{"id":108,"date":"2021-03-25T02:21:04","date_gmt":"2021-03-25T02:21:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/alternating-series-2\/"},"modified":"2021-12-09T05:03:32","modified_gmt":"2021-12-09T05:03:32","slug":"problem-set-alternating-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-alternating-series\/","title":{"raw":"Problem Set: Alternating Series","rendered":"Problem Set: Alternating Series"},"content":{"raw":"

State whether each of the following series converges absolutely, conditionally, or not at all.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{n}{n+3}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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2.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{\\sqrt{n}+1}{\\sqrt{n}+3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"132302\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"132302\"]Does not converge by divergence test. Terms do not tend to zero.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{1}{\\sqrt{n+3}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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4.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{\\sqrt{n+3}}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"115442\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"115442\"]Converges conditionally by alternating series test, since [latex]\\frac{\\sqrt{n+3}}{n}[\/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\\frac{1}{2}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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5.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{1}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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6.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{3}^{n}}{n\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"935744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"935744\"]Converges absolutely by limit comparison to [latex]\\frac{{3}^{n}}{{4}^{n}}[\/latex], for example.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\frac{n - 1}{n}\\right)}^{n}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\frac{n+1}{n}\\right)}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"932756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932756\"]Diverges by divergence test since [latex]\\underset{n\\to \\infty }{\\text{lim}}|{a}_{n}|=e[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\sin}^{2}n[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}n[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"615544\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"615544\"]Does not converge. Terms do not tend to zero.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\sin}^{2}\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"275524\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"275524\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{\\cos}^{2}\\left(\\frac{1}{n}\\right)=1[\/latex]. Diverges by divergence test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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13.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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14.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"502136\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"502136\"]Converges by alternating series test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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15.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{n}^{2}}{1+{n}^{4}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{n}^{e}}{1+{n}^{\\pi }}[\/latex]<\/p>\r\n[reveal-answer q=\"430237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"430237\"]Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\\pi -e[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n

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17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{2}^{\\frac{1}{n}}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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18. <\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{n}^{\\frac{1}{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"504427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"504427\"]Diverges; terms do not tend to zero.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\left(1-{n}^{\\frac{1}{n}}\\right)[\/latex] (<\/span>Hint:<\/em> [latex]{n}^{\\frac{1}{n}}\\approx 1+\\frac{\\text{ln}\\left(n\\right)}{n}[\/latex] for large [latex]n.[\/latex])<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left(1-\\cos\\left(\\frac{1}{n}\\right)\\right)[\/latex] (Hint:<\/em> [latex]\\cos\\left(\\frac{1}{n}\\right)\\approx 1 - \\frac{1}{{n}^{2}}[\/latex] for large [latex]n.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"527095\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"527095\"]Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\sqrt{n+1}-\\sqrt{n}\\right)[\/latex] (Hint:<\/em> Rationalize the numerator.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\frac{1}{\\sqrt{n}}-\\frac{1}{\\sqrt{n+1}}\\right)[\/latex] (Hint:<\/em> Find common denominator then rationalize numerator.)<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"168572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"168572\"]Converges absolutely by limit comparison with p-series, [latex]p=\\frac{3}{2}[\/latex], after applying the hint.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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23.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\text{ln}\\left(n+1\\right)-\\text{ln}n\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left({\\tan}^{-1}\\left(n+1\\right)-{\\tan}^{-1}n\\right)[\/latex] (Hint:<\/em> Use Mean Value Theorem.)<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"869924\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869924\"]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.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left({\\left(n+1\\right)}^{2}-{n}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"506711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"506711\"]Converges absolutely, since [latex]{a}_{n}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex] are terms of a telescoping series.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\pi \\right)}{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\pi \\right)}{{n}^{\\frac{1}{n}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"272021\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"272021\"]Terms do not tend to zero. Series diverges by divergence test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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29.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}\\sin\\left(\\frac{n\\pi }{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\sin\\left(\\frac{n\\pi}{2}\\right)\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"232924\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"232924\"]Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In each of the following problems, use the estimate [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a value of [latex]N[\/latex] that guarantees that the sum of the first [latex]N[\/latex] terms of the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}[\/latex] differs from the infinite sum by at most the given error. Calculate the partial sum [latex]{S}_{N}[\/latex] for this [latex]N[\/latex].<\/p>\r\n\r\n

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31. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{n}[\/latex], error [latex]<{10}^{-5}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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32. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{ln}\\left(n\\right)}[\/latex], [latex]n\\ge 2[\/latex], error [latex]<{10}^{-1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"578706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"578706\"][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{$\\ldots$ }[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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33. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{\\sqrt{n}}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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34. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{2}^{n}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"971174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"971174\"][latex]{2}^{N+1}>{10}^{6}[\/latex] or [latex]N+1>\\frac{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{$\\ldots$ }[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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35. [T]<\/strong> [latex]{b}_{n}=\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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36. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"115554\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"115554\"][latex]{\\left(N+1\\right)}^{2}>{10}^{6}[\/latex] or [latex]N>999[\/latex]; [latex]{S}_{1000}\\approx 0.822466[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.<\/p>\r\n\r\n

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37.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{2n - 1}-{b}_{2n}\\right)[\/latex] converges absolutely.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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38.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{2n - 1}-{b}_{2n}\\right)[\/latex] converges absolutely.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"48464\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"48464\"]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].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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39.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex] then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{2}\\left({b}_{3n - 2}+{b}_{3n - 1}\\right)\\text{-}{b}_{3n}\\right)[\/latex] converges.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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40.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{3n - 2}+{b}_{3n - 1}-{b}_{3n}\\right)[\/latex] converges then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{3n - 2}[\/latex] converges.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"93013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"93013\"]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.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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41.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges conditionally but not absolutely, then [latex]{b}_{n}[\/latex] does not tend to zero.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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42.\u00a0<\/strong>Let [latex]{a}_{n}^{+}={a}_{n}[\/latex] if [latex]{a}_{n}\\ge 0[\/latex] and [latex]{a}_{n}^{-}=\\text{-}{a}_{n}[\/latex] if [latex]{a}_{n}<0[\/latex]. (Also, [latex]{a}_{n}^{+}=0\\text{ if }{a}_{n}<0[\/latex] and [latex]{a}_{n}^{-}=0\\text{ if }{a}_{n}\\ge 0.[\/latex]) If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges conditionally but not absolutely, then neither [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{+}[\/latex] nor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{-}[\/latex] converge.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"967529\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967529\"]True. If one converges, then so must the other, implying absolute convergence.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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43.\u00a0<\/strong>Suppose that [latex]{a}_{n}[\/latex] is a sequence of positive real numbers and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\r\n

Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of ones and minus ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily converge?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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44.\u00a0<\/strong>Suppose that [latex]{a}_{n}[\/latex] is a sequence such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] converges for every possible sequence [latex]{b}_{n}[\/latex] of zeros and ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converge absolutely?<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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[reveal-answer q=\"63319\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"63319\"]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.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

The following series do not satisfy the hypotheses of the alternating series test as stated.<\/p>\r\n

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.<\/p>\r\n\r\n

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45.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\sin}^{2}n}{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\cos}^{2}n}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"341371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"341371\"]Not decreasing. Does not converge absolutely.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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47.\u00a0<\/strong>[latex]1+\\frac{1}{2}-\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}-\\frac{1}{7}-\\frac{1}{8}+\\text{$\\cdots$ }[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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48.\u00a0<\/strong>[latex]1+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}-\\frac{1}{9}+\\text{$\\cdots$ }[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"407343\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"407343\"]Not alternating. Can be expressed as [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{3n - 2}+\\frac{1}{3n - 1}-\\frac{1}{3n}\\right)[\/latex], which diverges by comparison with [latex]\\displaystyle\\sum \\frac{1}{3n - 2}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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49.\u00a0<\/strong>Show that the alternating series [latex]1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{3}-\\frac{1}{6}+\\frac{1}{4}-\\frac{1}{8}+\\text{$\\cdots$ }[\/latex] does not converge. What hypothesis of the alternating series test is not met?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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50.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum {a}_{n}[\/latex] converges absolutely. Show that the series consisting of the positive terms [latex]{a}_{n}[\/latex] also converges.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"30772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"30772\"]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.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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51.\u00a0<\/strong>Show that the alternating series [latex]\\frac{2}{3}-\\frac{3}{5}+\\frac{4}{7}-\\frac{5}{9}+\\text{$\\cdots$ }[\/latex] does not converge. What hypothesis of the alternating series test is not met?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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52.\u00a0<\/strong>The formula [latex]\\cos\\theta =1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}-\\frac{{\\theta }^{6}}{6\\text{!}}+\\text{$\\cdots$ }[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\cos\\theta [\/latex] by the fifth partial sum [latex]1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}\\frac{\\text{-}{\\theta }^{6}}{6\\text{!}}+\\frac{{\\theta }^{8}}{8\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\frac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi [\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"910727\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"910727\"]For [latex]N=5[\/latex] one has [latex]|{R}_{N}|{b}_{6}=\\frac{{\\theta }^{10}}{10\\text{!}}[\/latex]. When [latex]\\theta =1[\/latex], [latex]{R}_{5}\\le \\frac{1}{10\\text{!}}\\approx 2.75\\times {10}^{-7}[\/latex]. When [latex]\\theta =\\frac{\\pi}{6}[\/latex], [latex]{R}_{5}\\le {\\left(\\frac{\\pi}{6}\\right)}^\\frac{{10}}{10\\text{!}}\\approx 4.26\\times {10}^{-10}[\/latex]. When [latex]\\theta =\\pi [\/latex], [latex]{R}_{5}\\le \\frac{{\\pi }^{10}}{10\\text{!}}=0.0258[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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53.\u00a0<\/strong>The formula [latex]\\sin\\theta =\\theta -\\frac{{\\theta }^{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}-\\frac{{\\theta }^{7}}{7\\text{!}}+\\text{$\\cdots$ }[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\sin\\theta [\/latex] by the fifth partial sum [latex]\\theta -{\\theta }^\\frac{{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}\\frac{\\text{-}{\\theta }^{7}}{7\\text{!}}+\\frac{{\\theta }^{9}}{9\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\frac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi [\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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54.\u00a0<\/strong>How many terms in [latex]\\cos\\theta =1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}-\\frac{{\\theta }^{6}}{6\\text{!}}+\\text{$\\cdots$ }[\/latex] are needed to approximate [latex]\\cos1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"981014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981014\"]Let [latex]{b}_{n}=\\frac{1}{\\left(2n - 2\\right)}\\text{!}[\/latex]. Then [latex]{R}_{N}\\le \\frac{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-\\frac{1}{2\\text{!}}+\\frac{1}{4\\text{!}}-\\frac{1}{6\\text{!}}+\\frac{1}{8\\text{!}}=0.540325\\text{$\\ldots$ }[\/latex], whereas [latex]\\cos1=0.5403023\\text{$\\ldots$ }[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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55.\u00a0<\/strong>How many terms in [latex]\\sin\\theta =\\theta -\\frac{{\\theta }^{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}-\\frac{{\\theta }^{7}}{7\\text{!}}+\\text{$\\cdots$ }[\/latex] are needed to approximate [latex]\\sin1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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56.\u00a0<\/strong>Sometimes the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges to a certain fraction of an absolutely convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] at a faster rate. Given that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}=\\frac{{\\pi }^{2}}{6}[\/latex], find [latex]12=1-\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}-\\frac{1}{{4}^{2}}+\\text{$\\cdots$ }[\/latex]. Which of the series [latex]6\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] and [latex]S\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}[\/latex] gives a better estimation of [latex]{\\pi }^{2}[\/latex] using [latex]1000[\/latex] terms?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"467157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467157\"]Let [latex]T=\\displaystyle\\sum \\frac{1}{{n}^{2}}[\/latex]. Then [latex]T-S=\\frac{1}{2}T[\/latex], so [latex]S=\\frac{T}{2}[\/latex]. [latex]\\sqrt{6\\times \\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}}=3.140638\\text{$\\ldots$ }[\/latex]; [latex]\\sqrt{12\\times \\displaystyle\\sum _{n=1}^{1000}\\frac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}}=3.141591\\text{$\\ldots$ }[\/latex]; [latex]\\pi =3.141592\\text{$\\ldots$ }[\/latex]. The alternating series is more accurate for [latex]1000[\/latex] terms.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

The following alternating series converge to given multiples of [latex]\\pi [\/latex]. Find the value of [latex]N[\/latex] predicted by the remainder estimate such that the [latex]N\\text{th}[\/latex] partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum [latex]N[\/latex] for which the error bound holds, and give the desired approximate value in each case. Up to [latex]15[\/latex] decimals places, [latex]\\pi =3.141592653589793\\text{$\\ldots$ }[\/latex].<\/p>\r\n\r\n

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57. [T]<\/strong> [latex]\\frac{\\pi }{4}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{2n+1}[\/latex], error [latex]<0.0001[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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58. [T]<\/strong> [latex]\\frac{\\pi }{\\sqrt{12}}=\\displaystyle\\sum _{k=0}^{\\infty }\\frac{{\\left(-3\\right)}^{\\text{-}k}}{2k+1}[\/latex], error [latex]<0.0001[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"295428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"295428\"][latex]N=6[\/latex], [latex]{S}_{N}=0.9068[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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59. [T]<\/strong> The series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\sin\\left(x+\\pi n\\right)}{x+\\pi n}[\/latex] plays an important role in signal processing. Show that [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\sin\\left(x+\\pi n\\right)}{x+\\pi n}[\/latex] converges whenever [latex]0<x<\\pi [\/latex]. (Hint:<\/em> Use the formula for the sine of a sum of angles.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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60. [T]<\/strong> If [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\frac{1}{n}\\to \\text{ln}2[\/latex], what is [latex]1+\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{2}-\\frac{1}{4}-\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{9}+\\frac{1}{11}-\\frac{1}{8}-\\frac{1}{10}-\\frac{1}{12}+\\text{$\\cdots$ }\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"372042\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"372042\"][latex]\\text{ln}\\left(2\\right)[\/latex]. The [latex]3n\\text{th}[\/latex] partial sum is the same as that for the alternating harmonic series.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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61. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{n}[\/latex] for [latex]0\\le x<1[\/latex]. Explain why [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{n}[\/latex] diverges when [latex]x=0,1[\/latex]. How does the series behave for other [latex]x\\text{?}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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62. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\sin\\left(2\\pi nx\\right)}{n}[\/latex] for [latex]0\\le x<1[\/latex] and comment on its behavior<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"924256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924256\"]The series jumps rapidly near the endpoints. For [latex]x[\/latex] away from the endpoints, the graph looks like [latex]\\pi \\left(\\frac{1}{2}-x\\right)[\/latex]. \r\n<\/span><\/p>\r\n\r\n\"This[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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63. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{{n}^{2}}[\/latex] for [latex]0\\le x<1[\/latex] and describe its graph.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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64. [T]<\/strong> The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{S}_{n}}{n}[\/latex], where [latex]{s}_{n}[\/latex] is a randomly generated sequence of [latex]\\pm 1\\text{'s}[\/latex] in which the values [latex]\\pm 1[\/latex] are equally likely to occur. Use a random number generator to produce [latex]1000[\/latex] random [latex]\\pm 1\\text{s}[\/latex] and plot the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\frac{{s}_{n}}{n}[\/latex] of your random harmonic sequence for [latex]N=1[\/latex] to [latex]1000[\/latex]. Compare to a plot of the first [latex]1000[\/latex] partial sums of the harmonic series.<\/p>\r\n\r\n<\/div>\r\n

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\r\n[reveal-answer q=\"183555\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"183555\"]Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.<\/span>\"This[\/hidden-answer]<\/span><\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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65. [T]<\/strong> Estimates of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] can be accelerated<\/em> by writing its partial sums as [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{{n}^{2}}=\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n\\left(n+1\\right)}+\\displaystyle\\sum _{n=1}^{N}\\frac{1}{{n}^{2}\\left(n+1\\right)}[\/latex] and recalling that [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n\\left(n+1\\right)}=1-\\frac{1}{N+1}[\/latex] converges to one as [latex]N\\to \\infty [\/latex]. Compare the estimate of [latex]\\frac{{\\pi }^{2}}{6}[\/latex] using the sums [latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}[\/latex] with the estimate using [latex]1+\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}\\left(n+1\\right)}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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66. [T]<\/strong> The Euler transform<\/em><\/span> rewrites [latex]S=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex] as [latex]S=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{\\text{-}n - 1}\\displaystyle\\sum _{m=0}^{n}\\left(\\begin{array}{c}n\\\\ m\\end{array}\\right){b}_{n-m}[\/latex]. For the alternating harmonic series, it takes the form [latex]\\text{ln}\\left(2\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{n}=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex]. Compute partial sums of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex] until they approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]0.0001[\/latex]. How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate [latex]\\text{ln}\\left(2\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"42432\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"42432\"]By the alternating series test, [latex]|{S}_{n}-S|\\le {b}_{n+1}[\/latex], so one needs [latex]{10}^{4}[\/latex] terms of the alternating harmonic series to estimate [latex]\\text{ln}\\left(2\\right)[\/latex] to within [latex]0.0001[\/latex]. The first [latex]10[\/latex] partial sums of the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex] are (up to four decimals) [latex]0.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.6931[\/latex] and the tenth partial sum is within [latex]0.0001[\/latex] of [latex]\\text{ln}\\left(2\\right)=0.6931\\text{$\\ldots$ }[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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67. [T]<\/strong> In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If [latex]{a}_{n}\\ge 0[\/latex] is such that [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges, then, given any number [latex]A[\/latex] there is a sequence [latex]{s}_{n}[\/latex] of [latex]\\pm 1\\text{'s}[\/latex] such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{s}_{n}\\to A[\/latex]. Show this for [latex]A>0[\/latex] as follows.<\/p>\r\n\r\n

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  1. Recursively define [latex]{s}_{n}[\/latex] by [latex]{s}_{n}=1[\/latex] if [latex]{S}_{n - 1}=\\displaystyle\\sum _{k=1}^{n - 1}{a}_{k}{s}_{k}<A[\/latex] and [latex]{s}_{n}=-1[\/latex] otherwise.<\/li>\r\n \t
  2. Explain why eventually [latex]{S}_{n}\\ge A[\/latex], and for any [latex]m[\/latex] larger than this [latex]n[\/latex], [latex]A-{a}_{m}\\le {S}_{m}\\le A+{a}_{m}[\/latex].<\/li>\r\n \t
  3. Explain why this implies that [latex]{S}_{n}\\to A[\/latex] as [latex]n\\to \\infty [\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
    <\/div>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>","rendered":"

    State whether each of the following series converges absolutely, conditionally, or not at all.<\/p>\n

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    1.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{n}{n+3}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    2.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{\\sqrt{n}+1}{\\sqrt{n}+3}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Does not converge by divergence test. Terms do not tend to zero.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    3.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{1}{\\sqrt{n+3}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    4.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{\\sqrt{n+3}}{n}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges conditionally by alternating series test, since [latex]\\frac{\\sqrt{n+3}}{n}[\/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\\frac{1}{2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    5.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{1}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    6.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{3}^{n}}{n\\text{!}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges absolutely by limit comparison to [latex]\\frac{{3}^{n}}{{4}^{n}}[\/latex], for example.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\frac{n - 1}{n}\\right)}^{n}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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    8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\left(\\frac{n+1}{n}\\right)}^{n}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Diverges by divergence test since [latex]\\underset{n\\to \\infty }{\\text{lim}}|{a}_{n}|=e[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\sin}^{2}n[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}n[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Does not converge. Terms do not tend to zero.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\sin}^{2}\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{\\cos}^{2}\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\n<\/div>\n

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    [latex]\\underset{n\\to \\infty }{\\text{lim}}{\\cos}^{2}\\left(\\frac{1}{n}\\right)=1[\/latex]. Diverges by divergence test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    13.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    14.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex]<\/p>\n<\/div>\n

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    Converges by alternating series test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    15.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{n}^{2}}{1+{n}^{4}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{n}^{e}}{1+{n}^{\\pi }}[\/latex]<\/p>\n

    Show Solution<\/span><\/p>\n
    Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\\pi -e[\/latex]<\/div>\n<\/div>\n<\/div>\n
    <\/div>\n<\/div>\n<\/div>\n
    17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{2}^{\\frac{1}{n}}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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    18. <\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{n}^{\\frac{1}{n}}[\/latex]<\/p>\n<\/div>\n

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    Diverges; terms do not tend to zero.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\left(1-{n}^{\\frac{1}{n}}\\right)[\/latex] (<\/span>Hint:<\/em> [latex]{n}^{\\frac{1}{n}}\\approx 1+\\frac{\\text{ln}\\left(n\\right)}{n}[\/latex] for large [latex]n.[\/latex])<\/span><\/div>\n<\/div>\n<\/div>\n
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    20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left(1-\\cos\\left(\\frac{1}{n}\\right)\\right)[\/latex] (Hint:<\/em> [latex]\\cos\\left(\\frac{1}{n}\\right)\\approx 1 - \\frac{1}{{n}^{2}}[\/latex] for large [latex]n.[\/latex])<\/p>\n<\/div>\n

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    Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\sqrt{n+1}-\\sqrt{n}\\right)[\/latex] (Hint:<\/em> Rationalize the numerator.)<\/div>\n<\/div>\n<\/div>\n
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    22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\frac{1}{\\sqrt{n}}-\\frac{1}{\\sqrt{n+1}}\\right)[\/latex] (Hint:<\/em> Find common denominator then rationalize numerator.)<\/p>\n<\/div>\n

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    Converges absolutely by limit comparison with p-series, [latex]p=\\frac{3}{2}[\/latex], after applying the hint.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    23.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\text{ln}\\left(n+1\\right)-\\text{ln}n\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}n\\left({\\tan}^{-1}\\left(n+1\\right)-{\\tan}^{-1}n\\right)[\/latex] (Hint:<\/em> Use Mean Value Theorem.)<\/p>\n<\/div>\n

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    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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left({\\left(n+1\\right)}^{2}-{n}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)[\/latex]<\/p>\n<\/div>\n

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    Converges absolutely, since [latex]{a}_{n}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex] are terms of a telescoping series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\pi \\right)}{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\pi \\right)}{{n}^{\\frac{1}{n}}}[\/latex]<\/p>\n<\/div>\n

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    Terms do not tend to zero. Series diverges by divergence test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    29.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}\\sin\\left(\\frac{n\\pi }{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\sin\\left(\\frac{n\\pi}{2}\\right)\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\n<\/div>\n

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    Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In each of the following problems, use the estimate [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a value of [latex]N[\/latex] that guarantees that the sum of the first [latex]N[\/latex] terms of the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}[\/latex] differs from the infinite sum by at most the given error. Calculate the partial sum [latex]{S}_{N}[\/latex] for this [latex]N[\/latex].<\/p>\n

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    31. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{n}[\/latex], error [latex]<{10}^{-5}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    32. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{ln}\\left(n\\right)}[\/latex], [latex]n\\ge 2[\/latex], error [latex]<{10}^{-1}[\/latex]<\/p>\n<\/div>\n

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    [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{$\\ldots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    33. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{\\sqrt{n}}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    34. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{2}^{n}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/p>\n<\/div>\n

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    [latex]{2}^{N+1}>{10}^{6}[\/latex] or [latex]N+1>\\frac{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{$\\ldots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    35. [T]<\/strong> [latex]{b}_{n}=\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    36. [T]<\/strong> [latex]{b}_{n}=\\frac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-6}[\/latex]<\/p>\n<\/div>\n

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    [latex]{\\left(N+1\\right)}^{2}>{10}^{6}[\/latex] or [latex]N>999[\/latex]; [latex]{S}_{1000}\\approx 0.822466[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.<\/p>\n

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    37.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{2n - 1}-{b}_{2n}\\right)[\/latex] converges absolutely.<\/div>\n<\/div>\n<\/div>\n
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    38.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{2n - 1}-{b}_{2n}\\right)[\/latex] converges absolutely.<\/p>\n<\/div>\n

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    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].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    39.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] and [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex] then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{2}\\left({b}_{3n - 2}+{b}_{3n - 1}\\right)\\text{-}{b}_{3n}\\right)[\/latex] converges.<\/div>\n<\/div>\n<\/div>\n
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    40.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({b}_{3n - 2}+{b}_{3n - 1}-{b}_{3n}\\right)[\/latex] converges then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{3n - 2}[\/latex] converges.<\/p>\n<\/div>\n

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    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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    41.\u00a0<\/strong>If [latex]{b}_{n}\\ge 0[\/latex] is decreasing and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges conditionally but not absolutely, then [latex]{b}_{n}[\/latex] does not tend to zero.<\/div>\n<\/div>\n<\/div>\n
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    42.\u00a0<\/strong>Let [latex]{a}_{n}^{+}={a}_{n}[\/latex] if [latex]{a}_{n}\\ge 0[\/latex] and [latex]{a}_{n}^{-}=\\text{-}{a}_{n}[\/latex] if [latex]{a}_{n}<0[\/latex]. (Also, [latex]{a}_{n}^{+}=0\\text{ if }{a}_{n}<0[\/latex] and [latex]{a}_{n}^{-}=0\\text{ if }{a}_{n}\\ge 0.[\/latex]) If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges conditionally but not absolutely, then neither [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{+}[\/latex] nor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{-}[\/latex] converge.<\/p>\n<\/div>\n

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    True. If one converges, then so must the other, implying absolute convergence.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    43.\u00a0<\/strong>Suppose that [latex]{a}_{n}[\/latex] is a sequence of positive real numbers and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\n

    Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of ones and minus ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily converge?<\/p>\n<\/div>\n<\/div>\n<\/div>\n

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    44.\u00a0<\/strong>Suppose that [latex]{a}_{n}[\/latex] is a sequence such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] converges for every possible sequence [latex]{b}_{n}[\/latex] of zeros and ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converge absolutely?<\/span><\/p>\n<\/div>\n<\/div>\n

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    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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    The following series do not satisfy the hypotheses of the alternating series test as stated.<\/p>\n

    In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.<\/p>\n

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    45.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\sin}^{2}n}{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\cos}^{2}n}{n}[\/latex]<\/p>\n<\/div>\n

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    Not decreasing. Does not converge absolutely.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    47.\u00a0<\/strong>[latex]1+\\frac{1}{2}-\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}-\\frac{1}{7}-\\frac{1}{8}+\\text{$\\cdots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    48.\u00a0<\/strong>[latex]1+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}-\\frac{1}{9}+\\text{$\\cdots$ }[\/latex]<\/p>\n<\/div>\n

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    Not alternating. Can be expressed as [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{3n - 2}+\\frac{1}{3n - 1}-\\frac{1}{3n}\\right)[\/latex], which diverges by comparison with [latex]\\displaystyle\\sum \\frac{1}{3n - 2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    49.\u00a0<\/strong>Show that the alternating series [latex]1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{3}-\\frac{1}{6}+\\frac{1}{4}-\\frac{1}{8}+\\text{$\\cdots$ }[\/latex] does not converge. What hypothesis of the alternating series test is not met?<\/div>\n<\/div>\n<\/div>\n
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    50.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum {a}_{n}[\/latex] converges absolutely. Show that the series consisting of the positive terms [latex]{a}_{n}[\/latex] also converges.<\/p>\n<\/div>\n

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    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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    51.\u00a0<\/strong>Show that the alternating series [latex]\\frac{2}{3}-\\frac{3}{5}+\\frac{4}{7}-\\frac{5}{9}+\\text{$\\cdots$ }[\/latex] does not converge. What hypothesis of the alternating series test is not met?<\/div>\n<\/div>\n<\/div>\n
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    52.\u00a0<\/strong>The formula [latex]\\cos\\theta =1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}-\\frac{{\\theta }^{6}}{6\\text{!}}+\\text{$\\cdots$ }[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\cos\\theta[\/latex] by the fifth partial sum [latex]1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}\\frac{\\text{-}{\\theta }^{6}}{6\\text{!}}+\\frac{{\\theta }^{8}}{8\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\frac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi[\/latex].<\/p>\n<\/div>\n

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    For [latex]N=5[\/latex] one has [latex]|{R}_{N}|{b}_{6}=\\frac{{\\theta }^{10}}{10\\text{!}}[\/latex]. When [latex]\\theta =1[\/latex], [latex]{R}_{5}\\le \\frac{1}{10\\text{!}}\\approx 2.75\\times {10}^{-7}[\/latex]. When [latex]\\theta =\\frac{\\pi}{6}[\/latex], [latex]{R}_{5}\\le {\\left(\\frac{\\pi}{6}\\right)}^\\frac{{10}}{10\\text{!}}\\approx 4.26\\times {10}^{-10}[\/latex]. When [latex]\\theta =\\pi[\/latex], [latex]{R}_{5}\\le \\frac{{\\pi }^{10}}{10\\text{!}}=0.0258[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    53.\u00a0<\/strong>The formula [latex]\\sin\\theta =\\theta -\\frac{{\\theta }^{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}-\\frac{{\\theta }^{7}}{7\\text{!}}+\\text{$\\cdots$ }[\/latex] will be derived in the next chapter. Use the remainder [latex]|{R}_{N}|\\le {b}_{N+1}[\/latex] to find a bound for the error in estimating [latex]\\sin\\theta[\/latex] by the fifth partial sum [latex]\\theta -{\\theta }^\\frac{{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}\\frac{\\text{-}{\\theta }^{7}}{7\\text{!}}+\\frac{{\\theta }^{9}}{9\\text{!}}[\/latex] for [latex]\\theta =1[\/latex], [latex]\\theta =\\frac{\\pi}{6}[\/latex], and [latex]\\theta =\\pi[\/latex].<\/div>\n<\/div>\n<\/div>\n
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    54.\u00a0<\/strong>How many terms in [latex]\\cos\\theta =1-\\frac{{\\theta }^{2}}{2\\text{!}}+\\frac{{\\theta }^{4}}{4\\text{!}}-\\frac{{\\theta }^{6}}{6\\text{!}}+\\text{$\\cdots$ }[\/latex] are needed to approximate [latex]\\cos1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/p>\n<\/div>\n

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    Let [latex]{b}_{n}=\\frac{1}{\\left(2n - 2\\right)}\\text{!}[\/latex]. Then [latex]{R}_{N}\\le \\frac{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-\\frac{1}{2\\text{!}}+\\frac{1}{4\\text{!}}-\\frac{1}{6\\text{!}}+\\frac{1}{8\\text{!}}=0.540325\\text{$\\ldots$ }[\/latex], whereas [latex]\\cos1=0.5403023\\text{$\\ldots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    55.\u00a0<\/strong>How many terms in [latex]\\sin\\theta =\\theta -\\frac{{\\theta }^{3}}{3\\text{!}}+\\frac{{\\theta }^{5}}{5\\text{!}}-\\frac{{\\theta }^{7}}{7\\text{!}}+\\text{$\\cdots$ }[\/latex] are needed to approximate [latex]\\sin1[\/latex] accurate to an error of at most [latex]0.00001\\text{?}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    56.\u00a0<\/strong>Sometimes the alternating series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}{b}_{n}[\/latex] converges to a certain fraction of an absolutely convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] at a faster rate. Given that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}=\\frac{{\\pi }^{2}}{6}[\/latex], find [latex]12=1-\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}-\\frac{1}{{4}^{2}}+\\text{$\\cdots$ }[\/latex]. Which of the series [latex]6\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] and [latex]S\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}[\/latex] gives a better estimation of [latex]{\\pi }^{2}[\/latex] using [latex]1000[\/latex] terms?<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Let [latex]T=\\displaystyle\\sum \\frac{1}{{n}^{2}}[\/latex]. Then [latex]T-S=\\frac{1}{2}T[\/latex], so [latex]S=\\frac{T}{2}[\/latex]. [latex]\\sqrt{6\\times \\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}}=3.140638\\text{$\\ldots$ }[\/latex]; [latex]\\sqrt{12\\times \\displaystyle\\sum _{n=1}^{1000}\\frac{{\\left(-1\\right)}^{n - 1}}{{n}^{2}}}=3.141591\\text{$\\ldots$ }[\/latex]; [latex]\\pi =3.141592\\text{$\\ldots$ }[\/latex]. The alternating series is more accurate for [latex]1000[\/latex] terms.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    The following alternating series converge to given multiples of [latex]\\pi[\/latex]. Find the value of [latex]N[\/latex] predicted by the remainder estimate such that the [latex]N\\text{th}[\/latex] partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum [latex]N[\/latex] for which the error bound holds, and give the desired approximate value in each case. Up to [latex]15[\/latex] decimals places, [latex]\\pi =3.141592653589793\\text{$\\ldots$ }[\/latex].<\/p>\n

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    57. [T]<\/strong> [latex]\\frac{\\pi }{4}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{2n+1}[\/latex], error [latex]<0.0001[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    58. [T]<\/strong> [latex]\\frac{\\pi }{\\sqrt{12}}=\\displaystyle\\sum _{k=0}^{\\infty }\\frac{{\\left(-3\\right)}^{\\text{-}k}}{2k+1}[\/latex], error [latex]<0.0001[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]N=6[\/latex], [latex]{S}_{N}=0.9068[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    59. [T]<\/strong> The series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\sin\\left(x+\\pi n\\right)}{x+\\pi n}[\/latex] plays an important role in signal processing. Show that [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\sin\\left(x+\\pi n\\right)}{x+\\pi n}[\/latex] converges whenever [latex]0Hint:<\/em> Use the formula for the sine of a sum of angles.)<\/div>\n<\/div>\n<\/div>\n
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    60. [T]<\/strong> If [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\frac{1}{n}\\to \\text{ln}2[\/latex], what is [latex]1+\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{2}-\\frac{1}{4}-\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{9}+\\frac{1}{11}-\\frac{1}{8}-\\frac{1}{10}-\\frac{1}{12}+\\text{$\\cdots$ }\\text{?}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\text{ln}\\left(2\\right)[\/latex]. The [latex]3n\\text{th}[\/latex] partial sum is the same as that for the alternating harmonic series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    61. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{n}[\/latex] for [latex]0\\le x<1[\/latex]. Explain why [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{n}[\/latex] diverges when [latex]x=0,1[\/latex]. How does the series behave for other [latex]x\\text{?}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    62. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\sin\\left(2\\pi nx\\right)}{n}[\/latex] for [latex]0\\le x<1[\/latex] and comment on its behavior<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    The series jumps rapidly near the endpoints. For [latex]x[\/latex] away from the endpoints, the graph looks like [latex]\\pi \\left(\\frac{1}{2}-x\\right)[\/latex].
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    \n\"This<\/div>\n<\/div>\n

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    63. [T]<\/strong> Plot the series [latex]\\displaystyle\\sum _{n=1}^{100}\\frac{\\cos\\left(2\\pi nx\\right)}{{n}^{2}}[\/latex] for [latex]0\\le x<1[\/latex] and describe its graph.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    64. [T]<\/strong> The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{S}_{n}}{n}[\/latex], where [latex]{s}_{n}[\/latex] is a randomly generated sequence of [latex]\\pm 1\\text{'s}[\/latex] in which the values [latex]\\pm 1[\/latex] are equally likely to occur. Use a random number generator to produce [latex]1000[\/latex] random [latex]\\pm 1\\text{s}[\/latex] and plot the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\frac{{s}_{n}}{n}[\/latex] of your random harmonic sequence for [latex]N=1[\/latex] to [latex]1000[\/latex]. Compare to a plot of the first [latex]1000[\/latex] partial sums of the harmonic series.<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.<\/span>\"This<\/div>\n<\/div>\n

    <\/span><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    65. [T]<\/strong> Estimates of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] can be accelerated<\/em> by writing its partial sums as [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{{n}^{2}}=\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n\\left(n+1\\right)}+\\displaystyle\\sum _{n=1}^{N}\\frac{1}{{n}^{2}\\left(n+1\\right)}[\/latex] and recalling that [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n\\left(n+1\\right)}=1-\\frac{1}{N+1}[\/latex] converges to one as [latex]N\\to \\infty[\/latex]. Compare the estimate of [latex]\\frac{{\\pi }^{2}}{6}[\/latex] using the sums [latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}[\/latex] with the estimate using [latex]1+\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}\\left(n+1\\right)}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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    66. [T]<\/strong> The Euler transform<\/em><\/span> rewrites [latex]S=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex] as [latex]S=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{\\text{-}n - 1}\\displaystyle\\sum _{m=0}^{n}\\left(\\begin{array}{c}n\\\\ m\\end{array}\\right){b}_{n-m}[\/latex]. For the alternating harmonic series, it takes the form [latex]\\text{ln}\\left(2\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{n}=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex]. Compute partial sums of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex] until they approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]0.0001[\/latex]. How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate [latex]\\text{ln}\\left(2\\right)[\/latex].<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    By the alternating series test, [latex]|{S}_{n}-S|\\le {b}_{n+1}[\/latex], so one needs [latex]{10}^{4}[\/latex] terms of the alternating harmonic series to estimate [latex]\\text{ln}\\left(2\\right)[\/latex] to within [latex]0.0001[\/latex]. The first [latex]10[\/latex] partial sums of the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex] are (up to four decimals) [latex]0.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.6931[\/latex] and the tenth partial sum is within [latex]0.0001[\/latex] of [latex]\\text{ln}\\left(2\\right)=0.6931\\text{$\\ldots$ }[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    67. [T]<\/strong> In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If [latex]{a}_{n}\\ge 0[\/latex] is such that [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges, then, given any number [latex]A[\/latex] there is a sequence [latex]{s}_{n}[\/latex] of [latex]\\pm 1\\text{'s}[\/latex] such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{s}_{n}\\to A[\/latex]. Show this for [latex]A>0[\/latex] as follows.<\/p>\n

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    1. Recursively define [latex]{s}_{n}[\/latex] by [latex]{s}_{n}=1[\/latex] if [latex]{S}_{n - 1}=\\displaystyle\\sum _{k=1}^{n - 1}{a}_{k}{s}_{k}\n
    2. Explain why eventually [latex]{S}_{n}\\ge A[\/latex], and for any [latex]m[\/latex] larger than this [latex]n[\/latex], [latex]A-{a}_{m}\\le {S}_{m}\\le A+{a}_{m}[\/latex].<\/li>\n
    3. Explain why this implies that [latex]{S}_{n}\\to A[\/latex] as [latex]n\\to \\infty[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
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