{"id":114,"date":"2021-03-25T02:21:06","date_gmt":"2021-03-25T02:21:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/power-series-and-functions-2\/"},"modified":"2021-11-17T23:43:04","modified_gmt":"2021-11-17T23:43:04","slug":"power-series-and-functions-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/power-series-and-functions-2\/","title":{"raw":"Problem Set: Power Series and Functions","rendered":"Problem Set: Power Series and Functions"},"content":{"raw":"

In the following exercises, state whether each statement is true, or give an example to show that it is false.<\/p>\r\n\r\n

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1.\u00a0<\/strong>If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges, then [latex]{a}_{n}{x}^{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"506099\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"506099\"]True. If a series converges then its terms tend to zero.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges at [latex]x=0[\/latex] for any real numbers [latex]{a}_{n}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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3.\u00a0<\/strong>Given any sequence [latex]{a}_{n}[\/latex], there is always some [latex]R>0[\/latex], possibly very small, such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(\\text{-}R,R\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"478313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478313\"]False. It would imply that [latex]{a}_{n}{x}^{n}\\to 0[\/latex] for [latex]|x|<R[\/latex]. If [latex]{a}_{n}={n}^{n}[\/latex], then [latex]{a}_{n}{x}^{n}={\\left(nx\\right)}^{n}[\/latex] does not tend to zero for any [latex]x\\ne 0[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

4.\u00a0<\/strong>If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] has radius of convergence [latex]R>0[\/latex] and if [latex]|{b}_{n}|\\le |{a}_{n}|[\/latex] for all <\/span>n<\/em>, then the radius of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}{x}^{n}[\/latex] is greater than or equal to <\/span>R<\/em>.<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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5.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(x - 3\\right)}^{n}[\/latex] converges at [latex]x=6[\/latex]. At which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at x<\/em>, then it converges at any point closer to c<\/em> than x<\/em>.<\/p>\r\n\r\n

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  1. [latex]x=1[\/latex]<\/li>\r\n \t
  2. [latex]x=2[\/latex]<\/li>\r\n \t
  3. [latex]x=3[\/latex]<\/li>\r\n \t
  4. [latex]x=0[\/latex]<\/li>\r\n \t
  5. [latex]x=5.99[\/latex]<\/li>\r\n \t
  6. [latex]x=0.000001[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    [reveal-answer q=\"944061\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"944061\"]It must converge on [latex]\\left(0,6\\right][\/latex] and hence at: a. [latex]x=1[\/latex]; b. [latex]x=2[\/latex]; c. [latex]x=3[\/latex]; d. [latex]x=0[\/latex]; e. [latex]x=5.99[\/latex]; and f. [latex]x=0.000001[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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    6. <\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(x+1\\right)}^{n}[\/latex] converges at [latex]x=-2[\/latex].\r\nAt which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at x<\/em>, then it converges at any point closer to c<\/em> than x<\/em>.<\/p>\r\n\r\n

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    1. [latex]x=2[\/latex]<\/li>\r\n \t
    2. [latex]x=-1[\/latex]<\/li>\r\n \t
    3. [latex]x=-3[\/latex]<\/li>\r\n \t
    4. [latex]x=0[\/latex]<\/li>\r\n \t
    5. [latex]x=0.99[\/latex]<\/li>\r\n \t
    6. [latex]x=0.000001[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      In the following exercises, suppose that [latex]|\\frac{{a}_{n+1}}{{a}_{n}}|\\to 1[\/latex] as [latex]n\\to \\infty [\/latex]. Find the radius of convergence for each series.<\/p>\r\n\r\n

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      7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{2}^{n}{x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"746544\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746544\"][latex]|\\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\\frac{{a}_{n+1}}{{a}_{n}}|\\to 2|x|[\/latex] so [latex]R=\\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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      8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{x}^{n}}{{2}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{\\pi }^{n}{x}^{n}}{{e}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"143208\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"143208\"][latex]|\\frac{{a}_{n+1}{\\left(\\frac{\\pi }{e}\\right)}^{n+1}{x}^{n+1}}{{a}_{n}{\\left(\\frac{\\pi }{e}\\right)}^{n}{x}^{n}}|=\\frac{\\pi |x|}{e}|\\frac{{a}_{n+1}}{{a}_{n}}|\\to \\frac{\\pi |x|}{e}[\/latex] so [latex]R=\\frac{e}{\\pi }[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{\\left(-1\\right)}^{n}{x}^{n}}{{10}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(-1\\right)}^{n}{x}^{2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"944521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"944521\"][latex]|\\frac{{a}_{n+1}{\\left(-1\\right)}^{n+1}{x}^{2n+2}}{{a}_{n}{\\left(-1\\right)}^{n}{x}^{2n}}|=|{x}^{2}||\\frac{{a}_{n+1}}{{a}_{n}}|\\to |{x}^{2}|[\/latex] so [latex]R=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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      12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(-4\\right)}^{n}{x}^{2n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

      In the following exercises, find the radius of convergence R<\/em> and interval of convergence for [latex]\\displaystyle\\sum {a}_{n}{x}^{n}[\/latex] with the given coefficients [latex]{a}_{n}[\/latex].<\/p>\r\n\r\n

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

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      [reveal-answer q=\"586421\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"586421\"][latex]{a}_{n}=\\frac{{2}^{n}}{n}[\/latex] so [latex]\\frac{{a}_{n+1}x}{{a}_{n}}\\to 2x[\/latex]. so [latex]R=\\frac{1}{2}[\/latex]. When [latex]x=\\frac{1}{2}[\/latex] the series is harmonic and diverges. When [latex]x=-\\frac{1}{2}[\/latex] the series is alternating harmonic and converges. The interval of convergence is [latex]I=\\left[-\\frac{1}{2},\\frac{1}{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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      14.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\sqrt{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      15.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n{x}^{n}}{{2}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"54191\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"54191\"][latex]{a}_{n}=\\frac{n}{{2}^{n}}[\/latex] so [latex]\\frac{{a}_{n+1}x}{{a}_{n}}\\to \\frac{x}{2}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm2[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\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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      16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n{x}^{n}}{{e}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}{x}^{n}}{{2}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"182832\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"182832\"][latex]{a}_{n}=\\frac{{n}^{2}}{{2}^{n}}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\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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      18. <\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{k}^{e}{x}^{k}}{{e}^{k}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{\\pi }^{k}{x}^{k}}{{k}^{\\pi }}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"947912\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947912\"][latex]{a}_{k}=\\frac{{\\pi }^{k}}{{k}^{\\pi }}[\/latex] so [latex]R=\\frac{1}{\\pi }[\/latex]. When [latex]x=\\pm\\frac{1}{\\pi }[\/latex] the series is an absolutely convergent p-series. The interval of convergence is [latex]I=\\left[-\\frac{1}{\\pi },\\frac{1}{\\pi }\\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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      20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{10}^{n}{x}^{n}}{n\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"172843\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"172843\"][latex]{a}_{n}=\\frac{{10}^{n}}{n\\text{!}},\\frac{{a}_{n+1}x}{{a}_{n}}=\\frac{10x}{n+1}\\to 0<1[\/latex] so the series converges for all x by the ratio test and [latex]I=\\left(\\text{-}\\infty ,\\infty \\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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      22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\text{ln}\\left(2n\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

      In the following exercises, find the radius of convergence of each series.<\/p>\r\n\r\n

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

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      [reveal-answer q=\"222741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"222741\"][latex]{a}_{k}=\\frac{{\\left(k\\text{!}\\right)}^{2}}{\\left(2k\\right)\\text{!}}[\/latex] so [latex]\\frac{{a}_{k+1}}{{a}_{k}}=\\frac{{\\left(k+1\\right)}^{2}}{\\left(2k+2\\right)\\left(2k+1\\right)}\\to \\frac{1}{4}[\/latex] so [latex]R=4[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\left(2n\\right)\\text{!}{x}^{n}}{{n}^{2n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{k\\text{!}}{1\\cdot 3\\cdot 5\\text{$\\cdots$ }\\left(2k - 1\\right)}{x}^{k}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"92721\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"92721\"][latex]{a}_{k}=\\frac{k\\text{!}}{1\\cdot 3\\cdot 5\\cdots\\left(2k - 1\\right)}[\/latex] so [latex]\\frac{{a}_{k+1}}{{a}_{k}}=\\frac{k+1}{2k+1}\\to \\frac{1}{2}[\/latex] so [latex]R=2[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{2\\cdot 4\\cdot 6\\text{$\\cdots$ }2k}{\\left(2k\\right)\\text{!}}{x}^{k}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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      27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{\\left(\\begin{array}{c}2n\\\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right)=\\frac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"517268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"517268\"][latex]{a}_{n}=\\frac{1}{\\left(\\begin{array}{c}2n\\\\ n\\end{array}\\right)}[\/latex] so [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2n+2\\right)\\text{!}}\\frac{2n\\text{!}}{{\\left(n\\text{!}\\right)}^{2}}=\\frac{{\\left(n+1\\right)}^{2}}{\\left(2n+2\\right)\\left(2n+1\\right)}\\to \\frac{1}{4}[\/latex] so [latex]R=4[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\sin}^{2}n{x}^{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

      In the following exercises, use the ratio test to determine the radius of convergence of each series.<\/p>\r\n\r\n

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

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      [reveal-answer q=\"557244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"557244\"][latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{{\\left(n+1\\right)}^{3}}{\\left(3n+3\\right)\\left(3n+2\\right)\\left(3n+1\\right)}\\to \\frac{1}{27}[\/latex] so [latex]R=27[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{3n}{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}{x}^{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      31.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n\\text{!}}{{n}^{n}}{x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"611107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"611107\"][latex]{a}_{n}=\\frac{n\\text{!}}{{n}^{n}}[\/latex] so [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{\\left(n+1\\right)\\text{!}}{n\\text{!}}\\frac{{n}^{n}}{{\\left(n+1\\right)}^{n+1}}={\\left(\\frac{n}{n+1}\\right)}^{n}\\to \\frac{1}{e}[\/latex] so [latex]R=e[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

      In the following exercises, given that [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] with convergence in [latex]\\left(-1,1\\right)[\/latex], find the power series for each function with the given center a<\/em>, and identify its interval of convergence.<\/p>\r\n\r\n

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      33.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x};a=1[\/latex] (Hint:<\/em> [latex]\\frac{1}{x}=\\frac{1}{1-\\left(1-x\\right)}[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"533692\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"533692\"][latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex] on [latex]I=\\left(0,2\\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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      34.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1-{x}^{2}};a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      35. <\/strong>[latex]f\\left(x\\right)=\\frac{x}{1-{x}^{2}};a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"806284\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"806284\"][latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n+1}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      36.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1+{x}^{2}};a=0[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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      37.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{1+{x}^{2}};a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"486627\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"486627\"][latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n+2}[\/latex] on [latex]I=\\left(-1,1\\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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      38.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{2-x};a=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      39.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1 - 2x};a=0[\/latex].<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"58148\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"58148\"][latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex] on [latex]\\left(-\\frac{1}{2},\\frac{1}{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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      40.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1 - 4{x}^{2}};a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      41.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{1 - 4{x}^{2}};a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"496831\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"496831\"][latex]\\displaystyle\\sum _{n=0}^{\\infty }{4}^{n}{x}^{2n+2}[\/latex] on [latex]\\left(-\\frac{1}{2},\\frac{1}{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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      42.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{5 - 4x+{x}^{2}};a=2[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

      Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.<\/p>\r\n\r\n

      \r\n\r\n43.\u00a0<\/strong>Explain why, if [latex]{|{a}_{n}|}^{\\frac{1}{n}}\\to r>0[\/latex], then [latex]{|{a}_{n}{x}^{n}|}^{\\frac{1}{n}}\\to |x|r<1[\/latex] whenever [latex]|x|<\\frac{1}{r}[\/latex] and, therefore, the radius of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] is [latex]R=\\frac{1}{r}[\/latex].<\/span>\r\n
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      [reveal-answer q=\"470307\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470307\"][latex]{|{a}_{n}{x}^{n}|}^{\\frac{1}{n}}={|{a}_{n}|}^{\\frac{1}{n}}|x|\\to |x|r[\/latex] as [latex]n\\to \\infty [\/latex] and [latex]|x|r<1[\/latex] when [latex]|x|<\\frac{1}{r}[\/latex]. Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges when [latex]|x|<\\frac{1}{r}[\/latex] by the nth root test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

      \r\n

      [reveal-answer q=\"387237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387237\"][latex]{a}_{k}={\\left(\\frac{k - 1}{2k+3}\\right)}^{k}[\/latex] so [latex]{\\left({a}_{k}\\right)}^{\\frac{1}{k}}\\to \\frac{1}{2}<1[\/latex] so [latex]R=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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      46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{\\left(\\frac{2{k}^{2}-1}{{k}^{2}+3}\\right)}^{k}{x}^{k}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      47.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={\\left({n}^{\\frac{1}{n}}-1\\right)}^{n}{x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"115619\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"115619\"][latex]{a}_{n}={\\left({n}^{\\frac{1}{n}}-1\\right)}^{n}[\/latex] so [latex]{\\left({a}_{n}\\right)}^{\\frac{1}{n}}\\to 0[\/latex] so [latex]R=\\infty [\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      48.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] such that [latex]{a}_{n}=0[\/latex] if n<\/em> is odd. Explain why [latex]p\\left(x\\right)=-p\\left(\\text{-}x\\right)[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      49.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] such that [latex]{a}_{n}=0[\/latex] if n<\/em> is even. Explain why [latex]p\\left(x\\right)=p\\left(\\text{-}x\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"738369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"738369\"]We can rewrite [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n+1}{x}^{2n+1}[\/latex] and [latex]p\\left(x\\right)=p\\left(\\text{-}x\\right)[\/latex] since [latex]{x}^{2n+1}=\\text{-}{\\left(\\text{-}x\\right)}^{2n+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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      50.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(-1,1\\right][\/latex].\r\nFind the interval of convergence of [latex]p\\left(Ax\\right)[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      51.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(-1,1\\right][\/latex]. Find the interval of convergence of [latex]p\\left(2x - 1\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"881211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"881211\"]If [latex]x\\in \\left[0,1\\right][\/latex], then [latex]y=2x - 1\\in \\left[-1,1\\right][\/latex] so [latex]p\\left(2x - 1\\right)=p\\left(y\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{y}^{n}[\/latex] converges.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      In the following exercises, suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] satisfies [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n+1}}{{a}_{n}}=1[\/latex] where [latex]{a}_{n}\\ge 0[\/latex] for each n<\/em>. State whether each series converges on the full interval [latex]\\left(-1,1\\right)[\/latex], or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.<\/p>\r\n\r\n

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      52.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{2n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      53.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n}{x}^{2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"116242\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"116242\"]Converges on [latex]\\left(-1,1\\right)[\/latex] by the ratio test[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      54.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n}{x}^{n}\\left(Hint\\text{:}x=\\pm \\sqrt{{x}^{2}}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      55.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{{n}^{2}}{x}^{{n}^{2}}[\/latex] (Hint:<\/em> Let [latex]{b}_{k}={a}_{k}[\/latex] if [latex]k={n}^{2}[\/latex] for some n<\/em>, otherwise [latex]{b}_{k}=0.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"986909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"986909\"]Consider the series [latex]\\displaystyle\\sum {b}_{k}{x}^{k}[\/latex] where [latex]{b}_{k}={a}_{k}[\/latex] if [latex]k={n}^{2}[\/latex] and [latex]{b}_{k}=0[\/latex] otherwise. Then [latex]{b}_{k}\\le {a}_{k}[\/latex] and so the series converges on [latex]\\left(-1,1\\right)[\/latex] by 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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      56.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)[\/latex] is a polynomial of degree N<\/em>. Find the radius and interval of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }p\\left(n\\right){x}^{n}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      57. [T]<\/strong> Plot the graphs of [latex]\\frac{1}{1-x}[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{x}^{n}[\/latex] for [latex]n=10,20,30[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the approximation of [latex]\\frac{1}{1-x}[\/latex] by [latex]{S}_{N}[\/latex] near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/p>\r\n\r\n<\/div>\r\n

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      [reveal-answer q=\"411465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"411465\"]\"This<\/p>\r\nThe approximation is more accurate near [latex]x=-1[\/latex]. The partial sums follow [latex]\\frac{1}{1-x}[\/latex] more closely as N increases but are never accurate near [latex]x=1[\/latex] since the series diverges there.<\/span>[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      58. [T]<\/strong> Plot the graphs of [latex]\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\frac{{x}^{n}}{n}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      59. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{n}=\\displaystyle\\sum _{n=1}^{N}\\frac{{x}^{n}}{{n}^{2}}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"573860\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"573860\"]\"This<\/p>\r\nThe approximation appears to stabilize quickly near both [latex]x=\\pm 1[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      60. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\sin{n}{x}^{n}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      61. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex] for [latex]n=3,5,10[\/latex] on the interval [latex]\\left[-2\\pi ,2\\pi \\right][\/latex]. Comment on how these plots approximate [latex]\\sin{x}[\/latex] as N<\/em> increases.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      [reveal-answer q=\"817387\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"817387\"]\"This<\/p>\r\nThe polynomial curves have roots close to those of [latex]\\sin{x}[\/latex] up to their degree and then the polynomials diverge from [latex]\\sin{x}[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      62. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] for [latex]n=3,5,10[\/latex] on the interval [latex]\\left[-2\\pi ,2\\pi \\right][\/latex]. Comment on how these plots approximate [latex]\\cos{x}[\/latex] as N<\/em> increases.<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

      In the following exercises, state whether each statement is true, or give an example to show that it is false.<\/p>\n

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      1.\u00a0<\/strong>If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges, then [latex]{a}_{n}{x}^{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\n<\/div>\n

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      Show Solution<\/span><\/p>\n
      True. If a series converges then its terms tend to zero.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      2.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges at [latex]x=0[\/latex] for any real numbers [latex]{a}_{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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      3.\u00a0<\/strong>Given any sequence [latex]{a}_{n}[\/latex], there is always some [latex]R>0[\/latex], possibly very small, such that [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(\\text{-}R,R\\right)[\/latex].<\/p>\n<\/div>\n

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      Show Solution<\/span><\/p>\n
      False. It would imply that [latex]{a}_{n}{x}^{n}\\to 0[\/latex] for [latex]|x|\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      4.\u00a0<\/strong>If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] has radius of convergence [latex]R>0[\/latex] and if [latex]|{b}_{n}|\\le |{a}_{n}|[\/latex] for all <\/span>n<\/em>, then the radius of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}{x}^{n}[\/latex] is greater than or equal to <\/span>R<\/em>.<\/span><\/div>\n<\/div>\n<\/div>\n
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      5.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(x - 3\\right)}^{n}[\/latex] converges at [latex]x=6[\/latex]. At which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at x<\/em>, then it converges at any point closer to c<\/em> than x<\/em>.<\/p>\n

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      1. [latex]x=1[\/latex]<\/li>\n
      2. [latex]x=2[\/latex]<\/li>\n
      3. [latex]x=3[\/latex]<\/li>\n
      4. [latex]x=0[\/latex]<\/li>\n
      5. [latex]x=5.99[\/latex]<\/li>\n
      6. [latex]x=0.000001[\/latex]<\/li>\n<\/ol>\n<\/div>\n
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        Show Solution<\/span><\/p>\n
        It must converge on [latex]\\left(0,6\\right][\/latex] and hence at: a. [latex]x=1[\/latex]; b. [latex]x=2[\/latex]; c. [latex]x=3[\/latex]; d. [latex]x=0[\/latex]; e. [latex]x=5.99[\/latex]; and f. [latex]x=0.000001[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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        6. <\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(x+1\\right)}^{n}[\/latex] converges at [latex]x=-2[\/latex].
        \nAt which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at x<\/em>, then it converges at any point closer to c<\/em> than x<\/em>.<\/p>\n

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        1. [latex]x=2[\/latex]<\/li>\n
        2. [latex]x=-1[\/latex]<\/li>\n
        3. [latex]x=-3[\/latex]<\/li>\n
        4. [latex]x=0[\/latex]<\/li>\n
        5. [latex]x=0.99[\/latex]<\/li>\n
        6. [latex]x=0.000001[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, suppose that [latex]|\\frac{{a}_{n+1}}{{a}_{n}}|\\to 1[\/latex] as [latex]n\\to \\infty[\/latex]. Find the radius of convergence for each series.<\/p>\n

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          7.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{2}^{n}{x}^{n}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]|\\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\\frac{{a}_{n+1}}{{a}_{n}}|\\to 2|x|[\/latex] so [latex]R=\\frac{1}{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{x}^{n}}{{2}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{\\pi }^{n}{x}^{n}}{{e}^{n}}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]|\\frac{{a}_{n+1}{\\left(\\frac{\\pi }{e}\\right)}^{n+1}{x}^{n+1}}{{a}_{n}{\\left(\\frac{\\pi }{e}\\right)}^{n}{x}^{n}}|=\\frac{\\pi |x|}{e}|\\frac{{a}_{n+1}}{{a}_{n}}|\\to \\frac{\\pi |x|}{e}[\/latex] so [latex]R=\\frac{e}{\\pi }[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{a}_{n}{\\left(-1\\right)}^{n}{x}^{n}}{{10}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          11.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(-1\\right)}^{n}{x}^{2n}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]|\\frac{{a}_{n+1}{\\left(-1\\right)}^{n+1}{x}^{2n+2}}{{a}_{n}{\\left(-1\\right)}^{n}{x}^{2n}}|=|{x}^{2}||\\frac{{a}_{n+1}}{{a}_{n}}|\\to |{x}^{2}|[\/latex] so [latex]R=1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          12.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{\\left(-4\\right)}^{n}{x}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, find the radius of convergence R<\/em> and interval of convergence for [latex]\\displaystyle\\sum {a}_{n}{x}^{n}[\/latex] with the given coefficients [latex]{a}_{n}[\/latex].<\/p>\n

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

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          [latex]{a}_{n}=\\frac{{2}^{n}}{n}[\/latex] so [latex]\\frac{{a}_{n+1}x}{{a}_{n}}\\to 2x[\/latex]. so [latex]R=\\frac{1}{2}[\/latex]. When [latex]x=\\frac{1}{2}[\/latex] the series is harmonic and diverges. When [latex]x=-\\frac{1}{2}[\/latex] the series is alternating harmonic and converges. The interval of convergence is [latex]I=\\left[-\\frac{1}{2},\\frac{1}{2}\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          14.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\sqrt{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          15.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n{x}^{n}}{{2}^{n}}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}=\\frac{n}{{2}^{n}}[\/latex] so [latex]\\frac{{a}_{n+1}x}{{a}_{n}}\\to \\frac{x}{2}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm2[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          16.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n{x}^{n}}{{e}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          17.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}{x}^{n}}{{2}^{n}}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}=\\frac{{n}^{2}}{{2}^{n}}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          18. <\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{k}^{e}{x}^{k}}{{e}^{k}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          19.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{\\pi }^{k}{x}^{k}}{{k}^{\\pi }}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{k}=\\frac{{\\pi }^{k}}{{k}^{\\pi }}[\/latex] so [latex]R=\\frac{1}{\\pi }[\/latex]. When [latex]x=\\pm\\frac{1}{\\pi }[\/latex] the series is an absolutely convergent p-series. The interval of convergence is [latex]I=\\left[-\\frac{1}{\\pi },\\frac{1}{\\pi }\\right][\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          20.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          21.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{10}^{n}{x}^{n}}{n\\text{!}}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}=\\frac{{10}^{n}}{n\\text{!}},\\frac{{a}_{n+1}x}{{a}_{n}}=\\frac{10x}{n+1}\\to 0<1[\/latex] so the series converges for all x by the ratio test and [latex]I=\\left(\\text{-}\\infty ,\\infty \\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          22.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\text{ln}\\left(2n\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, find the radius of convergence of each series.<\/p>\n

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

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          [latex]{a}_{k}=\\frac{{\\left(k\\text{!}\\right)}^{2}}{\\left(2k\\right)\\text{!}}[\/latex] so [latex]\\frac{{a}_{k+1}}{{a}_{k}}=\\frac{{\\left(k+1\\right)}^{2}}{\\left(2k+2\\right)\\left(2k+1\\right)}\\to \\frac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          24.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\left(2n\\right)\\text{!}{x}^{n}}{{n}^{2n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          25.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{k\\text{!}}{1\\cdot 3\\cdot 5\\text{$\\cdots$ }\\left(2k - 1\\right)}{x}^{k}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{k}=\\frac{k\\text{!}}{1\\cdot 3\\cdot 5\\cdots\\left(2k - 1\\right)}[\/latex] so [latex]\\frac{{a}_{k+1}}{{a}_{k}}=\\frac{k+1}{2k+1}\\to \\frac{1}{2}[\/latex] so [latex]R=2[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          26.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{2\\cdot 4\\cdot 6\\text{$\\cdots$ }2k}{\\left(2k\\right)\\text{!}}{x}^{k}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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          27.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{\\left(\\begin{array}{c}2n\\\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right)=\\frac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}=\\frac{1}{\\left(\\begin{array}{c}2n\\\\ n\\end{array}\\right)}[\/latex] so [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2n+2\\right)\\text{!}}\\frac{2n\\text{!}}{{\\left(n\\text{!}\\right)}^{2}}=\\frac{{\\left(n+1\\right)}^{2}}{\\left(2n+2\\right)\\left(2n+1\\right)}\\to \\frac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          28.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\sin}^{2}n{x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, use the ratio test to determine the radius of convergence of each series.<\/p>\n

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

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          [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{{\\left(n+1\\right)}^{3}}{\\left(3n+3\\right)\\left(3n+2\\right)\\left(3n+1\\right)}\\to \\frac{1}{27}[\/latex] so [latex]R=27[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          30.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{3n}{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}{x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          31.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n\\text{!}}{{n}^{n}}{x}^{n}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}=\\frac{n\\text{!}}{{n}^{n}}[\/latex] so [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{\\left(n+1\\right)\\text{!}}{n\\text{!}}\\frac{{n}^{n}}{{\\left(n+1\\right)}^{n+1}}={\\left(\\frac{n}{n+1}\\right)}^{n}\\to \\frac{1}{e}[\/latex] so [latex]R=e[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          32.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\left(2n\\right)\\text{!}}{{n}^{2n}}{x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, given that [latex]\\frac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] with convergence in [latex]\\left(-1,1\\right)[\/latex], find the power series for each function with the given center a<\/em>, and identify its interval of convergence.<\/p>\n

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          33.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x};a=1[\/latex] (Hint:<\/em> [latex]\\frac{1}{x}=\\frac{1}{1-\\left(1-x\\right)}[\/latex])<\/p>\n<\/div>\n

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          [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex] on [latex]I=\\left(0,2\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          34.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1-{x}^{2}};a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          35. <\/strong>[latex]f\\left(x\\right)=\\frac{x}{1-{x}^{2}};a=0[\/latex]<\/p>\n<\/div>\n

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          [latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n+1}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          36.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1+{x}^{2}};a=0[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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          37.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{1+{x}^{2}};a=0[\/latex]<\/p>\n<\/div>\n

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          [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n+2}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          38.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{2-x};a=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          39.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1 - 2x};a=0[\/latex].<\/p>\n<\/div>\n

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          [latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex] on [latex]\\left(-\\frac{1}{2},\\frac{1}{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          40.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1 - 4{x}^{2}};a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          41.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{1 - 4{x}^{2}};a=0[\/latex]<\/p>\n<\/div>\n

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          [latex]\\displaystyle\\sum _{n=0}^{\\infty }{4}^{n}{x}^{2n+2}[\/latex] on [latex]\\left(-\\frac{1}{2},\\frac{1}{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          42.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{{x}^{2}}{5 - 4x+{x}^{2}};a=2[\/latex]<\/div>\n<\/div>\n<\/div>\n

          Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.<\/p>\n

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          43.\u00a0<\/strong>Explain why, if [latex]{|{a}_{n}|}^{\\frac{1}{n}}\\to r>0[\/latex], then [latex]{|{a}_{n}{x}^{n}|}^{\\frac{1}{n}}\\to |x|r<1[\/latex] whenever [latex]|x|<\\frac{1}{r}[\/latex] and, therefore, the radius of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] is [latex]R=\\frac{1}{r}[\/latex].<\/span><\/p>\n

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          [latex]{|{a}_{n}{x}^{n}|}^{\\frac{1}{n}}={|{a}_{n}|}^{\\frac{1}{n}}|x|\\to |x|r[\/latex] as [latex]n\\to \\infty[\/latex] and [latex]|x|r<1[\/latex] when [latex]|x|<\\frac{1}{r}[\/latex]. Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges when [latex]|x|<\\frac{1}{r}[\/latex] by the nth root test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          44.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{{n}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          45.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{\\left(\\frac{k - 1}{2k+3}\\right)}^{k}{x}^{k}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{k}={\\left(\\frac{k - 1}{2k+3}\\right)}^{k}[\/latex] so [latex]{\\left({a}_{k}\\right)}^{\\frac{1}{k}}\\to \\frac{1}{2}<1[\/latex] so [latex]R=2[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{\\left(\\frac{2{k}^{2}-1}{{k}^{2}+3}\\right)}^{k}{x}^{k}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          47.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={\\left({n}^{\\frac{1}{n}}-1\\right)}^{n}{x}^{n}[\/latex]<\/p>\n<\/div>\n

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          [latex]{a}_{n}={\\left({n}^{\\frac{1}{n}}-1\\right)}^{n}[\/latex] so [latex]{\\left({a}_{n}\\right)}^{\\frac{1}{n}}\\to 0[\/latex] so [latex]R=\\infty[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          48.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] such that [latex]{a}_{n}=0[\/latex] if n<\/em> is odd. Explain why [latex]p\\left(x\\right)=-p\\left(\\text{-}x\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n
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          49.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] such that [latex]{a}_{n}=0[\/latex] if n<\/em> is even. Explain why [latex]p\\left(x\\right)=p\\left(\\text{-}x\\right)[\/latex].<\/p>\n<\/div>\n

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          We can rewrite [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n+1}{x}^{2n+1}[\/latex] and [latex]p\\left(x\\right)=p\\left(\\text{-}x\\right)[\/latex] since [latex]{x}^{2n+1}=\\text{-}{\\left(\\text{-}x\\right)}^{2n+1}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          50.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(-1,1\\right][\/latex].
          \nFind the interval of convergence of [latex]p\\left(Ax\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n
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          51.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges on [latex]\\left(-1,1\\right][\/latex]. Find the interval of convergence of [latex]p\\left(2x - 1\\right)[\/latex].<\/p>\n<\/div>\n

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          If [latex]x\\in \\left[0,1\\right][\/latex], then [latex]y=2x - 1\\in \\left[-1,1\\right][\/latex] so [latex]p\\left(2x - 1\\right)=p\\left(y\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{y}^{n}[\/latex] converges.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          In the following exercises, suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] satisfies [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n+1}}{{a}_{n}}=1[\/latex] where [latex]{a}_{n}\\ge 0[\/latex] for each n<\/em>. State whether each series converges on the full interval [latex]\\left(-1,1\\right)[\/latex], or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.<\/p>\n

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          52.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          53.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n}{x}^{2n}[\/latex]<\/p>\n<\/div>\n

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          Converges on [latex]\\left(-1,1\\right)[\/latex] by the ratio test<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          54.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{2n}{x}^{n}\\left(Hint\\text{:}x=\\pm \\sqrt{{x}^{2}}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          55.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{{n}^{2}}{x}^{{n}^{2}}[\/latex] (Hint:<\/em> Let [latex]{b}_{k}={a}_{k}[\/latex] if [latex]k={n}^{2}[\/latex] for some n<\/em>, otherwise [latex]{b}_{k}=0.[\/latex])<\/p>\n<\/div>\n

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          Consider the series [latex]\\displaystyle\\sum {b}_{k}{x}^{k}[\/latex] where [latex]{b}_{k}={a}_{k}[\/latex] if [latex]k={n}^{2}[\/latex] and [latex]{b}_{k}=0[\/latex] otherwise. Then [latex]{b}_{k}\\le {a}_{k}[\/latex] and so the series converges on [latex]\\left(-1,1\\right)[\/latex] by the comparison test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          56.\u00a0<\/strong>Suppose that [latex]p\\left(x\\right)[\/latex] is a polynomial of degree N<\/em>. Find the radius and interval of convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }p\\left(n\\right){x}^{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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          57. [T]<\/strong> Plot the graphs of [latex]\\frac{1}{1-x}[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{x}^{n}[\/latex] for [latex]n=10,20,30[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the approximation of [latex]\\frac{1}{1-x}[\/latex] by [latex]{S}_{N}[\/latex] near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/p>\n<\/div>\n

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          \"This<\/p>\n

          The approximation is more accurate near [latex]x=-1[\/latex]. The partial sums follow [latex]\\frac{1}{1-x}[\/latex] more closely as N increases but are never accurate near [latex]x=1[\/latex] since the series diverges there.<\/span><\/div>\n<\/div>\n

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

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          58. [T]<\/strong> Plot the graphs of [latex]\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\frac{{x}^{n}}{n}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/div>\n<\/div>\n<\/div>\n
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          59. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{n}=\\displaystyle\\sum _{n=1}^{N}\\frac{{x}^{n}}{{n}^{2}}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          \"This<\/p>\n

          The approximation appears to stabilize quickly near both [latex]x=\\pm 1[\/latex].<\/div>\n<\/div>\n

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

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          60. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=1}^{N}\\sin{n}{x}^{n}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N<\/em> increases.<\/div>\n<\/div>\n<\/div>\n
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          61. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex] for [latex]n=3,5,10[\/latex] on the interval [latex]\\left[-2\\pi ,2\\pi \\right][\/latex]. Comment on how these plots approximate [latex]\\sin{x}[\/latex] as N<\/em> increases.<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          \"This<\/p>\n

          The polynomial curves have roots close to those of [latex]\\sin{x}[\/latex] up to their degree and then the polynomials diverge from [latex]\\sin{x}[\/latex].<\/div>\n<\/div>\n

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

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          62. [T]<\/strong> Plot the graphs of the partial sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] for [latex]n=3,5,10[\/latex] on the interval [latex]\\left[-2\\pi ,2\\pi \\right][\/latex]. Comment on how these plots approximate [latex]\\cos{x}[\/latex] as N<\/em> increases.<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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