{"id":107,"date":"2021-03-25T02:21:03","date_gmt":"2021-03-25T02:21:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/comparison-tests-2\/"},"modified":"2021-11-17T03:08:38","modified_gmt":"2021-11-17T03:08:38","slug":"comparison-tests-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/comparison-tests-2\/","title":{"raw":"Problem Set: Comparison Tests","rendered":"Problem Set: Comparison Tests"},"content":{"raw":"

Use the comparison test to determine whether the following series converge.<\/p>\r\n\r\n

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

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[reveal-answer q=\"319426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"319426\"]Converges by comparison with [latex]\\frac{1}{{n}^{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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3.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2\\left(n+1\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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4.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2n - 1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"906619\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"906619\"]Diverges by comparison with harmonic series, since [latex]2n - 1\\ge n[\/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=2}^{\\infty }\\frac{1}{{\\left(n\\text{ln}n\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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6.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n\\text{!}}{\\left(n+2\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"840322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"840322\"][latex]{a}_{n}=\\frac{1}{\\left(n+1\\right)\\left(n+2\\right)}<\\frac{1}{{n}^{2}}[\/latex]. Converges by comparison with p-series, [latex]p=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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7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sin\\left(\\frac{1}{n}\\right)}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"657521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"657521\"][latex]\\sin\\left(\\frac{1}{n}\\right)\\le \\frac{1}{n}[\/latex], so converges by comparison with p-series, [latex]p=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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9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\sin}^{2}n}{{n}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sin\\left(\\frac{1}{n}\\right)}{\\sqrt{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"34800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"34800\"][latex]\\sin\\left(\\frac{1}{n}\\right)\\le 1[\/latex], so converges by comparison with p-series, [latex]p=\\frac{3}{2}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{1.2}-1}{{n}^{2.3}+1}[\/latex]<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sqrt{n+1}-\\sqrt{n}}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"785515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785515\"]Since [latex]\\sqrt{n+1}-\\sqrt{n}=\\frac{1}{\\left(\\sqrt{n+1}+\\sqrt{n}\\right)}\\le \\frac{2}{\\sqrt{n}}[\/latex], series converges by comparison with p-series for [latex]p=1.5[\/latex].[\/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 }\\frac{\\sqrt[4]{n}}{\\sqrt[3]{{n}^{4}+{n}^{2}}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

Use the limit comparison test to determine whether each of the following series converges or diverges.<\/p>\r\n\r\n

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

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[reveal-answer q=\"89074\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"89074\"]Converges by limit comparison with p-series for [latex]p>1[\/latex].[\/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(\\frac{\\text{ln}n}{{n}^{0.6}}\\right)}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(1+\\frac{1}{n}\\right)}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"83228\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"83228\"]Converges by limit comparison with p-series, [latex]p=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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17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\text{ln}\\left(1+\\frac{1}{{n}^{2}}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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18.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{4}^{n}-{3}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"548521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"548521\"]Converges by limit comparison with [latex]{4}^{\\text{-}n}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}-n\\sin{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{e}^{\\left(1.1\\right)n}-{3}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"884210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884210\"]Converges by limit comparison with [latex]\\frac{1}{{e}^{1.1n}}[\/latex].[\/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 }\\frac{1}{{e}^{\\left(1.01\\right)n}-{3}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{1+\\frac{1}{n}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"179824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"179824\"]Diverges by limit comparison with harmonic series.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

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[reveal-answer q=\"522204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522204\"]Converges by limit comparison with p-series, [latex]p=3[\/latex].[\/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-\\cos\\left(\\frac{1}{n}\\right)\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}\\left(\\frac{\\pi }{2}-{\\tan}^{-1}n\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"68835\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68835\"]Converges by limit comparison with p-series, [latex]p=3[\/latex].[\/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 }{\\left(1-\\frac{1}{n}\\right)}^{n.n}[\/latex] (Hint:<\/em> [latex]{\\left(1-\\frac{1}{n}\\right)}^{n}\\to \\frac{1}{e}.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n
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28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(1-{e}^{-\\frac{1}{n}}\\right)[\/latex] (Hint:<\/em> [latex]\\frac{1}{e}\\approx {\\left(1 - \\frac{1}{n}\\right)}^{n}[\/latex], so [latex]1-{e}^{-\\frac{1}{n}}\\approx \\frac{1}{n}.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"625493\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625493\"]Diverges by limit comparison with [latex]\\frac{1}{n}[\/latex].[\/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>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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30.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{\\left(\\text{ln}n\\right)}{n}\\right)}^{p}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"598486\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598486\"]Converges for [latex]p>1[\/latex] by comparison with a [latex]p[\/latex] series for slightly smaller [latex]p[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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31.\u00a0<\/strong>For which [latex]p[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{pn}}{{3}^{n}}[\/latex] converge?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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32.\u00a0<\/strong>For which [latex]p>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{p}}{{2}^{n}}[\/latex] converge?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"524760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"524760\"]Converges for all [latex]p>0[\/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.\u00a0<\/strong>For which [latex]r>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{r}^{{n}^{2}}}{{2}^{n}}[\/latex] converge?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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34.\u00a0<\/strong>For which [latex]r>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{r}^{{n}^{2}}}[\/latex] converge?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"956570\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"956570\"]Converges for all [latex]r>1[\/latex]. If [latex]r>1[\/latex] then [latex]{r}^{n}>4[\/latex], say, once [latex]n>\\frac{\\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]\\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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35.\u00a0<\/strong>Find all values of [latex]p[\/latex] and [latex]q[\/latex] such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{p}}{{\\left(n\\text{!}\\right)}^{q}}[\/latex] converges.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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36.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\sin}^{2}\\left(\\frac{nr}{2}\\right)}{n}[\/latex] converge or diverge? Explain.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"288445\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"288445\"]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 }\\frac{1}{2n+1}[\/latex], which diverges by limit comparison with the 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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37.\u00a0<\/strong>Explain why, for each [latex]n[\/latex], at least one of [latex]\\left\\{|\\sin{n}|,|\\sin\\left(n+1\\right)|\\text{,...},|\\sin{n}+6|\\right\\}[\/latex] is larger than [latex]\\frac{1}{2}[\/latex]. Use this relation to test convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{|\\sin{n}|}{\\sqrt{n}}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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38.\u00a0<\/strong>Suppose that [latex]{a}_{n}\\ge 0[\/latex] and [latex]{b}_{n}\\ge 0[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}^{2}{}_{n}[\/latex] converge. Prove that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}\\le \\frac{1}{2}\\left(\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{2}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}^{2}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"529833\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"529833\"][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.[\/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>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Write [latex]{2}^{\\text{ln}\\text{ln}n}[\/latex] as a power of [latex]\\text{ln}n.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n
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40.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Use [latex]n={e}^{\\text{ln}\\left(n\\right)}[\/latex] to compare to a [latex]p-\\text{series}\\text{.}[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"410723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410723\"][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}<\\frac{1}{{n}^{2}}[\/latex], and the series converges by comparison to a [latex]p-\\text{series}\\text{.}[\/latex][\/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>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Compare [latex]{a}_{n}[\/latex] to [latex]\\frac{1}{n}.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n
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42.\u00a0<\/strong>Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges, does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] necessarily converge?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"351133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"351133\"][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 \\frac{1}{{n}^{2}}[\/latex] converges, but [latex]\\displaystyle\\sum \\frac{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.[\/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}>0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of zeros and ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily converge?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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44.\u00a0<\/strong>Suppose that [latex]{a}_{n}>0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges. Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily diverge?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"811230\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"811230\"]No. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{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}\\frac{{b}_{k}}{k}=\\displaystyle\\sum \\frac{1}{{k}^{2}}[\/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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45.\u00a0<\/strong>Complete the details of the following argument: If [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] converges to a finite sum [latex]s[\/latex], then [latex]\\frac{1}{2}s=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\text{$\\cdots$ }[\/latex] and [latex]s-\\frac{1}{2}s=1+\\frac{1}{3}+\\frac{1}{5}+\\text{$\\cdots$ }[\/latex]. Why does this lead to a contradiction?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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46.\u00a0<\/strong>Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\sin}^{2}\\left({a}_{n}\\right)[\/latex] converges.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"707811\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"707811\"][latex]|\\sin{t}|\\le |t|[\/latex], so the result 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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47.\u00a0<\/strong>Suppose that [latex]\\frac{{a}_{n}}{{b}_{n}}\\to 0[\/latex] in the comparison test, where [latex]{a}_{n}\\ge 0[\/latex] and [latex]{b}_{n}\\ge 0[\/latex]. Prove that if [latex]\\displaystyle\\sum {b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum {a}_{n}[\/latex] converges.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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48.\u00a0<\/strong>Let [latex]{b}_{n}[\/latex] be an infinite sequence of zeros and ones. What is the largest possible value of [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"829494\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"829494\"]By the comparison test, [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\le \\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}}=1[\/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>Let [latex]{d}_{n}[\/latex] be an infinite sequence of digits, meaning [latex]{d}_{n}[\/latex] takes values in [latex]\\left\\{0,1\\text{,$\\ldots$ },9\\right\\}[\/latex]. What is the largest possible value of [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{d}_{n}}{{10}^{n}}[\/latex] that converges?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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50.\u00a0<\/strong>Explain why, if [latex]x>\\frac{1}{2}[\/latex], then [latex]x[\/latex] cannot be written [latex]x=\\displaystyle\\sum _{n=2}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\left({b}_{n}=0\\text{ or }1,{b}_{1}=0\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"282762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"282762\"]If [latex]{b}_{1}=0[\/latex], then, by comparison, [latex]x\\le \\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{2}^{n}}=\\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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51. [T]<\/strong> Evelyn has a perfect balancing scale, an unlimited number of [latex]1\\text{-kg}[\/latex] weights, and one each of [latex]\\frac{1}{2}\\text{-kg},\\frac{1}{4}\\text{-kg},\\frac{1}{8}\\text{-kg}[\/latex], and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?<\/div>\r\n<\/div>\r\n<\/div>\r\n
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52. [T]<\/strong> Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of [latex]1\\text{-kg}[\/latex] weights, and nine each of [latex]0.1\\text{-kg,}[\/latex] [latex]0.01\\text{-kg},0.001\\text{-kg,}[\/latex] and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?<\/p>\r\n[reveal-answer q=\"7181\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7181\"]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}\\frac{{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]\\frac{1}{{10}^{n}}\\text{-kg}[\/latex] weights that were added. If at some state this series is Robert\u2019s exact weight, the process will stop. Otherwise it represents the [latex]N\\text{th}[\/latex] partial sum of an infinite series that gives Robert\u2019s exact weight, and the error of this sum is at most [latex]\\frac{1}{{10}^{N}}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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53.\u00a0<\/strong>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2n}[\/latex] is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which [latex]n[\/latex] is odd. Let [latex]m>1[\/latex] be fixed. Show, more generally, that deleting all terms [latex]\\frac{1}{n}[\/latex] where [latex]n=mk[\/latex] for some integer [latex]k[\/latex] also results in a divergent series.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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54.\u00a0<\/strong>In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series<\/em> is a series obtained from [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] by removing any term [latex]\\frac{1}{n}[\/latex] if a given digit, say [latex]9[\/latex], appears in the decimal expansion of [latex]n[\/latex]. Argue that this depleted harmonic series converges by answering the following questions.<\/p>\r\n\r\n

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  1. How many whole numbers [latex]n[\/latex] have [latex]d[\/latex] digits?<\/li>\r\n \t
  2. How many [latex]d\\text{-digit}[\/latex] whole numbers [latex]h\\left(d\\right)[\/latex]. do not contain [latex]9[\/latex] as one or more of their digits?<\/li>\r\n \t
  3. What is the smallest [latex]d\\text{-digit}[\/latex] number [latex]m\\left(d\\right)\\text{?}[\/latex]<\/li>\r\n \t
  4. Explain why the deleted harmonic series is bounded by [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}[\/latex].<\/li>\r\n \t
  5. Show that [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}[\/latex] converges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    [reveal-answer q=\"39343\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"39343\"]a. [latex]{10}^{d}-{10}^{d - 1}<{10}^{d}[\/latex] b. [latex]h\\left(d\\right)<{9}^{d}[\/latex] c. [latex]m\\left(d\\right)={10}^{d - 1}+1[\/latex] d. Group the terms in the deleted harmonic series together by number of digits. [latex]h\\left(d\\right)[\/latex] bounds the number of terms, and each term is at most [latex]\\frac{1}{m}\\left(d\\right)[\/latex]. [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}\\le \\displaystyle\\sum _{d=1}^{\\infty }\\frac{{9}^{d}}{{\\left(10\\right)}^{d - 1}}\\le 90[\/latex]. One can actually use comparison to estimate the value to smaller than [latex]80[\/latex]. The actual value is smaller than [latex]23[\/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>Suppose that a sequence of numbers [latex]{a}_{n}>0[\/latex] has the property that [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{1}{n+1}{S}_{n}[\/latex], where [latex]{S}_{n}={a}_{1}+\\cdots +{a}_{n}[\/latex]. Can you determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges? (Hint:<\/em> [latex]{S}_{n}[\/latex] is monotone.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    56.\u00a0<\/strong>Suppose that a sequence of numbers [latex]{a}_{n}>0[\/latex] has the property that [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{1}{{\\left(n+1\\right)}^{2}}{S}_{n}[\/latex], where [latex]{S}_{n}={a}_{1}+\\text{$\\cdots$ }+{a}_{n}[\/latex]. Can you determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges? (Hint:<\/em> [latex]{S}_{2}={a}_{2}+{a}_{1}={a}_{2}+{S}_{1}={a}_{2}+1=1+\\frac{1}{4}=\\left(1+\\frac{1}{4}\\right){S}_{1}[\/latex], [latex]{S}_{3}=\\frac{1}{{3}^{2}}{S}_{2}+{S}_{2}=\\left(1+\\frac{1}{9}\\right){S}_{2}=\\left(1+\\frac{1}{9}\\right)\\left(1+\\frac{1}{4}\\right){S}_{1}[\/latex], etc. Look at [latex]\\text{ln}\\left({S}_{n}\\right)[\/latex], and use [latex]\\text{ln}\\left(1+t\\right)\\le t[\/latex], [latex]t>0.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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    [reveal-answer q=\"673495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673495\"]Continuing the hint gives [latex]{S}_{N}=\\left(1+\\frac{1}{{N}^{2}}\\right)\\left(1+\\frac{1}{{\\left(N - 1\\right)}^{2}}\\text{$\\ldots$ }\\left(1+\\frac{1}{4}\\right)\\right)[\/latex]. Then [latex]\\text{ln}\\left({S}_{N}\\right)=\\text{ln}\\left(1+\\frac{1}{{N}^{2}}\\right)+\\text{ln}\\left(1+\\frac{1}{{\\left(N - 1\\right)}^{2}}\\right)+\\text{$\\cdots$ }+\\text{ln}\\left(1+\\frac{1}{4}\\right)[\/latex]. Since [latex]\\text{ln}\\left(1+t\\right)[\/latex] is bounded by a constant times [latex]t[\/latex], when [latex]0<t<1[\/latex] one has [latex]\\text{ln}\\left({S}_{N}\\right)\\le C\\displaystyle\\sum _{n=1}^{N}\\frac{1}{{n}^{2}}[\/latex], which converges by comparison to the p-series for [latex]p=2[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

    Use the comparison test to determine whether the following series converge.<\/p>\n

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

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    Show Solution<\/span><\/p>\n
    Converges by comparison with [latex]\\frac{1}{{n}^{2}}[\/latex].<\/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 }\\frac{1}{2\\left(n+1\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    4.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2n - 1}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Diverges by comparison with harmonic series, since [latex]2n - 1\\ge n[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    5.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{\\left(n\\text{ln}n\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    6.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n\\text{!}}{\\left(n+2\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{a}_{n}=\\frac{1}{\\left(n+1\\right)\\left(n+2\\right)}<\\frac{1}{{n}^{2}}[\/latex]. Converges by comparison with p-series, [latex]p=2[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sin\\left(\\frac{1}{n}\\right)}{n}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\sin\\left(\\frac{1}{n}\\right)\\le \\frac{1}{n}[\/latex], so converges by comparison with p-series, [latex]p=2[\/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 }\\frac{{\\sin}^{2}n}{{n}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sin\\left(\\frac{1}{n}\\right)}{\\sqrt{n}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\sin\\left(\\frac{1}{n}\\right)\\le 1[\/latex], so converges by comparison with p-series, [latex]p=\\frac{3}{2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{1.2}-1}{{n}^{2.3}+1}[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\sqrt{n+1}-\\sqrt{n}}{n}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Since [latex]\\sqrt{n+1}-\\sqrt{n}=\\frac{1}{\\left(\\sqrt{n+1}+\\sqrt{n}\\right)}\\le \\frac{2}{\\sqrt{n}}[\/latex], series converges by comparison with p-series for [latex]p=1.5[\/latex].<\/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 }\\frac{\\sqrt[4]{n}}{\\sqrt[3]{{n}^{4}+{n}^{2}}}[\/latex]<\/div>\n<\/div>\n<\/div>\n

    Use the limit comparison test to determine whether each of the following series converges or diverges.<\/p>\n

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

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with p-series for [latex]p>1[\/latex].<\/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(\\frac{\\text{ln}n}{{n}^{0.6}}\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(1+\\frac{1}{n}\\right)}{n}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with p-series, [latex]p=2[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\text{ln}\\left(1+\\frac{1}{{n}^{2}}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    18.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{4}^{n}-{3}^{n}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with [latex]{4}^{\\text{-}n}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}-n\\sin{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{e}^{\\left(1.1\\right)n}-{3}^{n}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with [latex]\\frac{1}{{e}^{1.1n}}[\/latex].<\/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 }\\frac{1}{{e}^{\\left(1.01\\right)n}-{3}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{1+\\frac{1}{n}}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Diverges by limit comparison with harmonic series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    23.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{1+\\frac{1}{n}}{n}^{1+\\frac{1}{n}}}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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    24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{n}-\\sin\\left(\\frac{1}{n}\\right)\\right)[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/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-\\cos\\left(\\frac{1}{n}\\right)\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}\\left(\\frac{\\pi }{2}-{\\tan}^{-1}n\\right)[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges by limit comparison with p-series, [latex]p=3[\/latex].<\/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 }{\\left(1-\\frac{1}{n}\\right)}^{n.n}[\/latex] (Hint:<\/em> [latex]{\\left(1-\\frac{1}{n}\\right)}^{n}\\to \\frac{1}{e}.[\/latex])<\/div>\n<\/div>\n<\/div>\n
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    28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(1-{e}^{-\\frac{1}{n}}\\right)[\/latex] (Hint:<\/em> [latex]\\frac{1}{e}\\approx {\\left(1 - \\frac{1}{n}\\right)}^{n}[\/latex], so [latex]1-{e}^{-\\frac{1}{n}}\\approx \\frac{1}{n}.[\/latex])<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Diverges by limit comparison with [latex]\\frac{1}{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    29.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    30.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{\\left(\\text{ln}n\\right)}{n}\\right)}^{p}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges for [latex]p>1[\/latex] by comparison with a [latex]p[\/latex] series for slightly smaller [latex]p[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    31.\u00a0<\/strong>For which [latex]p[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{pn}}{{3}^{n}}[\/latex] converge?<\/div>\n<\/div>\n<\/div>\n
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    32.\u00a0<\/strong>For which [latex]p>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{p}}{{2}^{n}}[\/latex] converge?<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges for all [latex]p>0[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    33.\u00a0<\/strong>For which [latex]r>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{r}^{{n}^{2}}}{{2}^{n}}[\/latex] converge?<\/div>\n<\/div>\n<\/div>\n
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    34.\u00a0<\/strong>For which [latex]r>0[\/latex] does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{r}^{{n}^{2}}}[\/latex] converge?<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Converges for all [latex]r>1[\/latex]. If [latex]r>1[\/latex] then [latex]{r}^{n}>4[\/latex], say, once [latex]n>\\frac{\\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]\\frac{1}{2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    35.\u00a0<\/strong>Find all values of [latex]p[\/latex] and [latex]q[\/latex] such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{p}}{{\\left(n\\text{!}\\right)}^{q}}[\/latex] converges.<\/div>\n<\/div>\n<\/div>\n
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    36.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\sin}^{2}\\left(\\frac{nr}{2}\\right)}{n}[\/latex] converge or diverge? Explain.<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    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 }\\frac{1}{2n+1}[\/latex], which diverges by limit comparison with the harmonic series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    37.\u00a0<\/strong>Explain why, for each [latex]n[\/latex], at least one of [latex]\\left\\{|\\sin{n}|,|\\sin\\left(n+1\\right)|\\text{,...},|\\sin{n}+6|\\right\\}[\/latex] is larger than [latex]\\frac{1}{2}[\/latex]. Use this relation to test convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{|\\sin{n}|}{\\sqrt{n}}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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    38.\u00a0<\/strong>Suppose that [latex]{a}_{n}\\ge 0[\/latex] and [latex]{b}_{n}\\ge 0[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}^{2}{}_{n}[\/latex] converge. Prove that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}\\le \\frac{1}{2}\\left(\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}^{2}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}^{2}\\right)[\/latex].<\/p>\n<\/div>\n

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    [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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    39.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Write [latex]{2}^{\\text{ln}\\text{ln}n}[\/latex] as a power of [latex]\\text{ln}n.[\/latex])<\/div>\n<\/div>\n<\/div>\n
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    40.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Use [latex]n={e}^{\\text{ln}\\left(n\\right)}[\/latex] to compare to a [latex]p-\\text{series}\\text{.}[\/latex])<\/p>\n<\/div>\n

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    [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}<\\frac{1}{{n}^{2}}[\/latex], and the series converges by comparison to a [latex]p-\\text{series}\\text{.}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    41.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }{\\left(\\text{ln}n\\right)}^{\\text{-}\\text{ln}\\text{ln}n}[\/latex] converge? (Hint:<\/em> Compare [latex]{a}_{n}[\/latex] to [latex]\\frac{1}{n}.[\/latex])<\/div>\n<\/div>\n<\/div>\n
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    42.\u00a0<\/strong>Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges, does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] necessarily converge?<\/p>\n<\/div>\n

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    [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 \\frac{1}{{n}^{2}}[\/latex] converges, but [latex]\\displaystyle\\sum \\frac{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.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    43.\u00a0<\/strong>Suppose that [latex]{a}_{n}>0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of zeros and ones. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily converge?<\/div>\n<\/div>\n<\/div>\n
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    44.\u00a0<\/strong>Suppose that [latex]{a}_{n}>0[\/latex] for all [latex]n[\/latex] and that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges. Suppose that [latex]{b}_{n}[\/latex] is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{b}_{n}[\/latex] necessarily diverge?<\/p>\n<\/div>\n

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    No. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{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}\\frac{{b}_{k}}{k}=\\displaystyle\\sum \\frac{1}{{k}^{2}}[\/latex] converges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    45.\u00a0<\/strong>Complete the details of the following argument: If [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] converges to a finite sum [latex]s[\/latex], then [latex]\\frac{1}{2}s=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\text{$\\cdots$ }[\/latex] and [latex]s-\\frac{1}{2}s=1+\\frac{1}{3}+\\frac{1}{5}+\\text{$\\cdots$ }[\/latex]. Why does this lead to a contradiction?<\/div>\n<\/div>\n<\/div>\n
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    46.\u00a0<\/strong>Show that if [latex]{a}_{n}\\ge 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}^{2}{}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\sin}^{2}\\left({a}_{n}\\right)[\/latex] converges.<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]|\\sin{t}|\\le |t|[\/latex], so the result follows from the comparison test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    47.\u00a0<\/strong>Suppose that [latex]\\frac{{a}_{n}}{{b}_{n}}\\to 0[\/latex] in the comparison test, where [latex]{a}_{n}\\ge 0[\/latex] and [latex]{b}_{n}\\ge 0[\/latex]. Prove that if [latex]\\displaystyle\\sum {b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum {a}_{n}[\/latex] converges.<\/div>\n<\/div>\n<\/div>\n
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    48.\u00a0<\/strong>Let [latex]{b}_{n}[\/latex] be an infinite sequence of zeros and ones. What is the largest possible value of [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\text{?}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    By the comparison test, [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\le \\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}}=1[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    49.\u00a0<\/strong>Let [latex]{d}_{n}[\/latex] be an infinite sequence of digits, meaning [latex]{d}_{n}[\/latex] takes values in [latex]\\left\\{0,1\\text{,$\\ldots$ },9\\right\\}[\/latex]. What is the largest possible value of [latex]x=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{d}_{n}}{{10}^{n}}[\/latex] that converges?<\/div>\n<\/div>\n<\/div>\n
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    50.\u00a0<\/strong>Explain why, if [latex]x>\\frac{1}{2}[\/latex], then [latex]x[\/latex] cannot be written [latex]x=\\displaystyle\\sum _{n=2}^{\\infty }\\frac{{b}_{n}}{{2}^{n}}\\left({b}_{n}=0\\text{ or }1,{b}_{1}=0\\right)[\/latex].<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    If [latex]{b}_{1}=0[\/latex], then, by comparison, [latex]x\\le \\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{2}^{n}}=\\frac{1}{2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    51. [T]<\/strong> Evelyn has a perfect balancing scale, an unlimited number of [latex]1\\text{-kg}[\/latex] weights, and one each of [latex]\\frac{1}{2}\\text{-kg},\\frac{1}{4}\\text{-kg},\\frac{1}{8}\\text{-kg}[\/latex], and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?<\/div>\n<\/div>\n<\/div>\n
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    52. [T]<\/strong> Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of [latex]1\\text{-kg}[\/latex] weights, and nine each of [latex]0.1\\text{-kg,}[\/latex] [latex]0.01\\text{-kg},0.001\\text{-kg,}[\/latex] and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?<\/p>\n

    Show Solution<\/span><\/p>\n
    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}\\frac{{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]\\frac{1}{{10}^{n}}\\text{-kg}[\/latex] weights that were added. If at some state this series is Robert\u2019s exact weight, the process will stop. Otherwise it represents the [latex]N\\text{th}[\/latex] partial sum of an infinite series that gives Robert\u2019s exact weight, and the error of this sum is at most [latex]\\frac{1}{{10}^{N}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    53.\u00a0<\/strong>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2n}[\/latex] is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which [latex]n[\/latex] is odd. Let [latex]m>1[\/latex] be fixed. Show, more generally, that deleting all terms [latex]\\frac{1}{n}[\/latex] where [latex]n=mk[\/latex] for some integer [latex]k[\/latex] also results in a divergent series.<\/div>\n<\/div>\n<\/div>\n
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    54.\u00a0<\/strong>In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series<\/em> is a series obtained from [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] by removing any term [latex]\\frac{1}{n}[\/latex] if a given digit, say [latex]9[\/latex], appears in the decimal expansion of [latex]n[\/latex]. Argue that this depleted harmonic series converges by answering the following questions.<\/p>\n

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    1. How many whole numbers [latex]n[\/latex] have [latex]d[\/latex] digits?<\/li>\n
    2. How many [latex]d\\text{-digit}[\/latex] whole numbers [latex]h\\left(d\\right)[\/latex]. do not contain [latex]9[\/latex] as one or more of their digits?<\/li>\n
    3. What is the smallest [latex]d\\text{-digit}[\/latex] number [latex]m\\left(d\\right)\\text{?}[\/latex]<\/li>\n
    4. Explain why the deleted harmonic series is bounded by [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}[\/latex].<\/li>\n
    5. Show that [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}[\/latex] converges.<\/li>\n<\/ol>\n<\/div>\n
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      a. [latex]{10}^{d}-{10}^{d - 1}<{10}^{d}[\/latex] b. [latex]h\\left(d\\right)<{9}^{d}[\/latex] c. [latex]m\\left(d\\right)={10}^{d - 1}+1[\/latex] d. Group the terms in the deleted harmonic series together by number of digits. [latex]h\\left(d\\right)[\/latex] bounds the number of terms, and each term is at most [latex]\\frac{1}{m}\\left(d\\right)[\/latex]. [latex]\\displaystyle\\sum _{d=1}^{\\infty }\\frac{h\\left(d\\right)}{m\\left(d\\right)}\\le \\displaystyle\\sum _{d=1}^{\\infty }\\frac{{9}^{d}}{{\\left(10\\right)}^{d - 1}}\\le 90[\/latex]. One can actually use comparison to estimate the value to smaller than [latex]80[\/latex]. The actual value is smaller than [latex]23[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      55.\u00a0<\/strong>Suppose that a sequence of numbers [latex]{a}_{n}>0[\/latex] has the property that [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{1}{n+1}{S}_{n}[\/latex], where [latex]{S}_{n}={a}_{1}+\\cdots +{a}_{n}[\/latex]. Can you determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges? (Hint:<\/em> [latex]{S}_{n}[\/latex] is monotone.)<\/div>\n<\/div>\n<\/div>\n
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      56.\u00a0<\/strong>Suppose that a sequence of numbers [latex]{a}_{n}>0[\/latex] has the property that [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{1}{{\\left(n+1\\right)}^{2}}{S}_{n}[\/latex], where [latex]{S}_{n}={a}_{1}+\\text{$\\cdots$ }+{a}_{n}[\/latex]. Can you determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges? (Hint:<\/em> [latex]{S}_{2}={a}_{2}+{a}_{1}={a}_{2}+{S}_{1}={a}_{2}+1=1+\\frac{1}{4}=\\left(1+\\frac{1}{4}\\right){S}_{1}[\/latex], [latex]{S}_{3}=\\frac{1}{{3}^{2}}{S}_{2}+{S}_{2}=\\left(1+\\frac{1}{9}\\right){S}_{2}=\\left(1+\\frac{1}{9}\\right)\\left(1+\\frac{1}{4}\\right){S}_{1}[\/latex], etc. Look at [latex]\\text{ln}\\left({S}_{n}\\right)[\/latex], and use [latex]\\text{ln}\\left(1+t\\right)\\le t[\/latex], [latex]t>0.[\/latex])<\/p>\n<\/div>\n

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      Show Solution<\/span><\/p>\n
      Continuing the hint gives [latex]{S}_{N}=\\left(1+\\frac{1}{{N}^{2}}\\right)\\left(1+\\frac{1}{{\\left(N - 1\\right)}^{2}}\\text{$\\ldots$ }\\left(1+\\frac{1}{4}\\right)\\right)[\/latex]. Then [latex]\\text{ln}\\left({S}_{N}\\right)=\\text{ln}\\left(1+\\frac{1}{{N}^{2}}\\right)+\\text{ln}\\left(1+\\frac{1}{{\\left(N - 1\\right)}^{2}}\\right)+\\text{$\\cdots$ }+\\text{ln}\\left(1+\\frac{1}{4}\\right)[\/latex]. Since [latex]\\text{ln}\\left(1+t\\right)[\/latex] is bounded by a constant times [latex]t[\/latex], when [latex]0\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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