\dfrac{1}{n}} \\to 1[\/latex], so [latex]a_n \\to 0[\/latex]<\/li>\n
              25. Since [latex]\\left(1 + \\dfrac{1}{n}\\right)^n \\to e[\/latex], one has [latex]\\left(1 - \\dfrac{2}{n}\\right)^n \\approx \\left(1 + k\\right)^{-2k} \\to e^{-2}[\/latex] as [latex]k \\to \\infty[\/latex].<\/li>\n
              26. [latex]2^n + 3^n \\le 2 \\cdot 3^n[\/latex] and [latex]\\dfrac{3^n}{4^n} \\to 0[\/latex] as [latex]n \\to \\infty[\/latex], so [latex]a_n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\n
              27. [latex]0[\/latex]<\/li>\n
              28. [latex]\\dfrac{a_{n+1}}{a_n} = \\dfrac{n!}{(n+1)(n+2) \\ldots (2n)} = \\dfrac{1 \\cdot 2 \\cdot 3 \\ldots n}{(n+1)(n+2) \\ldots (2n)} < \\dfrac{1}{2^n}[\/latex]. In particular, [latex]\\dfrac{a_{n+1}}{a_n} \\le \\dfrac{1}{2}[\/latex], so [latex]a_n \\to 0[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\n
              29. a. Without losses, the population would obey [latex]P_n = 1.06 P_{n-1}[\/latex]. The subtraction of [latex]150[\/latex] accounts for fish losses.
                \nb. After [latex]12[\/latex] months, we have [latex]P_{12} \\approx 1494[\/latex].<\/li>\n
              30. a. The student owes [latex]$9383[\/latex] after [latex]12[\/latex] months.
                \nb. The loan will be paid in full after [latex]139[\/latex] months or eleven and a half years.<\/li>\n<\/ol>\n

                Introduction to Series<\/span><\/h2>\n
                  \n
                1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n}[\/latex]<\/li>\n
                2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}[\/latex]<\/li>\n
                3. [latex]1,3,6,10[\/latex]<\/li>\n
                4. [latex]1,1,0,0[\/latex]<\/li>\n
                5. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\\dfrac{1}{n - 1}-\\dfrac{1}{n}[\/latex]. Series converges to [latex]S=1[\/latex].<\/li>\n
                6. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\\sqrt{n}-\\sqrt{n - 1}=\\dfrac{1}{\\sqrt{n - 1}+\\sqrt{n}}[\/latex]. Series diverges because partial sums are unbounded.<\/li>\n
                7. [latex]{S}{1}=\\dfrac{1}{3}[\/latex], [latex]{S}{2}=\\dfrac{1}{3}+\\dfrac{2}{4}>\\dfrac{1}{3}+\\dfrac{1}{3}=\\dfrac{2}{3}[\/latex], [latex]{S}{3}=\\dfrac{1}{3}+\\dfrac{2}{4}+\\dfrac{3}{5}>3\\cdot \\left(\\dfrac{1}{3}\\right)=1[\/latex]. In general [latex]{S}{k}>\\dfrac{k}{3}[\/latex]. Series diverges.<\/li>\n
                8. [latex]\\begin{array}{l}{S}{1}=\\dfrac{1}{\\left(2.3\\right)}=\\dfrac{1}{6}=\\dfrac{2}{3} - \\dfrac{1}{2},\\hfill \\\\ {S}{2}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}=\\dfrac{2}{12}+\\dfrac{1}{12}=\\dfrac{1}{4}=\\dfrac{3}{4} - \\dfrac{1}{2},\\hfill \\\\ {S}{3}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}+\\dfrac{1}{\\left(4.5\\right)}=\\dfrac{10}{60}+\\dfrac{5}{60}+\\dfrac{3}{60}=\\dfrac{3}{10}=\\dfrac{4}{5} - \\dfrac{1}{2},\\hfill \\\\ {S}{4}=\\dfrac{1}{\\left(2.3\\right)}+\\dfrac{1}{\\left(3.4\\right)}+\\dfrac{1}{\\left(4.5\\right)}+\\dfrac{1}{\\left(5.6\\right)}=\\dfrac{10}{60}+\\dfrac{5}{60}+\\dfrac{3}{60}+\\dfrac{2}{60}=\\dfrac{1}{3}=\\dfrac{5}{6} - \\dfrac{1}{2}.\\hfill \\end{array}[\/latex]The pattern is [latex]{S}_{k}=\\dfrac{\\left(k+1\\right)}{\\left(k+2\\right)}-\\dfrac{1}{2}[\/latex] and the series converges to [latex]\\dfrac{1}{2}[\/latex].<\/li>\n
                9. [latex]0[\/latex]<\/li>\n
                10. [latex]-3[\/latex]<\/li>\n
                11. diverges, [latex]\\displaystyle\\sum _{n=1001}^{\\infty }\\dfrac{1}{n}[\/latex]<\/li>\n
                12. convergent geometric series, [latex]r=\\dfrac{1}{10}<1[\/latex]<\/li>\n
                13. convergent geometric series, [latex]r=\\dfrac{\\pi}{{e}^{2}}<1[\/latex]<\/li>\n
                14. [latex]\\displaystyle\\sum _{n=1}^{\\infty }5\\cdot {\\left(-\\dfrac{1}{5}\\right)}^{n}[\/latex], converges to [latex]-\\dfrac{5}{6}[\/latex]<\/li>\n
                15. [latex]\\displaystyle\\sum _{n=1}^{\\infty }100\\cdot {\\left(\\dfrac{1}{10}\\right)}^{n}[\/latex], converges to [latex]\\dfrac{100}{9}[\/latex]<\/li>\n
                16. [latex]x\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\text{-}x\\right)}^{n}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{x}^{n}[\/latex]<\/li>\n
                17. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\sin}^{2n}\\left(x\\right)[\/latex]<\/li>\n
                18. [latex]{S}_{k}=2-{2}^{\\dfrac{1}{\\left(k+1\\right)}}\\to 1[\/latex] as [latex]k\\to \\infty[\/latex].<\/li>\n
                19. [latex]{S}_{k}=1-\\sqrt{k+1}[\/latex] diverges<\/li>\n
                20. [latex]\\displaystyle\\sum {n=1}^{\\infty }\\text{ln}n-\\text{ln}\\left(n+1\\right),{S}{k}=\\text{-}\\text{ln}\\left(k+1\\right)[\/latex]<\/li>\n
                21. [latex]{a}{n}=\\dfrac{1}{\\text{ln}n}-\\dfrac{1}{\\text{ln}\\left(n+1\\right)}[\/latex] and [latex]{S}{k}=\\dfrac{1}{\\text{ln}\\left(2\\right)}-\\dfrac{1}{\\text{ln}\\left(k+1\\right)}\\to \\dfrac{1}{\\text{ln}\\left(2\\right)}[\/latex]<\/li>\n
                22. [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}=f\\left(1\\right)-f\\left(2\\right)[\/latex]<\/li>\n
                23. [latex]{c}{0}+{c}{1}+{c}{2}+{c}{3}+{c}_{4}=0[\/latex]<\/li>\n
                24. [latex]\\dfrac{2}{{n}^{3}-1}=\\dfrac{1}{n - 1}-\\dfrac{2}{n}+\\dfrac{1}{n+1}[\/latex], [latex]{S}_{n}=\\left(1 - 1+\\dfrac{1}{3}\\right)+\\left(\\dfrac{1}{2} - \\dfrac{2}{3}+\\dfrac{1}{4}\\right)[\/latex] [latex]+\\left(\\dfrac{1}{3} - \\dfrac{2}{4}+\\dfrac{1}{5}\\right)+\\left(\\dfrac{1}{4} - \\dfrac{2}{5}+\\dfrac{1}{6}\\right)+\\text{\u22ef }=\\dfrac{1}{2}[\/latex]<\/li>\n
                25. [latex]{t}_{k}[\/latex] converges to [latex]0.57721\\text{\u22ef }{t}_{k}[\/latex] is a sum of rectangles of height [latex]\\dfrac{1}{k}[\/latex] over the interval [latex]\\left[k,k+1\\right][\/latex] which lie above the graph of [latex]\\dfrac{1}{x}[\/latex].<\/li>\n
                26. [latex]N=22[\/latex], [latex]{S}_{N}=6.1415[\/latex]<\/li>\n
                27. [latex]N=3[\/latex], [latex]{S}_{N}=1.559877597243667..[\/latex].<\/li>\n
                28. a. The probability of any given ordered sequence of outcomes for [latex]n[\/latex] coin flips is [latex]\\dfrac{1}{{2}^{n}}[\/latex]. b. The probability of coming up heads for the first time on the [latex]n[\/latex] th flip is the probability of the sequence [latex]TT\\text{$\\ldots$\u00a0}TH[\/latex] which is [latex]\\dfrac{1}{{2}^{n}}[\/latex]. The probability of coming up heads for the first time on an even flip is [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{2}^{2n}}[\/latex] or [latex]\\dfrac{1}{3}[\/latex].<\/li>\n
                29. [latex]\\dfrac{5}{9}[\/latex]<\/li>\n
                30. The part of the first dose after [latex]n[\/latex] hours is [latex]d{r}^{n}[\/latex], the part of the second dose is [latex]d{r}^{n-N}[\/latex], and, in general, the part remaining of the [latex]m\\text{th}[\/latex] dose is [latex]d{r}^{n-mN}[\/latex], so [latex]A\\left(n\\right)=\\displaystyle\\sum _{l=0}^{m}d{r}^{n-lN}=\\displaystyle\\sum _{l=0}^{m}d{r}^{k+\\left(m-l\\right)N}=\\displaystyle\\sum _{q=0}^{m}d{r}^{k+qN}=d{r}^{k}\\displaystyle\\sum _{q=0}^{m}{r}^{Nq}=d{r}^{k}\\dfrac{1-{r}^{\\left(m+1\\right)N}}{1-{r}^{N}},n=k+mN[\/latex].<\/li>\n<\/ol>\n

                  The Divergence and Integral Tests<\/span><\/h2>\n
                    \n
                  1. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/li>\n
                  2. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=2[\/latex]. Series diverges.<\/li>\n
                  3. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\infty[\/latex] (does not exist). Series diverges.<\/li>\n
                  4. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex]. Series diverges.<\/li>\n
                  5. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist. Series diverges.<\/li>\n
                  6. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\dfrac{1}{{e}^{2}}[\/latex]. Series diverges.<\/li>\n
                  7. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/li>\n
                  8. Series converges, [latex]p>1[\/latex].<\/li>\n
                  9. Series converges, [latex]p=\\dfrac{4}{3}>1[\/latex].<\/li>\n
                  10. Series converges, [latex]p=2e-\\pi >1[\/latex].<\/li>\n
                  11. Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{dx}{{\\left(x+5\\right)}^{\\frac{1}{3}}}[\/latex].<\/li>\n
                  12. Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{x}{1+{x}^{2}}dx[\/latex].<\/li>\n
                  13. Series converges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\dfrac{2x}{1+{x}^{4}}dx[\/latex].<\/li>\n
                  14. [latex]{2}^{\\text{-}\\text{ln}n}=\\dfrac{1}{{n}^{\\text{ln}2}}[\/latex]. Since [latex]\\text{ln}2<1[\/latex], diverges by [latex]p[\/latex] -series.<\/li>\n
                  15. [latex]{2}^{-2\\text{ln}n}=\\dfrac{1}{{n}^{2\\text{ln}2}}[\/latex]. Since [latex]2\\text{ln}2 - 1<1[\/latex], diverges by [latex]p[\/latex] -series.<\/li>\n
                  16. [latex]{R}{1000}\\le {\\displaystyle\\int }{1000}^{\\infty }\\dfrac{dt}{{t}^{2}}=-\\dfrac{1}{t}{|}_{1000}^{\\infty }=0.001[\/latex]<\/li>\n
                  17. [latex]{R}{1000}\\le {\\displaystyle\\int }{1000}^{\\infty }\\dfrac{dt}{1+{t}^{2}}={\\tan}^{-1}\\infty -{\\tan}^{-1}\\left(1000\\right)=\\dfrac{\\pi}{2}-{\\tan}^{-1}\\left(1000\\right)\\approx 0.000999[\/latex]<\/li>\n
                  18. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{x}^{2}}=\\dfrac{1}{N},N>{10}^{4}[\/latex]<\/li>\n
                  19. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{x}^{1.01}}=100{N}^{-0.01},N>{10}^{600}[\/latex]<\/li>\n
                  20. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{1+{x}^{2}}=\\dfrac{\\pi}{2}-{\\tan}^{-1}\\left(N\\right),N>\\tan\\left(\\dfrac{\\pi}{2}-{10}^{-3}\\right)\\approx 1000[\/latex]<\/li>\n
                  21. [latex]{R}{N}<{\\displaystyle\\int }{N}^{\\infty }\\dfrac{dx}{{e}^{x}}={e}^{\\text{-}N},N>5\\text{ln}\\left(10\\right)[\/latex], okay if [latex]N=12;\\displaystyle\\sum _{n=1}^{12}{e}^{\\text{-}n}=0.581973...[\/latex]. Estimate agrees with [latex]\\dfrac{1}{\\left(e - 1\\right)}[\/latex] to five decimal places.<\/li>\n
                  22. [latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\dfrac{dx}{{x}^{4}}=\\dfrac{4}{{N}^{3}},N>{\\left({4.10}^{4}\\right)}^{\\frac{1}{3}}[\/latex], okay if [latex]N=35[\/latex];[latex]\\displaystyle\\sum _{n=1}^{35}\\dfrac{1}{{n}^{4}}=1.08231\\text{...}[\/latex]. Estimate agrees with the sum to four decimal places.<\/li>\n
                  23. [latex]\\text{ln}\\left(2\\right)[\/latex]<\/li>\n
                  24. [latex]T=0.5772..[\/latex].<\/li>\n
                  25. The expected number of random insertions to get [latex]B[\/latex] to the top is [latex]n+\\dfrac{n}{2}+\\dfrac{n}{3}+\\text{...}+\\dfrac{n}{\\left(n - 1\\right)}[\/latex]. Then one more insertion puts [latex]B[\/latex] back in at random. Thus, the expected number of shuffles to randomize the deck is [latex]n\\left(1+\\dfrac{1}{2}+\\text{...}+\\dfrac{1}{n}\\right)[\/latex].<\/li>\n
                  26. Set [latex]{b}{n}={a}{n+N}[\/latex] and [latex]g\\left(t\\right)=f\\left(t+N\\right)[\/latex] such that [latex]f[\/latex] is decreasing on [latex]\\left[t,\\infty \\right)[\/latex].<\/li>\n
                  27. The series converges for [latex]p>1[\/latex] by integral test using change of variable.<\/li>\n
                  28. [latex]N={e}^{{e}^{100}}\\approx {e}^{{10}^{43}}[\/latex] terms are needed.<\/li>\n<\/ol>\n

                    Comparison Tests<\/span><\/h2>\n
                      \n
                    1. Converges by comparison with [latex]\\dfrac{1}{{n}^{2}}[\/latex].<\/li>\n
                    2. Diverges by comparison with harmonic series, since [latex]2n - 1\\ge n[\/latex].<\/li>\n
                    3. [latex]{a}_{n}=\\dfrac{1}{\\left(n+1\\right)\\left(n+2\\right)}<\\dfrac{1}{{n}^{2}}[\/latex]. Converges by comparison with p-series, [latex]p=2[\/latex].<\/li>\n
                    4. [latex]\\sin\\left(\\dfrac{1}{n}\\right)\\le \\dfrac{1}{n}[\/latex], so converges by comparison with p-series, [latex]p=2[\/latex].<\/li>\n
                    5. [latex]\\sin\\left(\\dfrac{1}{n}\\right)\\le 1[\/latex], so converges by comparison with p-series, [latex]p=\\dfrac{3}{2}[\/latex].<\/li>\n
                    6. Since [latex]\\sqrt{n+1}-\\sqrt{n}=\\dfrac{1}{\\left(\\sqrt{n+1}+\\sqrt{n}\\right)}\\le \\dfrac{2}{\\sqrt{n}}[\/latex], series converges by comparison with p-series for [latex]p=1.5[\/latex].<\/li>\n
                    7. Converges by limit comparison with p-series for [latex]p>1[\/latex].<\/li>\n
                    8. Converges by limit comparison with p-series, [latex]p=2[\/latex].<\/li>\n
                    9. Converges by limit comparison with [latex]{4}^{\\text{-}n}[\/latex].<\/li>\n
                    10. Converges by limit comparison with [latex]\\dfrac{1}{{e}^{1.1n}}[\/latex].<\/li>\n
                    11. Diverges by limit comparison with harmonic series.<\/li>\n
                    12. Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/li>\n
                    13. Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/li>\n
                    14. Diverges by limit comparison with [latex]\\dfrac{1}{n}[\/latex].<\/li>\n
                    15. Converges for [latex]p>1[\/latex] by comparison with a [latex]p[\/latex] series for slightly smaller [latex]p[\/latex].<\/li>\n
                    16. Converges for all [latex]p>0[\/latex].<\/li>\n
                    17. Converges for all [latex]r>1[\/latex]. If [latex]r>1[\/latex] then [latex]{r}^{n}>4[\/latex], say, once [latex]n>\\dfrac{\\text{ln}\\left(2\\right)}{\\text{ln}\\left(r\\right)}[\/latex] and then the series converges by limit comparison with a geometric series with ratio [latex]\\dfrac{1}{2}[\/latex].<\/li>\n
                    18. The numerator is equal to [latex]1[\/latex] when [latex]n[\/latex] is odd and [latex]0[\/latex] when [latex]n[\/latex] is even, so the series can be rewritten [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{2n+1}[\/latex], which diverges by limit comparison with the harmonic series.<\/li>\n
                    19. [latex]{\\left(a-b\\right)}^{2}={a}^{2}-2ab+{b}^{2}[\/latex] or [latex]{a}^{2}+{b}^{2}\\ge 2ab[\/latex], so convergence follows from comparison of [latex]2{a}{n}{b}{n}[\/latex] with [latex]{a}^{2}{}{n}+{b}^{2}{}{n}[\/latex]. Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.<\/li>\n
                    20. [latex]{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}={e}^{\\text{-}\\text{ln}\\left(n\\right)\\text{ln}\\text{ln}\\left(n\\right)}[\/latex]. If [latex]n[\/latex] is sufficiently large, then [latex]\\text{ln}\\text{ln}n>2[\/latex], so [latex]{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}<\\dfrac{1}{{n}^{2}}[\/latex], and the series converges by comparison to a [latex]p-\\text{series}\\text{.}[\/latex]<\/li>\n
                    21. [latex]{a}{n}\\to 0[\/latex], so [latex]{a}^{2}{}{n}\\le |{a}_{n}|[\/latex] for large [latex]n[\/latex]. Convergence follows from limit comparison. [latex]\\displaystyle\\sum \\dfrac{1}{{n}^{2}}[\/latex] converges, but [latex]\\displaystyle\\sum \\dfrac{1}{n}[\/latex] does not, so the fact that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{2}{}{n}[\/latex] converges does not imply that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}[\/latex] converges.<\/li>\n
                    22. No. [latex]\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{1}{n}[\/latex] diverges. Let [latex]{b}{k}=0[\/latex] unless [latex]k={n}^{2}[\/latex] for some [latex]n[\/latex]. Then [latex]\\displaystyle\\sum {k}\\dfrac{{b}{k}}{k}=\\displaystyle\\sum \\dfrac{1}{{k}^{2}}[\/latex] converges.<\/li>\n
                    23. [latex]|\\sin{t}|\\le |t|[\/latex], so the result follows from the comparison test.<\/li>\n
                    24. By the comparison test, [latex]x=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{{b}{n}}{{2}^{n}}\\le \\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{2}^{n}}=1[\/latex].<\/li>\n
                    25. If [latex]{b}_{1}=0[\/latex], then, by comparison, [latex]x\\le \\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{1}{{2}^{n}}=\\dfrac{1}{2}[\/latex].<\/li>\n
                    26. Yes. Keep adding [latex]1\\text{-kg}[\/latex] weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the [latex]1\\text{-kg}[\/latex] weights, and add [latex]0.1\\text{-kg}[\/latex] weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last [latex]0.1\\text{-kg}[\/latex] weight. Start adding [latex]0.01\\text{-kg}[\/latex] weights. If it balances, stop. If it tips to the side with the weights, remove the last [latex]0.01\\text{-kg}[\/latex] weight that was added. Continue in this way for the [latex]0.001\\text{-kg}[\/latex] weights, and so on. After a finite number of steps, one has a finite series of the form [latex]A+\\displaystyle\\sum {n=1}^{N}\\dfrac{{s}{n}}{{10}^{n}}[\/latex] where [latex]A[\/latex] is the number of full kg weights and [latex]{d}_{n}[\/latex] is the number of [latex]\\dfrac{1}{{10}^{n}}\\text{-kg}[\/latex] weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the [latex]N\\text{th}[\/latex] partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most [latex]\\dfrac{1}{{10}^{N}}[\/latex].<\/li>\n<\/ol>\n

                      Alternating Series<\/span><\/h2>\n
                        \n
                      1. Does not converge by divergence test. Terms do not tend to zero.<\/li>\n
                      2. Converges conditionally by alternating series test, since [latex]\\dfrac{\\sqrt{n+3}}{n}[\/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\\dfrac{1}{2}[\/latex].<\/li>\n
                      3. Converges absolutely by limit comparison to [latex]\\dfrac{{3}^{n}}{{4}^{n}}[\/latex], for example.<\/li>\n
                      4. Diverges by divergence test since [latex]\\underset{n\\to \\infty }{\\text{lim}}|{a}_{n}|=e[\/latex].<\/li>\n
                      5. Does not converge. Terms do not tend to zero.<\/li>\n
                      6. [latex]\\underset{n\\to \\infty }{\\text{lim}}{\\cos}^{2}\\left(\\dfrac{1}{n}\\right)=1[\/latex]. Diverges by divergence test.<\/li>\n
                      7. Converges by alternating series test.<\/li>\n
                      8. Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\\pi -e[\/latex]<\/li>\n
                      9. Diverges; terms do not tend to zero.<\/li>\n
                      10. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/li>\n
                      11. Converges absolutely by limit comparison with p-series, [latex]p=\\dfrac{3}{2}[\/latex], after applying the hint.<\/li>\n
                      12. Converges by alternating series test since [latex]n\\left({\\tan}^{-1}\\left(n+1\\right)\\text{-}{\\tan}^{-1}n\\right)[\/latex] is decreasing to zero for large [latex]n[\/latex]. Does not converge absolutely by limit comparison with harmonic series after applying hint.<\/li>\n
                      13. Converges absolutely, since [latex]{a}_{n}=\\dfrac{1}{n}-\\dfrac{1}{n+1}[\/latex] are terms of a telescoping series.<\/li>\n
                      14. Terms do not tend to zero. Series diverges by divergence test.<\/li>\n
                      15. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/li>\n
                      16. [latex]\\text{ln}\\left(N+1\\right)>10[\/latex], [latex]N+1>{e}^{10}[\/latex], [latex]N\\ge 22026[\/latex]; [latex]{S}_{22026}=0.0257\\text{...}[\/latex]<\/li>\n
                      17. [latex]{2}^{N+1}>{10}^{6}[\/latex] or [latex]N+1>\\dfrac{6\\text{ln}\\left(10\\right)}{\\text{ln}\\left(2\\right)}=19.93[\/latex]. or [latex]N\\ge 19[\/latex]; [latex]{S}_{19}=0.333333969\\text{...}[\/latex]<\/li>\n
                      18. [latex]{\\left(N+1\\right)}^{2}>{10}^{6}[\/latex] or [latex]N>999[\/latex]; [latex]{S}_{1000}\\approx 0.822466[\/latex].<\/li>\n
                      19. True. [latex]{b}{n}[\/latex] need not tend to zero since if [latex]{c}{n}={b}{n}-\\text{lim}{b}{n}[\/latex], then [latex]{c}{2n - 1}-{c}{2n}={b}{2n - 1}-{b}{2n}[\/latex].<\/li>\n
                      20. True. [latex]{b}{3n - 1}-{b}{3n}\\ge 0[\/latex], so convergence of [latex]\\displaystyle\\sum {b}_{3n - 2}[\/latex] follows from the comparison test.<\/li>\n
                      21. True. If one converges, then so must the other, implying absolute convergence.<\/li>\n
                      22. Yes. Take [latex]{b}{n}=1[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{b}{n}=0[\/latex] if [latex]{a}{n}<0[\/latex]. Then [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{b}{n}=\\displaystyle\\sum {n:{a}{n}\\ge 0}{a}{n}[\/latex] converges. Similarly, one can show [latex]\\displaystyle\\sum {n:{a}{n}<0}{a}_{n}[\/latex] converges. Since both series converge, the series must converge absolutely.<\/li>\n
                      23. Not decreasing. Does not converge absolutely.<\/li>\n
                      24. Not alternating. Can be expressed as [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\dfrac{1}{3n - 2}+\\dfrac{1}{3n - 1}-\\dfrac{1}{3n}\\right)[\/latex], which diverges by comparison with [latex]\\displaystyle\\sum \\dfrac{1}{3n - 2}[\/latex].<\/li>\n
                      25. Let [latex]{a}^{+}{}{n}={a}{n}[\/latex] if [latex]{a}{n}\\ge 0[\/latex] and [latex]{a}^{+}{}{n}=0[\/latex] if [latex]{a}{n}<0[\/latex]. Then [latex]{a}^{+}{}{n}\\le |{a}{n}|[\/latex] for all [latex]n[\/latex] so the sequence of partial sums of [latex]{a}^{+}{}{n}[\/latex] is increasing and bounded above by the sequence of partial sums of [latex]|{a}_{n}|[\/latex], which converges; hence, [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}^{+}{}{n}[\/latex] converges.<\/li>\n
                      26. For [latex]N=5[\/latex] one has [latex]|{R}{N}|{b}{6}=\\dfrac{{\\theta }^{10}}{10\\text{!}}[\/latex]. When [latex]\\theta =1[\/latex], [latex]{R}{5}\\le \\dfrac{1}{10\\text{!}}\\approx 2.75\\times {10}^{-7}[\/latex]. When [latex]\\theta =\\dfrac{\\pi}{6}[\/latex], [latex]{R}{5}\\le {\\left(\\dfrac{\\pi}{6}\\right)}^\\dfrac{{10}}{10\\text{!}}\\approx 4.26\\times {10}^{-10}[\/latex]. When [latex]\\theta =\\pi[\/latex], [latex]{R}_{5}\\le \\dfrac{{\\pi }^{10}}{10\\text{!}}=0.0258[\/latex].<\/li>\n
                      27. Let [latex]{b}_{n}=\\dfrac{1}{\\left(2n - 2\\right)}\\text{!}[\/latex]. Then [latex]{R}_{N}\\le \\dfrac{1}{\\left(2N\\right)\\text{!}}<0.00001[\/latex] when [latex]\\left(2N\\right)\\text{!}>{10}^{5}[\/latex] or [latex]N=5[\/latex] and [latex]1-\\dfrac{1}{2\\text{!}}+\\dfrac{1}{4\\text{!}}-\\dfrac{1}{6\\text{!}}+\\dfrac{1}{8\\text{!}}=0.540325\\text{...}[\/latex], whereas [latex]\\cos1=0.5403023\\text{...}[\/latex]<\/li>\n
                      28. Let [latex]T=\\displaystyle\\sum \\dfrac{1}{{n}^{2}}[\/latex]. Then [latex]T-S=\\dfrac{1}{2}T[\/latex], so [latex]S=\\dfrac{T}{2}[\/latex]. [latex]\\sqrt{6\\times \\displaystyle\\sum _{n=1}^{1000}\\dfrac{1}{{n}^{2}}}=3.140638\\text{...}[\/latex]; [latex]\\sqrt{12\\times \\displaystyle\\sum _{n=1}^{1000}\\dfrac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}}=3.141591\\text{...}[\/latex]; [latex]\\pi =3.141592\\text{...}[\/latex]. The alternating series is more accurate for [latex]1000[\/latex] terms.<\/li>\n<\/ol>\n

                        Ratio and Root Tests<\/span><\/h2>\n
                          \n
                        1. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to 0[\/latex]. Converges.<\/li>\n
                        2. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=\\dfrac{1}{2}{\\left(\\dfrac{n+1}{n}\\right)}^{2}\\to \\dfrac{1}{2}<1[\/latex]. Converges.<\/li>\n
                        3. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to \\dfrac{1}{27}<1[\/latex]. Converges.<\/li>\n
                        4. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to \\dfrac{4}{{e}^{2}}<1[\/latex]. Converges.<\/li>\n
                        5. [latex]\\dfrac{{a}{n+1}}{{a}{n}}\\to 1[\/latex]. Ratio test is inconclusive.<\/li>\n
                        6. [latex]\\dfrac{{a}{n}}{{a}{n+1}}\\to \\dfrac{1}{{e}^{2}}[\/latex]. Converges.<\/li>\n
                        7. [latex]{\\left({a}_{k}\\right)}^{\\dfrac{1}{k}}\\to 2>1[\/latex]. Diverges.<\/li>\n
                        8. [latex]{\\left({a}_{n}\\right)}^{\\dfrac{1}{n}}\\to \\dfrac{1}{2}<1[\/latex]. Converges.<\/li>\n
                        9. [latex]{\\left({a}_{k}\\right)}^{\\dfrac{1}{k}}\\to \\dfrac{1}{e}<1[\/latex]. Converges.<\/li>\n
                        10. [latex]{a}_{n}^{\\dfrac{1}{n}}=\\dfrac{1}{e}+\\dfrac{1}{n}\\to \\dfrac{1}{e}<1[\/latex]. Converges.<\/li>\n
                        11. [latex]{a}_{n}^{\\dfrac{1}{n}}=\\dfrac{\\left(\\text{ln}\\left(1+\\text{ln}n\\right)\\right)}{\\left(\\text{ln}n\\right)}\\to 0[\/latex] by L’H\u00f4pital’s rule. Converges.<\/li>\n
                        12. [latex]\\dfrac{{a}{k+1}}{{a}{k}}=\\dfrac{1}{2k+1}\\to 0[\/latex]. Converges by ratio test.<\/li>\n
                        13. [latex]{\\left({a}_{n}\\right)}^{\\dfrac{1}{n}}\\to \\dfrac{1}{e}[\/latex]. Converges by root test.<\/li>\n
                        14. [latex]{a}_{k}^{\\dfrac{1}{k}}\\to \\text{ln}\\left(3\\right)>1[\/latex]. Diverges by root test.<\/li>\n
                        15. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=[\/latex] [latex]\\dfrac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\\to 0[\/latex]. Converge.<\/li>\n
                        16. Converges by root test and limit comparison test since [latex]{x}_{n}\\to \\sqrt{2}[\/latex].<\/li>\n
                        17. Converges absolutely by limit comparison with [latex]p-\\text{series,}[\/latex] [latex]p=2[\/latex].<\/li>\n
                        18. [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\dfrac{1}{{e}^{2}}\\ne 0[\/latex]. Series diverges.<\/li>\n
                        19. Terms do not tend to zero: [latex]{a}_{k}\\ge \\dfrac{1}{2}[\/latex], since [latex]{\\sin}^{2}x\\le 1[\/latex].<\/li>\n
                        20. [latex]{a}_{n}=\\dfrac{2}{\\left(n+1\\right)\\left(n+2\\right)}[\/latex], which converges by comparison with [latex]p-\\text{series}[\/latex] for [latex]p=2[\/latex].<\/li>\n
                        21. [latex]{a}_{k}=\\dfrac{{2}^{k}1\\cdot 2\\text{...}k}{\\left(2k+1\\right)\\left(2k+2\\right)\\text{...}3k}\\le {\\left(\\dfrac{2}{3}\\right)}^{k}[\/latex] converges by comparison with geometric series.<\/li>\n
                        22. [latex]{a}_{k}\\approx {e}^{\\text{-}\\text{ln}{k}^{2}}=\\dfrac{1}{{k}^{2}}[\/latex]. Series converges by limit comparison with [latex]p-\\text{series,}[\/latex] [latex]p=2[\/latex].<\/li>\n
                        23. If [latex]{b}{k}=\\dfrac{{c}^{1-k}}{\\left(c - 1\\right)}[\/latex] and [latex]{a}{k}=k[\/latex], then [latex]{b}{k+1}-{b}{k}=\\text{-}{c}^{\\text{-}k}[\/latex] and [latex]\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{k}{{c}^{k}}={a}{1}{b}_{1}+\\dfrac{1}{c - 1}\\displaystyle\\sum _{k=1}^{\\infty }{c}^{\\text{-}k}=\\dfrac{c}{{\\left(c - 1\\right)}^{2}}[\/latex].<\/li>\n
                        24. [latex]6+4+1=11[\/latex]<\/li>\n
                        25. [latex]|x|\\le 1[\/latex]<\/li>\n
                        26. [latex]|x|<\\infty[\/latex]<\/li>\n
                        27. All real numbers [latex]p[\/latex] by the ratio test.<\/li>\n
                        28. [latex]r<\\dfrac{1}{p}[\/latex]<\/li>\n
                        29. [latex]0 < r < 1[\/latex]. Note that the ratio and root tests are inconclusive. Using the hint, there are [latex]2k[\/latex] terms [latex]{r}^{\\sqrt{n}}[\/latex] for [latex]{k}^{2}\\le n<{\\left(k+1\\right)}^{2}[\/latex], and for [latex]r<1[\/latex] each term is at least [latex]{r}^{k}[\/latex]. Thus, [latex]\\displaystyle\\sum _{n=1}^{\\infty }{r}^{\\sqrt{n}}=\\displaystyle\\sum _{k=1}^{\\infty }\\displaystyle\\sum _{n={k}^{2}}^{{\\left(k+1\\right)}^{2}-1}{r}^{\\sqrt{n}}[\/latex] [latex]\\ge \\displaystyle\\sum _{k=1}^{\\infty }2k{r}^{k}[\/latex], which converges by the ratio test for [latex]r<1[\/latex]. For [latex]r\\ge 1[\/latex] the series diverges by the divergence test.<\/li>\n
                        30. One has [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}={a}_{3}=\\dfrac{1}{2}\\text{...}{a}_{2n}={a}_{2n+1}=\\dfrac{1}{{2}^{n}}[\/latex]. The ratio test does not apply because [latex]\\dfrac{{a}_{n+1}}{{a}_{n}}=1[\/latex] if [latex]n[\/latex] is even. However, [latex]\\dfrac{{a}_{n+2}}{{a}_{n}}=\\dfrac{1}{2}[\/latex], so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":109,"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\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/183"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/183\/revisions"}],"predecessor-version":[{"id":194,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/183\/revisions\/194"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/109"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/183\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=183"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=183"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=183"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}