{"id":106,"date":"2021-03-25T02:21:03","date_gmt":"2021-03-25T02:21:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/the-divergence-and-integral-tests-2\/"},"modified":"2022-01-03T18:28:54","modified_gmt":"2022-01-03T18:28:54","slug":"the-divergence-and-integral-tests-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/the-divergence-and-integral-tests-2\/","title":{"raw":"Problem Set: The Divergence and Integral Tests","rendered":"Problem Set: The Divergence and Integral Tests"},"content":{"raw":"

For each of the following sequences, if the divergence test applies, either state that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist or find [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. If the divergence test does not apply, state why.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n}{n+2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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2.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n}{5{n}^{2}-3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"982130\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982130\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.[\/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]{a}_{n}=\\frac{n}{\\sqrt{3{n}^{2}+2n+1}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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4.\u00a0<\/strong>[latex]{a}_{n}=\\frac{\\left(2n+1\\right)\\left(n - 1\\right)}{{\\left(n+1\\right)}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"81165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81165\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=2[\/latex]. Series diverges.[\/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]{a}_{n}=\\frac{{\\left(2n+1\\right)}^{2n}}{{\\left(3{n}^{2}+1\\right)}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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6.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{2}^{n}}{{3}^{\\frac{n}{2}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"275238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"275238\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\infty [\/latex] (does not exist). Series diverges.[\/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]{a}_{n}=\\frac{{2}^{n}+{3}^{n}}{{10}^{\\frac{n}{2}}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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8.\u00a0<\/strong>[latex]{a}_{n}={e}^{\\frac{-2}{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"477038\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477038\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex]. Series diverges.[\/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]{a}_{n}=\\cos{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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10.\u00a0<\/strong>[latex]{a}_{n}=\\tan{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"596763\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"596763\"] [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist. Series diverges.[\/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]{a}_{n}=\\frac{1-{\\cos}^{2}\\left(\\frac{1}{n}\\right)}{{\\sin}^{2}\\left(\\frac{2}{n}\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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12.\u00a0<\/strong>[latex]{a}_{n}={\\left(1-\\frac{1}{n}\\right)}^{2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"219838\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"219838\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\frac{1}{{e}^{2}}[\/latex]. Series diverges.[\/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]{a}_{n}=\\frac{\\text{ln}n}{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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14.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(\\text{ln}n\\right)}^{2}}{\\sqrt{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"127477\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"127477\"][latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

State whether the given [latex]p[\/latex] -series converges.<\/p>\r\n\r\n

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

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[reveal-answer q=\"698030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698030\"]Series converges, [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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17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt[3]{{n}^{2}}}[\/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}{\\sqrt[3]{{n}^{4}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"337903\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"337903\"]Series converges, [latex]p=\\frac{4}{3}>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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19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{e}}{{n}^{\\pi }}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{\\pi }}{{n}^{2e}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"456831\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"456831\"]Series converges, [latex]p=2e-\\pi >1[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

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21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n+5}}[\/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}{\\sqrt[3]{n+5}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"54070\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"54070\"]Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{dx}{{\\left(x+5\\right)}^{\\frac{1}{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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23.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n\\text{ln}n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{1+{n}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"90014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"90014\"]Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{1+{x}^{2}}dx[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{1+{e}^{2n}}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{2n}{1+{n}^{4}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"293897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"293897\"]Series converges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{2x}{1+{x}^{4}}dx[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n{\\text{ln}}^{2}n}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n

Express the following sums as [latex]p[\/latex] -series and determine whether each converges.<\/p>\r\n\r\n

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28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}n}[\/latex] (Hint:<\/em> [latex]{2}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}2}}[\/latex] .)<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"131028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"131028\"][latex]{2}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}2}}[\/latex]. Since [latex]\\text{ln}2<1[\/latex], diverges by [latex]p[\/latex] -series.[\/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 }{3}^{\\text{-}\\text{ln}n}[\/latex] (Hint:<\/em> [latex]{3}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}3}}[\/latex] .)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{2}^{-2\\text{ln}n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"523187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"523187\"][latex]{2}^{-2\\text{ln}n}=\\frac{1}{{n}^{2\\text{ln}2}}[\/latex]. Since [latex]2\\text{ln}2 - 1<1[\/latex], diverges by [latex]p[\/latex] -series.[\/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>[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{3}^{-2\\text{ln}n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

Use the estimate [latex]{R}_{N}\\le {\\displaystyle\\int }_{N}^{\\infty }f\\left(t\\right)dt[\/latex] to find a bound for the remainder [latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] where [latex]{a}_{n}=f\\left(n\\right)[\/latex].<\/p>\r\n\r\n

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32.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"962691\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962691\"][latex]{R}_{1000}\\le {\\displaystyle\\int }_{1000}^{\\infty }\\frac{dt}{{t}^{2}}=-\\frac{1}{t}{|}_{1000}^{\\infty }=0.001[\/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>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{3}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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34.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{1+{n}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"163732\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"163732\"][latex]{R}_{1000}\\le {\\displaystyle\\int }_{1000}^{\\infty }\\frac{dt}{1+{t}^{2}}={\\tan}^{-1}\\infty -{\\tan}^{-1}\\left(1000\\right)=\\frac{\\pi}{2}-{\\tan}^{-1}\\left(1000\\right)\\approx 0.000999[\/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>[latex]\\displaystyle\\sum _{n=1}^{100}\\frac{n}{{2}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

[T]<\/strong> Find the minimum value of [latex]N[\/latex] such that the remainder estimate [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f[\/latex] guarantees that [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] estimates [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], accurate to within the given error.<\/p>\r\n\r\n

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36.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"493014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"493014\"][latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{2}}=\\frac{1}{N},N>{10}^{4}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

37.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{1.1}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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38.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{1.01}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"87600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"87600\"][latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{1.01}}=100{N}^{-0.01},N>{10}^{600}[\/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>[latex]{a}_{n}=\\frac{1}{n{\\text{ln}}^{2}n}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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40.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{1+{n}^{2}}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"136496\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"136496\"][latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{1+{x}^{2}}=\\frac{\\pi}{2}-{\\tan}^{-1}\\left(N\\right),N>\\tan\\left(\\frac{\\pi}{2}-{10}^{-3}\\right)\\approx 1000[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find a value of [latex]N[\/latex] such that [latex]{R}_{N}[\/latex] is smaller than the desired error. Compute the corresponding sum [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] and compare it to the given estimate of the infinite series.<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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41.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{11}}[\/latex], error [latex]<{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{11}}=1.000494\\text{$\\ldots$ }[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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42.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{e}^{n}}[\/latex], error [latex]<{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{e}^{n}}=\\frac{1}{e - 1}=0.581976\\text{$\\ldots$ }[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"320225\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"320225\"][latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{e}^{x}}={e}^{\\text{-}N},N>5\\text{ln}\\left(10\\right)[\/latex], okay if [latex]N=12;\\displaystyle\\sum _{n=1}^{12}{e}^{\\text{-}n}=0.581973...[\/latex]. Estimate agrees with [latex]\\frac{1}{\\left(e - 1\\right)}[\/latex] to five decimal places.[\/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>[latex]{a}_{n}=\\frac{1}{{e}^{{n}^{2}}}[\/latex], error [latex]<{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{e}^{n2}}=0.40488139857\\text{$\\ldots$ }[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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44.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{4}}[\/latex], error [latex]<{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}=\\frac{{\\pi }^{4}}{90}=1.08232...[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"955240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"955240\"][latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{4}}=\\frac{4}{{N}^{3}},N>{\\left({4.10}^{4}\\right)}^{\\frac{1}{3}}[\/latex], okay if [latex]N=35[\/latex];[latex]\\displaystyle\\sum _{n=1}^{35}\\frac{1}{{n}^{4}}=1.08231\\text{$\\ldots$ }[\/latex]. Estimate agrees with the sum to four decimal places.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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45.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{6}}[\/latex], error [latex]<{10}^{-6}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}={\\pi }^\\frac{{6}}{945}=1.01734306...[\/latex],<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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46.\u00a0<\/strong>Find the limit as [latex]n\\to \\infty [\/latex] of [latex]\\frac{1}{n}+\\frac{1}{n+1}+\\text{$\\cdots$ }+\\frac{1}{2n}[\/latex]. (Hint:<\/em> Compare to [latex]{\\displaystyle\\int }_{n}^{2n}\\frac{1}{t}dt.\\text{)}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"677793\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"677793\"][latex]\\text{ln}\\left(2\\right)[\/latex][\/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>Find the limit as [latex]n\\to \\infty [\/latex] of [latex]\\frac{1}{n}+\\frac{1}{n+1}+\\text{$\\cdots$ }+\\frac{1}{3n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.<\/p>\r\n\r\n

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48.\u00a0<\/strong>In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number [latex]{H}_{k}=\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\text{$\\cdots$ }+\\frac{1}{k}\\right)[\/latex]. Recall that [latex]{T}_{k}={H}_{k}-\\text{ln}k[\/latex] is decreasing. Compute [latex]T=\\underset{k\\to \\infty }{\\text{lim}}{T}_{k}[\/latex] to four decimal places. (Hint:<\/em> [latex]\\frac{1}{k+1}<{\\displaystyle\\int }_{k}^{k+1}\\frac{1}{x}dx[\/latex] .)<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"927582\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"927582\"][latex]T=0.5772..[\/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. [T]<\/strong> Complete sampling with replacement, sometimes called the coupon collector\u2019s problem<\/em><\/span>, is phrased as follows: Suppose you have [latex]N[\/latex] unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps [latex]E\\left(N\\right)[\/latex] that it takes to draw each unique item at least once. It turns out that [latex]E\\left(N\\right)=N.{H}_{N}=N\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\text{$\\cdots$ }+\\frac{1}{N}\\right)[\/latex]. Find [latex]E\\left(N\\right)[\/latex] for [latex]N=10,20,\\text{and }50[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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50. [T]<\/strong> The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has [latex]n[\/latex] cards, then the probability that the insertion will be below the card initially at the bottom (call this card [latex]B[\/latex]) is [latex]\\frac{1}{n}[\/latex]. Thus the expected number of top random insertions before [latex]B[\/latex] is no longer at the bottom is n<\/em>. Once one card is below [latex]B[\/latex], there are two places below [latex]B[\/latex] and the probability that a randomly inserted card will fall below [latex]B[\/latex] is [latex]\\frac{2}{n}[\/latex]. The expected number of top random insertions before this happens is [latex]\\frac{n}{2}[\/latex]. The two cards below [latex]B[\/latex] are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"197194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197194\"]The expected number of random insertions to get [latex]B[\/latex] to the top is [latex]n+\\frac{n}{2}+\\frac{n}{3}+\\text{$\\cdots$ }+\\frac{n}{\\left(n - 1\\right)}[\/latex]. Then one more insertion puts [latex]B[\/latex] back in at random. Thus, the expected number of shuffles to randomize the deck is [latex]n\\left(1+\\frac{1}{2}+\\text{$\\cdots$ }+\\frac{1}{n}\\right)[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

51.\u00a0<\/strong>Suppose a scooter can travel [latex]100[\/latex] km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel [latex]100{H}_{N}[\/latex] km, where [latex]{H}_{N}=1+\\frac{1}{2}+\\text{$\\cdots$ }+\\frac{1}{N}[\/latex].<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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52.\u00a0<\/strong>Show that for the remainder estimate to apply on [latex]\\left[N,\\infty \\right)[\/latex] it is sufficient that [latex]f\\left(x\\right)[\/latex] be decreasing on [latex]\\left[N,\\infty \\right)[\/latex], but [latex]f[\/latex] need not be decreasing on [latex]\\left[1,\\infty \\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"183062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"183062\"]Set [latex]{b}_{n}={a}_{n+N}[\/latex] and [latex]g\\left(t\\right)=f\\left(t+N\\right)[\/latex] such that [latex]f[\/latex] is decreasing on [latex]\\left[t,\\infty \\right)[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

53. [T]<\/strong> Use the remainder estimate and integration by parts to approximate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{e}^{n}}[\/latex] within an error smaller than [latex]0.0001[\/latex].<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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54.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"857018\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"857018\"]The series converges for [latex]p>1[\/latex] by integral test using change of variable.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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55. [T]<\/strong> Suppose a computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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56. [T]<\/strong> A fast computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=2}^{N}\\frac{1}{n\\text{ln}n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"12000\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"12000\"][latex]N={e}^{{e}^{100}}\\approx {e}^{{10}^{43}}[\/latex] terms are needed.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

For each of the following sequences, if the divergence test applies, either state that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist or find [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. If the divergence test does not apply, state why.<\/p>\n

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1.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n}{n+2}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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2.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n}{5{n}^{2}-3}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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3.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n}{\\sqrt{3{n}^{2}+2n+1}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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4.\u00a0<\/strong>[latex]{a}_{n}=\\frac{\\left(2n+1\\right)\\left(n - 1\\right)}{{\\left(n+1\\right)}^{2}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=2[\/latex]. Series diverges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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5.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(2n+1\\right)}^{2n}}{{\\left(3{n}^{2}+1\\right)}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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6.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{2}^{n}}{{3}^{\\frac{n}{2}}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\infty[\/latex] (does not exist). Series diverges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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7.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{2}^{n}+{3}^{n}}{{10}^{\\frac{n}{2}}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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8.\u00a0<\/strong>[latex]{a}_{n}={e}^{\\frac{-2}{n}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex]. Series diverges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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9.\u00a0<\/strong>[latex]{a}_{n}=\\cos{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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10.\u00a0<\/strong>[latex]{a}_{n}=\\tan{n}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist. Series diverges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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11.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1-{\\cos}^{2}\\left(\\frac{1}{n}\\right)}{{\\sin}^{2}\\left(\\frac{2}{n}\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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12.\u00a0<\/strong>[latex]{a}_{n}={\\left(1-\\frac{1}{n}\\right)}^{2n}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\frac{1}{{e}^{2}}[\/latex]. Series diverges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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13.\u00a0<\/strong>[latex]{a}_{n}=\\frac{\\text{ln}n}{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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14.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(\\text{ln}n\\right)}^{2}}{\\sqrt{n}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex]. Divergence test does not apply.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

State whether the given [latex]p[\/latex] -series converges.<\/p>\n

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

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Show Solution<\/span><\/p>\n
Series converges, [latex]p>1[\/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 }\\frac{1}{\\sqrt[3]{{n}^{2}}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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18.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt[3]{{n}^{4}}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Series converges, [latex]p=\\frac{4}{3}>1[\/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{{n}^{e}}{{n}^{\\pi }}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{\\pi }}{{n}^{2e}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Series converges, [latex]p=2e-\\pi >1[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

Use the integral test to determine whether the following sums converge.<\/p>\n

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21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n+5}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt[3]{n+5}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{dx}{{\\left(x+5\\right)}^{\\frac{1}{3}}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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23.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n\\text{ln}n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{1+{n}^{2}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Series diverges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{1+{x}^{2}}dx[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{1+{e}^{2n}}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{2n}{1+{n}^{4}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Series converges by comparison with [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{2x}{1+{x}^{4}}dx[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n{\\text{ln}}^{2}n}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n

Express the following sums as [latex]p[\/latex] -series and determine whether each converges.<\/p>\n

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28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{2}^{\\text{-}\\text{ln}n}[\/latex] (Hint:<\/em> [latex]{2}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}2}}[\/latex] .)<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{2}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}2}}[\/latex]. Since [latex]\\text{ln}2<1[\/latex], diverges by [latex]p[\/latex] -series.<\/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 }{3}^{\\text{-}\\text{ln}n}[\/latex] (Hint:<\/em> [latex]{3}^{\\text{-}\\text{ln}n}=\\frac{1}{{n}^{\\text{ln}3}}[\/latex] .)<\/div>\n<\/div>\n<\/div>\n
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30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{2}^{-2\\text{ln}n}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{2}^{-2\\text{ln}n}=\\frac{1}{{n}^{2\\text{ln}2}}[\/latex]. Since [latex]2\\text{ln}2 - 1<1[\/latex], diverges by [latex]p[\/latex] -series.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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31.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }n{3}^{-2\\text{ln}n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

Use the estimate [latex]{R}_{N}\\le {\\displaystyle\\int }_{N}^{\\infty }f\\left(t\\right)dt[\/latex] to find a bound for the remainder [latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] where [latex]{a}_{n}=f\\left(n\\right)[\/latex].<\/p>\n

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32.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{2}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{1000}\\le {\\displaystyle\\int }_{1000}^{\\infty }\\frac{dt}{{t}^{2}}=-\\frac{1}{t}{|}_{1000}^{\\infty }=0.001[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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33.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{{n}^{3}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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34.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{1000}\\frac{1}{1+{n}^{2}}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{1000}\\le {\\displaystyle\\int }_{1000}^{\\infty }\\frac{dt}{1+{t}^{2}}={\\tan}^{-1}\\infty -{\\tan}^{-1}\\left(1000\\right)=\\frac{\\pi}{2}-{\\tan}^{-1}\\left(1000\\right)\\approx 0.000999[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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35.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{100}\\frac{n}{{2}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n

[T]<\/strong> Find the minimum value of [latex]N[\/latex] such that the remainder estimate [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f[\/latex] guarantees that [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] estimates [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], accurate to within the given error.<\/p>\n

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36.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{2}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{2}}=\\frac{1}{N},N>{10}^{4}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
37.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{1.1}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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38.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{1.01}}[\/latex], error [latex]<{10}^{-4}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{1.01}}=100{N}^{-0.01},N>{10}^{600}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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39.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{n{\\text{ln}}^{2}n}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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40.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{1+{n}^{2}}[\/latex], error [latex]<{10}^{-3}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{1+{x}^{2}}=\\frac{\\pi}{2}-{\\tan}^{-1}\\left(N\\right),N>\\tan\\left(\\frac{\\pi}{2}-{10}^{-3}\\right)\\approx 1000[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, find a value of [latex]N[\/latex] such that [latex]{R}_{N}[\/latex] is smaller than the desired error. Compute the corresponding sum [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] and compare it to the given estimate of the infinite series.<\/span><\/p>\n<\/div>\n<\/div>\n

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41.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{11}}[\/latex], error [latex]<{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{11}}=1.000494\\text{$\\ldots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n
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42.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{e}^{n}}[\/latex], error [latex]<{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{e}^{n}}=\\frac{1}{e - 1}=0.581976\\text{$\\ldots$ }[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{e}^{x}}={e}^{\\text{-}N},N>5\\text{ln}\\left(10\\right)[\/latex], okay if [latex]N=12;\\displaystyle\\sum _{n=1}^{12}{e}^{\\text{-}n}=0.581973...[\/latex]. Estimate agrees with [latex]\\frac{1}{\\left(e - 1\\right)}[\/latex] to five decimal places.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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43.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{e}^{{n}^{2}}}[\/latex], error [latex]<{10}^{-5}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{e}^{n2}}=0.40488139857\\text{$\\ldots$ }[\/latex]<\/div>\n<\/div>\n<\/div>\n
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44.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{4}}[\/latex], error [latex]<{10}^{-4}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}=\\frac{{\\pi }^{4}}{90}=1.08232...[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{dx}{{x}^{4}}=\\frac{4}{{N}^{3}},N>{\\left({4.10}^{4}\\right)}^{\\frac{1}{3}}[\/latex], okay if [latex]N=35[\/latex];[latex]\\displaystyle\\sum _{n=1}^{35}\\frac{1}{{n}^{4}}=1.08231\\text{$\\ldots$ }[\/latex]. Estimate agrees with the sum to four decimal places.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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45.\u00a0<\/strong>[latex]{a}_{n}=\\frac{1}{{n}^{6}}[\/latex], error [latex]<{10}^{-6}[\/latex], [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}={\\pi }^\\frac{{6}}{945}=1.01734306...[\/latex],<\/span><\/div>\n<\/div>\n<\/div>\n
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46.\u00a0<\/strong>Find the limit as [latex]n\\to \\infty[\/latex] of [latex]\\frac{1}{n}+\\frac{1}{n+1}+\\text{$\\cdots$ }+\\frac{1}{2n}[\/latex]. (Hint:<\/em> Compare to [latex]{\\displaystyle\\int }_{n}^{2n}\\frac{1}{t}dt.\\text{)}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\text{ln}\\left(2\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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47.\u00a0<\/strong>Find the limit as [latex]n\\to \\infty[\/latex] of [latex]\\frac{1}{n}+\\frac{1}{n+1}+\\text{$\\cdots$ }+\\frac{1}{3n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.<\/p>\n

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48.\u00a0<\/strong>In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number [latex]{H}_{k}=\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\text{$\\cdots$ }+\\frac{1}{k}\\right)[\/latex]. Recall that [latex]{T}_{k}={H}_{k}-\\text{ln}k[\/latex] is decreasing. Compute [latex]T=\\underset{k\\to \\infty }{\\text{lim}}{T}_{k}[\/latex] to four decimal places. (Hint:<\/em> [latex]\\frac{1}{k+1}<{\\displaystyle\\int }_{k}^{k+1}\\frac{1}{x}dx[\/latex] .)<\/p>\n<\/div>\n

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[latex]T=0.5772..[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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49. [T]<\/strong> Complete sampling with replacement, sometimes called the coupon collector\u2019s problem<\/em><\/span>, is phrased as follows: Suppose you have [latex]N[\/latex] unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps [latex]E\\left(N\\right)[\/latex] that it takes to draw each unique item at least once. It turns out that [latex]E\\left(N\\right)=N.{H}_{N}=N\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\text{$\\cdots$ }+\\frac{1}{N}\\right)[\/latex]. Find [latex]E\\left(N\\right)[\/latex] for [latex]N=10,20,\\text{and }50[\/latex].<\/div>\n<\/div>\n<\/div>\n
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50. [T]<\/strong> The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has [latex]n[\/latex] cards, then the probability that the insertion will be below the card initially at the bottom (call this card [latex]B[\/latex]) is [latex]\\frac{1}{n}[\/latex]. Thus the expected number of top random insertions before [latex]B[\/latex] is no longer at the bottom is n<\/em>. Once one card is below [latex]B[\/latex], there are two places below [latex]B[\/latex] and the probability that a randomly inserted card will fall below [latex]B[\/latex] is [latex]\\frac{2}{n}[\/latex]. The expected number of top random insertions before this happens is [latex]\\frac{n}{2}[\/latex]. The two cards below [latex]B[\/latex] are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
The expected number of random insertions to get [latex]B[\/latex] to the top is [latex]n+\\frac{n}{2}+\\frac{n}{3}+\\text{$\\cdots$ }+\\frac{n}{\\left(n - 1\\right)}[\/latex]. Then one more insertion puts [latex]B[\/latex] back in at random. Thus, the expected number of shuffles to randomize the deck is [latex]n\\left(1+\\frac{1}{2}+\\text{$\\cdots$ }+\\frac{1}{n}\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
51.\u00a0<\/strong>Suppose a scooter can travel [latex]100[\/latex] km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel [latex]100{H}_{N}[\/latex] km, where [latex]{H}_{N}=1+\\frac{1}{2}+\\text{$\\cdots$ }+\\frac{1}{N}[\/latex].<\/span><\/div>\n<\/div>\n<\/div>\n
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52.\u00a0<\/strong>Show that for the remainder estimate to apply on [latex]\\left[N,\\infty \\right)[\/latex] it is sufficient that [latex]f\\left(x\\right)[\/latex] be decreasing on [latex]\\left[N,\\infty \\right)[\/latex], but [latex]f[\/latex] need not be decreasing on [latex]\\left[1,\\infty \\right)[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
Set [latex]{b}_{n}={a}_{n+N}[\/latex] and [latex]g\\left(t\\right)=f\\left(t+N\\right)[\/latex] such that [latex]f[\/latex] is decreasing on [latex]\\left[t,\\infty \\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
53. [T]<\/strong> Use the remainder estimate and integration by parts to approximate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{e}^{n}}[\/latex] within an error smaller than [latex]0.0001[\/latex].<\/span><\/div>\n<\/div>\n<\/div>\n
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54.\u00a0<\/strong>Does [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n{\\left(\\text{ln}n\\right)}^{p}}[\/latex] converge if [latex]p[\/latex] is large enough? If so, for which [latex]p\\text{?}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
The series converges for [latex]p>1[\/latex] by integral test using change of variable.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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55. [T]<\/strong> Suppose a computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=1}^{N}\\frac{1}{n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/div>\n<\/div>\n<\/div>\n
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56. [T]<\/strong> A fast computer can sum one million terms per second of the divergent series [latex]\\displaystyle\\sum _{n=2}^{N}\\frac{1}{n\\text{ln}n}[\/latex]. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed [latex]100[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]N={e}^{{e}^{100}}\\approx {e}^{{10}^{43}}[\/latex] terms are needed.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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Candela Citations<\/h3>\n\t\t\t\t\t
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