{"id":116,"date":"2021-03-25T02:21:07","date_gmt":"2021-03-25T02:21:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/properties-of-power-series-2\/"},"modified":"2021-11-17T23:43:20","modified_gmt":"2021-11-17T23:43:20","slug":"properties-of-power-series-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/properties-of-power-series-2\/","title":{"raw":"Problem Set: Properties of Power Series","rendered":"Problem Set: Properties of Power Series"},"content":{"raw":"
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1.\u00a0<\/strong>If [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] and [latex]g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{n\\text{!}}[\/latex], find the power series of [latex]\\frac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)[\/latex] and of [latex]\\frac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"907296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"907296\"][latex]\\frac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\frac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>If [latex]C\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]S\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], find the power series of [latex]C\\left(x\\right)+S\\left(x\\right)[\/latex] and of [latex]C\\left(x\\right)-S\\left(x\\right)[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, use partial fractions to find the power series of each function.<\/p>\r\n\r\n

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3.\u00a0<\/strong>[latex]\\frac{4}{\\left(x - 3\\right)\\left(x+1\\right)}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"410030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410030\"][latex]\\frac{4}{\\left(x - 3\\right)\\left(x+1\\right)}=\\frac{1}{x - 3}-\\frac{1}{x+1}=-\\frac{1}{3\\left(1-\\frac{x}{3}\\right)}-\\frac{1}{1-\\left(\\text{-}x\\right)}=-\\frac{1}{3}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{x}{3}\\right)}^{n}-\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left({\\left(-1\\right)}^{n+1}-\\frac{1}{{3}^{n+1}}\\right){x}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]\\frac{3}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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5.\u00a0<\/strong>[latex]\\frac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"838253\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"838253\"][latex]\\frac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}=\\frac{1}{{x}^{2}-1}-\\frac{1}{4}\\frac{1}{1+{\\left(\\frac{x}{2}\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n}-\\frac{1}{4}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(\\frac{x}{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\left(-1\\right)+{\\left(-1\\right)}^{n+1}\\frac{1}{{2n+2}}\\right){x}^{2n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]\\frac{30}{\\left({x}^{2}+1\\right)\\left({x}^{2}-9\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, express each series as a rational function.<\/p>\r\n\r\n

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

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[reveal-answer q=\"713244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"713244\"][latex]\\frac{1}{x}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{{x}^{n}}=\\frac{1}{x}\\frac{1}{1-\\frac{1}{x}}=\\frac{1}{x - 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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8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{x}^{2n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{\\left(x - 3\\right)}^{2n - 1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"167769\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"167769\"][latex]\\frac{1}{x - 3}\\frac{1}{1-\\frac{1}{{\\left(x - 3\\right)}^{2}}}=\\frac{x - 3}{{\\left(x - 3\\right)}^{2}-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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10.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{{\\left(x - 3\\right)}^{2n - 1}}-\\frac{1}{{\\left(x - 2\\right)}^{2n - 1}}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

The following exercises explore applications of annuities<\/span>.<\/p>\r\n\r\n

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11.\u00a0<\/strong>Calculate the present values P<\/em> of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"222779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"222779\"][latex]P={P}_{1}+\\cdots+{P}_{20}[\/latex] where [latex]{P}_{k}=10,000\\frac{1}{{\\left(1+r\\right)}^{k}}[\/latex]. Then [latex]P=10,000\\displaystyle\\sum _{k=1}^{20}\\frac{1}{{\\left(1+r\\right)}^{k}}=10,000\\frac{1-{\\left(1+r\\right)}^{-20}}{r}[\/latex]. When [latex]r=0.03,P\\approx 10,000\\times 14.8775=148,775[\/latex]. When [latex]r=0.05,P\\approx 10,000\\times 12.4622=124,622[\/latex]. When [latex]r=0.07,P\\approx 105,940[\/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. <\/strong>Calculate the present values P<\/em> of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of [latex]r=0.03,r=0.05[\/latex] and [latex]r=0.07[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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13.\u00a0<\/strong>Calculate the annual payouts C<\/em> to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"608125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"608125\"]In general, [latex]P=\\frac{C\\left(1-{\\left(1+r\\right)}^{\\text{-}N}\\right)}{r}[\/latex] for N years of payouts, or [latex]C=\\frac{Pr}{1-{\\left(1+r\\right)}^{\\text{-}N}}[\/latex]. For [latex]N=20[\/latex] and [latex]P=100,000[\/latex], one has [latex]C=6721.57[\/latex] when [latex]r=0.03;C=8024.26[\/latex] when [latex]r=0.05[\/latex]; and [latex]C\\approx 9439.29[\/latex] when [latex]r=0.07[\/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>Calculate the annual payouts C<\/em> to be given perpetually on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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15.\u00a0<\/strong>Suppose that an annuity has a present value [latex]P=1\\text{million dollars}[\/latex]. What interest rate r<\/em> would allow for perpetual annual payouts of $50,000?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"978426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"978426\"]In general, [latex]P=\\frac{C}{r}[\/latex]. Thus, [latex]r=\\frac{C}{P}=5\\times \\frac{{10}^{4}}{{10}^{6}}=0.05[\/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>Suppose that an annuity has a present value [latex]P=10\\text{million dollars}\\text{.}[\/latex] What interest rate r<\/em> would allow for perpetual annual payouts of $100,000?<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.<\/p>\r\n\r\n

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17.\u00a0<\/strong>[latex]x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>3k<\/em><\/sup>, [latex]{x}^{3k - 1}[\/latex], and [latex]{x}^{3k - 2}.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"968221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"968221\"][latex]\\left(x+{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\frac{x+{x}^{2}-{x}^{3}}{1-{x}^{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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18.\u00a0<\/strong>[latex]x+{x}^{2}-{x}^{3}-{x}^{4}+{x}^{5}+{x}^{6}-{x}^{7}-{x}^{8}+\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>4k<\/em><\/sup>, [latex]{x}^{4k - 1}[\/latex], etc.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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19.\u00a0<\/strong>[latex]x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>3k<\/em><\/sup>, [latex]{x}^{3k - 1}[\/latex], and [latex]{x}^{3k - 2}.[\/latex])<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"649927\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"649927\"][latex]\\left(x-{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\frac{x-{x}^{2}-{x}^{3}}{1-{x}^{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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20.\u00a0<\/strong>[latex]\\frac{x}{2}+\\frac{{x}^{2}}{4}-\\frac{{x}^{3}}{8}+\\frac{{x}^{4}}{16}+\\frac{{x}^{5}}{32}-\\frac{{x}^{6}}{64}+\\cdots[\/latex] (Hint:<\/em> Group powers [latex]{\\left(\\frac{x}{2}\\right)}^{3k},{\\left(\\frac{x}{2}\\right)}^{3k - 1}[\/latex], and [latex]{\\left(\\frac{x}{2}\\right)}^{3k - 2}.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the power series of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] given f<\/em> and g<\/em> as defined.<\/p>\r\n\r\n

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21.\u00a0<\/strong>[latex]f\\left(x\\right)=2\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }n{x}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"900083\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"900083\"][latex]{a}_{n}=2,{b}_{n}=n[\/latex] so [latex]{c}_{n}=\\displaystyle\\sum _{k=0}^{n}{b}_{k}{a}_{n-k}=2\\displaystyle\\sum _{k=0}^{n}k=\\left(n\\right)\\left(n+1\\right)[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n\\left(n+1\\right){x}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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22.\u00a0<\/strong>[latex]f\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}{x}^{n}[\/latex]. Express the coefficients of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] in terms of [latex]{H}_{n}=\\displaystyle\\sum _{k=1}^{n}\\frac{1}{k}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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23.\u00a0<\/strong>[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{x}{2}\\right)}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"843784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843784\"][latex]{a}_{n}={b}_{n}={2}^{\\text{-}n}[\/latex] so [latex]{c}_{n}=\\displaystyle\\sum _{k=1}^{n}{b}_{k}{a}_{n-k}={2}^{\\text{-}n}\\displaystyle\\sum _{k=1}^{n}1=\\frac{n}{{2}^{n}}[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{\\left(\\frac{x}{2}\\right)}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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24.\u00a0<\/strong>[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{x}^{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, differentiate the given series expansion of f<\/em> term-by-term to obtain the corresponding series expansion for the derivative of f<\/em>.<\/p>\r\n\r\n

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

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[reveal-answer q=\"217050\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"217050\"]The derivative of [latex]f[\/latex] is [latex]-\\frac{1}{{\\left(1+x\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

In the following exercises, integrate the given series expansion of [latex]f[\/latex] term-by-term from zero to x<\/em> to obtain the corresponding series expansion for the indefinite integral of [latex]f[\/latex].<\/p>\r\n\r\n

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

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[reveal-answer q=\"534241\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"534241\"]The indefinite integral of [latex]f[\/latex] is [latex]\\frac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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

In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.<\/p>\r\n\r\n

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29.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\frac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"248998\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"248998\"][latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=\\frac{1}{1-x};{f}^{\\prime }\\left(\\frac{1}{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n - 1}}={\\frac{d}{dx}{\\left(1-x\\right)}^{-1}|}_{x=\\frac{1}{2}}={\\frac{1}{{\\left(1-x\\right)}^{2}}|}_{x=\\frac{1}{2}}=4[\/latex] so [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n}}=2[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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30.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{3}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\frac{1}{3}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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31.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n}}[\/latex] as [latex]f''\\left(\\frac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"536346\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"536346\"][latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=\\frac{1}{1-x};f\\text{''}\\left(\\frac{1}{2}\\right)=\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n - 2}}={\\frac{{d}^{2}}{d{x}^{2}}{\\left(1-x\\right)}^{-1}|}_{x=\\frac{1}{2}}={\\frac{2}{{\\left(1-x\\right)}^{3}}|}_{x=\\frac{1}{2}}=16[\/latex] so [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n}}=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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32.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n+1}[\/latex] as [latex]{\\displaystyle\\int }_{0}^{1}f\\left(t\\right)dt[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}=\\frac{1}{1+{x}^{2}}[\/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], use term-by-term differentiation or integration to find power series for each function centered at the given point.<\/p>\r\n\r\n

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

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[reveal-answer q=\"718559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"718559\"][latex]\\displaystyle\\int \\displaystyle\\sum {\\left(1-x\\right)}^{n}dx=\\displaystyle\\int \\displaystyle\\sum {\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}dx=\\displaystyle\\sum \\frac{{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n+1}}{n+1}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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34.\u00a0<\/strong>[latex]\\text{ln}\\left(1-x\\right)[\/latex] at [latex]x=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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35.\u00a0<\/strong>[latex]\\text{ln}\\left(1-{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"659108\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"659108\"][latex]\\text{-}{\\displaystyle\\int }_{t=0}^{{x}^{2}}\\frac{1}{1-t}dt=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\displaystyle\\int }_{0}^{{x}^{2}}{t}^{n}dx-\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2\\left(n+1\\right)}}{n+1}=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{2n}}{n}[\/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{2x}{{\\left(1-{x}^{2}\\right)}^{2}}[\/latex] at [latex]x=0[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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37.\u00a0<\/strong>[latex]f\\left(x\\right)={\\tan}^{-1}\\left({x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"723261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"723261\"][latex]{\\displaystyle\\int }_{0}^{{x}^{2}}\\frac{dt}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\displaystyle\\int }_{0}^{{x}^{2}}{t}^{2n}dt={\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n+1}}{2n+1}|}_{t=0}^{{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+2}}{2n+1}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

38.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}\\left(1+{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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39.\u00a0<\/strong>[latex]f\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\text{ln}tdt[\/latex] where [latex]\\text{ln}\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{\\left(x - 1\\right)}^{n}}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"220373\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"220373\"]Term-by-term integration gives [latex]{\\displaystyle\\int }_{0}^{x}\\text{ln}tdt=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{\\left(x - 1\\right)}^{n+1}}{n\\left(n+1\\right)}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right){\\left(x - 1\\right)}^{n+1}=\\left(x - 1\\right)\\text{ln}x+\\displaystyle\\sum _{n=2}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{\\left(x - 1\\right)}^{n}}{n}= x\\text{ln}x-x[\/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. [T]<\/strong> Evaluate the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex] at [latex]x=1[\/latex] to show that [latex]\\text{ln}\\left(2\\right)[\/latex] is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within 0.001, and find such an approximation.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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41. [T]<\/strong> Subtract the infinite series of [latex]\\text{ln}\\left(1-x\\right)[\/latex] from [latex]\\text{ln}\\left(1+x\\right)[\/latex] to get a power series for [latex]\\text{ln}\\left(\\frac{1+x}{1-x}\\right)[\/latex]. Evaluate at [latex]x=\\frac{1}{3}[\/latex]. What is the smallest N<\/em> such that the N<\/em>th partial sum of this series approximates [latex]\\text{ln}\\left(2\\right)[\/latex] with an error less than 0.001?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"782414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"782414\"]We have [latex]\\text{ln}\\left(1-x\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{n}[\/latex] so [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex]. Thus, [latex]\\text{ln}\\left(\\frac{1+x}{1-x}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\left(1+{\\left(-1\\right)}^{n - 1}\\right)\\frac{{x}^{n}}{n}=2\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{2n - 1}}{2n - 1}[\/latex]. When [latex]x=\\frac{1}{3}[\/latex] we obtain [latex]\\text{ln}\\left(2\\right)=2\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}[\/latex]. We have [latex]2\\displaystyle\\sum _{n=1}^{3}\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}=0.69300\\ldots[\/latex], while [latex]2\\displaystyle\\sum _{n=1}^{4}\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}=0.69313\\ldots[\/latex] and [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=4[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.<\/p>\r\n\r\n

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42.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=0}^{\\infty }\\left({x}^{k}-{x}^{2k+1}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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43.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{3k}}{6k}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"3825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3825\"][latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{k}}{k}=\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] so [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{3k}}{6k}=-\\frac{1}{6}\\text{ln}\\left(1-{x}^{3}\\right)[\/latex]. The radius of convergence is equal to 1 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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44.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{\\left(1+{x}^{2}\\right)}^{\\text{-}k}[\/latex] using [latex]y=\\frac{1}{1+{x}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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45.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{2}^{\\text{-}kx}[\/latex] using [latex]y={2}^{\\text{-}x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"53097\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"53097\"]If [latex]y={2}^{\\text{-}x}[\/latex], then [latex]\\displaystyle\\sum _{k=1}^{\\infty }{y}^{k}=\\frac{y}{1-y}=\\frac{{2}^{\\text{-}x}}{1-{2}^{\\text{-}x}}=\\frac{1}{{2}^{x}-1}[\/latex]. If [latex]{a}_{k}={2}^{\\text{-}kx}[\/latex], then [latex]\\frac{{a}_{k+1}}{{a}_{k}}={2}^{\\text{-}x}<1[\/latex] when [latex]x>0[\/latex]. So the series converges for all [latex]x>0[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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46.\u00a0<\/strong>Show that, up to powers x<\/em>3<\/sup> and y<\/em>3<\/sup>, [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] satisfies [latex]E\\left(x+y\\right)=E\\left(x\\right)E\\left(y\\right)[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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47.\u00a0<\/strong>Differentiate the series [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] term-by-term to show that [latex]E\\left(x\\right)[\/latex] is equal to its derivative.<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"186453\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"186453\"]Answers will vary.[\/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>Show that if [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a sum of even powers, that is, [latex]{a}_{n}=0[\/latex] if n<\/em> is odd, then [latex]F={\\displaystyle\\int }_{0}^{x}f\\left(t\\right)dt[\/latex] is a sum of odd powers, while if f<\/em> is a sum of odd powers, then F<\/em> is a sum of even powers.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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49. [T]<\/strong> Suppose that the coefficients an<\/sub><\/em> of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}_{n}=\\frac{{a}_{n - 1}}{n}+\\frac{{a}_{n - 2}}{n\\left(n - 1\\right)}[\/latex]. For [latex]{a}_{0}=0[\/latex] and [latex]{a}_{1}=1[\/latex], compute and plot the sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{a}_{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"683294\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"683294\"]<\/span>\"This<\/p>\r\nThe solid curve is [latex]{S}_{5}[\/latex]. The dashed curve is [latex]{S}_{2}[\/latex], dotted is [latex]{S}_{3}[\/latex], and dash-dotted is [latex]{S}_{4}[\/latex][\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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50. [T]<\/strong> Suppose that the coefficients an<\/sub><\/em> of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}_{n}=\\frac{{a}_{n - 1}}{\\sqrt{n}}-\\frac{{a}_{n - 2}}{\\sqrt{n\\left(n - 1\\right)}}[\/latex]. For [latex]{a}_{0}=1[\/latex] and [latex]{a}_{1}=0[\/latex], compute and plot the sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{a}_{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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51. [T]<\/strong> Given the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex], determine how many terms N<\/em> of the sum evaluated at [latex]x=-\\frac{1}{2}[\/latex] are needed to approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]\\frac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"982979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982979\"]When [latex]x=-\\frac{1}{2},\\text{-}\\text{ln}\\left(2\\right)=\\text{ln}\\left(\\frac{1}{2}\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=11}^{\\infty }\\frac{1}{n{2}^{n}}<\\displaystyle\\sum _{n=11}^{\\infty }\\frac{1}{{2}^{n}}=\\frac{1}{{2}^{10}}[\/latex], one has [latex]\\displaystyle\\sum _{n=1}^{10}\\frac{1}{n{2}^{n}}=0.69306\\ldots[\/latex] whereas [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=10[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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52. [T]<\/strong> Given the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{\\infty }{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex], use the alternating series test to determine how many terms N<\/em> of the sum evaluated at [latex]x=1[\/latex] are needed to approximate [latex]{\\tan}^{-1}\\left(1\\right)=\\frac{\\pi }{4}[\/latex] accurate to within [latex]\\frac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{k=0}^{N}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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53. [T]<\/strong> Recall that [latex]{\\tan}^{-1}\\left(\\frac{1}{\\sqrt{3}}\\right)=\\frac{\\pi }{6}[\/latex]. Assuming an exact value of [latex]\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex], estimate [latex]\\frac{\\pi }{6}[\/latex] by evaluating partial sums [latex]{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex] of the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{\\infty }{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex] at [latex]x=\\frac{1}{\\sqrt{3}}[\/latex]. What is the smallest number N<\/em> such that [latex]6{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex] approximates \u03c0<\/em> accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"614647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"614647\"][latex]6{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)=2\\sqrt{3}\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{1}{{3}^{n}\\left(2n+1\\right)}[\/latex]. One has [latex]\\pi -6{S}_{4}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00101\\ldots[\/latex] and [latex]\\pi -6{S}_{5}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00028\\ldots[\/latex] so [latex]N=5[\/latex] is the smallest partial sum with accuracy to within 0.001. Also, [latex]\\pi -6{S}_{7}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00002\\ldots[\/latex] while [latex]\\pi -6{S}_{8}\\left(\\frac{1}{\\sqrt{3}}\\right)=-0.000007\\ldots[\/latex] so [latex]N=8[\/latex] is the smallest N to give accuracy to within 0.00001.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

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1.\u00a0<\/strong>If [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] and [latex]g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{n\\text{!}}[\/latex], find the power series of [latex]\\frac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)[\/latex] and of [latex]\\frac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\frac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\frac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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2.\u00a0<\/strong>If [latex]C\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]S\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], find the power series of [latex]C\\left(x\\right)+S\\left(x\\right)[\/latex] and of [latex]C\\left(x\\right)-S\\left(x\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n

In the following exercises, use partial fractions to find the power series of each function.<\/p>\n

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3.\u00a0<\/strong>[latex]\\frac{4}{\\left(x - 3\\right)\\left(x+1\\right)}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\frac{4}{\\left(x - 3\\right)\\left(x+1\\right)}=\\frac{1}{x - 3}-\\frac{1}{x+1}=-\\frac{1}{3\\left(1-\\frac{x}{3}\\right)}-\\frac{1}{1-\\left(\\text{-}x\\right)}=-\\frac{1}{3}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{x}{3}\\right)}^{n}-\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left({\\left(-1\\right)}^{n+1}-\\frac{1}{{3}^{n+1}}\\right){x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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4.\u00a0<\/strong>[latex]\\frac{3}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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5.\u00a0<\/strong>[latex]\\frac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}[\/latex]<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]\\frac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}=\\frac{1}{{x}^{2}-1}-\\frac{1}{4}\\frac{1}{1+{\\left(\\frac{x}{2}\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n}-\\frac{1}{4}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(\\frac{x}{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\left(-1\\right)+{\\left(-1\\right)}^{n+1}\\frac{1}{{2n+2}}\\right){x}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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6.\u00a0<\/strong>[latex]\\frac{30}{\\left({x}^{2}+1\\right)\\left({x}^{2}-9\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n

In the following exercises, express each series as a rational function.<\/p>\n

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

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Show Solution<\/span><\/p>\n
[latex]\\frac{1}{x}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{{x}^{n}}=\\frac{1}{x}\\frac{1}{1-\\frac{1}{x}}=\\frac{1}{x - 1}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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8.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{x}^{2n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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9.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{\\left(x - 3\\right)}^{2n - 1}}[\/latex]<\/p>\n<\/div>\n

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

The following exercises explore applications of annuities<\/span>.<\/p>\n

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11.\u00a0<\/strong>Calculate the present values P<\/em> of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]P={P}_{1}+\\cdots+{P}_{20}[\/latex] where [latex]{P}_{k}=10,000\\frac{1}{{\\left(1+r\\right)}^{k}}[\/latex]. Then [latex]P=10,000\\displaystyle\\sum _{k=1}^{20}\\frac{1}{{\\left(1+r\\right)}^{k}}=10,000\\frac{1-{\\left(1+r\\right)}^{-20}}{r}[\/latex]. When [latex]r=0.03,P\\approx 10,000\\times 14.8775=148,775[\/latex]. When [latex]r=0.05,P\\approx 10,000\\times 12.4622=124,622[\/latex]. When [latex]r=0.07,P\\approx 105,940[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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12. <\/strong>Calculate the present values P<\/em> of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of [latex]r=0.03,r=0.05[\/latex] and [latex]r=0.07[\/latex].<\/div>\n<\/div>\n<\/div>\n
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13.\u00a0<\/strong>Calculate the annual payouts C<\/em> to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
In general, [latex]P=\\frac{C\\left(1-{\\left(1+r\\right)}^{\\text{-}N}\\right)}{r}[\/latex] for N years of payouts, or [latex]C=\\frac{Pr}{1-{\\left(1+r\\right)}^{\\text{-}N}}[\/latex]. For [latex]N=20[\/latex] and [latex]P=100,000[\/latex], one has [latex]C=6721.57[\/latex] when [latex]r=0.03;C=8024.26[\/latex] when [latex]r=0.05[\/latex]; and [latex]C\\approx 9439.29[\/latex] when [latex]r=0.07[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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14.\u00a0<\/strong>Calculate the annual payouts C<\/em> to be given perpetually on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/div>\n<\/div>\n<\/div>\n
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15.\u00a0<\/strong>Suppose that an annuity has a present value [latex]P=1\\text{million dollars}[\/latex]. What interest rate r<\/em> would allow for perpetual annual payouts of $50,000?<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
In general, [latex]P=\\frac{C}{r}[\/latex]. Thus, [latex]r=\\frac{C}{P}=5\\times \\frac{{10}^{4}}{{10}^{6}}=0.05[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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16.\u00a0<\/strong>Suppose that an annuity has a present value [latex]P=10\\text{million dollars}\\text{.}[\/latex] What interest rate r<\/em> would allow for perpetual annual payouts of $100,000?<\/div>\n<\/div>\n<\/div>\n

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.<\/p>\n

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17.\u00a0<\/strong>[latex]x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>3k<\/em><\/sup>, [latex]{x}^{3k - 1}[\/latex], and [latex]{x}^{3k - 2}.[\/latex])<\/p>\n<\/div>\n

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[latex]\\left(x+{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\frac{x+{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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18.\u00a0<\/strong>[latex]x+{x}^{2}-{x}^{3}-{x}^{4}+{x}^{5}+{x}^{6}-{x}^{7}-{x}^{8}+\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>4k<\/em><\/sup>, [latex]{x}^{4k - 1}[\/latex], etc.)<\/div>\n<\/div>\n<\/div>\n
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19.\u00a0<\/strong>[latex]x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\\cdots[\/latex] (Hint:<\/em> Group powers x<\/em>3k<\/em><\/sup>, [latex]{x}^{3k - 1}[\/latex], and [latex]{x}^{3k - 2}.[\/latex])<\/p>\n<\/div>\n

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[latex]\\left(x-{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\frac{x-{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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20.\u00a0<\/strong>[latex]\\frac{x}{2}+\\frac{{x}^{2}}{4}-\\frac{{x}^{3}}{8}+\\frac{{x}^{4}}{16}+\\frac{{x}^{5}}{32}-\\frac{{x}^{6}}{64}+\\cdots[\/latex] (Hint:<\/em> Group powers [latex]{\\left(\\frac{x}{2}\\right)}^{3k},{\\left(\\frac{x}{2}\\right)}^{3k - 1}[\/latex], and [latex]{\\left(\\frac{x}{2}\\right)}^{3k - 2}.[\/latex])<\/div>\n<\/div>\n<\/div>\n

In the following exercises, find the power series of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] given f<\/em> and g<\/em> as defined.<\/p>\n

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21.\u00a0<\/strong>[latex]f\\left(x\\right)=2\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }n{x}^{n}[\/latex]<\/p>\n<\/div>\n

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[latex]{a}_{n}=2,{b}_{n}=n[\/latex] so [latex]{c}_{n}=\\displaystyle\\sum _{k=0}^{n}{b}_{k}{a}_{n-k}=2\\displaystyle\\sum _{k=0}^{n}k=\\left(n\\right)\\left(n+1\\right)[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n\\left(n+1\\right){x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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22.\u00a0<\/strong>[latex]f\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}{x}^{n}[\/latex]. Express the coefficients of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] in terms of [latex]{H}_{n}=\\displaystyle\\sum _{k=1}^{n}\\frac{1}{k}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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23.\u00a0<\/strong>[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{x}{2}\\right)}^{n}[\/latex]<\/p>\n<\/div>\n

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[latex]{a}_{n}={b}_{n}={2}^{\\text{-}n}[\/latex] so [latex]{c}_{n}=\\displaystyle\\sum _{k=1}^{n}{b}_{k}{a}_{n-k}={2}^{\\text{-}n}\\displaystyle\\sum _{k=1}^{n}1=\\frac{n}{{2}^{n}}[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{\\left(\\frac{x}{2}\\right)}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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24.\u00a0<\/strong>[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

In the following exercises, differentiate the given series expansion of f<\/em> term-by-term to obtain the corresponding series expansion for the derivative of f<\/em>.<\/p>\n

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25.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1+x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}[\/latex]<\/p>\n<\/div>\n

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The derivative of [latex]f[\/latex] is [latex]-\\frac{1}{{\\left(1+x\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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26.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{1-{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n

In the following exercises, integrate the given series expansion of [latex]f[\/latex] term-by-term from zero to x<\/em> to obtain the corresponding series expansion for the indefinite integral of [latex]f[\/latex].<\/p>\n

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

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The indefinite integral of [latex]f[\/latex] is [latex]\\frac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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28.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{2x}{1+{x}^{2}}=2\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n+1}[\/latex]<\/div>\n<\/div>\n<\/div>\n

In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.<\/p>\n

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29.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\frac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/p>\n<\/div>\n

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[latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=\\frac{1}{1-x};{f}^{\\prime }\\left(\\frac{1}{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n - 1}}={\\frac{d}{dx}{\\left(1-x\\right)}^{-1}|}_{x=\\frac{1}{2}}={\\frac{1}{{\\left(1-x\\right)}^{2}}|}_{x=\\frac{1}{2}}=4[\/latex] so [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{2}^{n}}=2[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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30.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{3}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\frac{1}{3}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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31.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n}}[\/latex] as [latex]f''\\left(\\frac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/p>\n<\/div>\n

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[latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}=\\frac{1}{1-x};f\\text{''}\\left(\\frac{1}{2}\\right)=\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n - 2}}={\\frac{{d}^{2}}{d{x}^{2}}{\\left(1-x\\right)}^{-1}|}_{x=\\frac{1}{2}}={\\frac{2}{{\\left(1-x\\right)}^{3}}|}_{x=\\frac{1}{2}}=16[\/latex] so [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{n\\left(n - 1\\right)}{{2}^{n}}=4[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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32.\u00a0<\/strong>Evaluate [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n+1}[\/latex] as [latex]{\\displaystyle\\int }_{0}^{1}f\\left(t\\right)dt[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}=\\frac{1}{1+{x}^{2}}[\/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], use term-by-term differentiation or integration to find power series for each function centered at the given point.<\/p>\n

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

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[latex]\\displaystyle\\int \\displaystyle\\sum {\\left(1-x\\right)}^{n}dx=\\displaystyle\\int \\displaystyle\\sum {\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}dx=\\displaystyle\\sum \\frac{{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n+1}}{n+1}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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34.\u00a0<\/strong>[latex]\\text{ln}\\left(1-x\\right)[\/latex] at [latex]x=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
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35.\u00a0<\/strong>[latex]\\text{ln}\\left(1-{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/p>\n<\/div>\n

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[latex]\\text{-}{\\displaystyle\\int }_{t=0}^{{x}^{2}}\\frac{1}{1-t}dt=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\displaystyle\\int }_{0}^{{x}^{2}}{t}^{n}dx-\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2\\left(n+1\\right)}}{n+1}=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{2n}}{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
36.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{2x}{{\\left(1-{x}^{2}\\right)}^{2}}[\/latex] at [latex]x=0[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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37.\u00a0<\/strong>[latex]f\\left(x\\right)={\\tan}^{-1}\\left({x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/p>\n<\/div>\n

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[latex]{\\displaystyle\\int }_{0}^{{x}^{2}}\\frac{dt}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\displaystyle\\int }_{0}^{{x}^{2}}{t}^{2n}dt={\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n+1}}{2n+1}|}_{t=0}^{{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+2}}{2n+1}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
38.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}\\left(1+{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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39.\u00a0<\/strong>[latex]f\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\text{ln}tdt[\/latex] where [latex]\\text{ln}\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{\\left(x - 1\\right)}^{n}}{n}[\/latex]<\/p>\n<\/div>\n

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Term-by-term integration gives [latex]{\\displaystyle\\int }_{0}^{x}\\text{ln}tdt=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{\\left(x - 1\\right)}^{n+1}}{n\\left(n+1\\right)}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right){\\left(x - 1\\right)}^{n+1}=\\left(x - 1\\right)\\text{ln}x+\\displaystyle\\sum _{n=2}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{\\left(x - 1\\right)}^{n}}{n}= x\\text{ln}x-x[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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40. [T]<\/strong> Evaluate the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex] at [latex]x=1[\/latex] to show that [latex]\\text{ln}\\left(2\\right)[\/latex] is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within 0.001, and find such an approximation.<\/div>\n<\/div>\n<\/div>\n
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41. [T]<\/strong> Subtract the infinite series of [latex]\\text{ln}\\left(1-x\\right)[\/latex] from [latex]\\text{ln}\\left(1+x\\right)[\/latex] to get a power series for [latex]\\text{ln}\\left(\\frac{1+x}{1-x}\\right)[\/latex]. Evaluate at [latex]x=\\frac{1}{3}[\/latex]. What is the smallest N<\/em> such that the N<\/em>th partial sum of this series approximates [latex]\\text{ln}\\left(2\\right)[\/latex] with an error less than 0.001?<\/p>\n<\/div>\n

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We have [latex]\\text{ln}\\left(1-x\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{n}}{n}[\/latex] so [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex]. Thus, [latex]\\text{ln}\\left(\\frac{1+x}{1-x}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\left(1+{\\left(-1\\right)}^{n - 1}\\right)\\frac{{x}^{n}}{n}=2\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{x}^{2n - 1}}{2n - 1}[\/latex]. When [latex]x=\\frac{1}{3}[\/latex] we obtain [latex]\\text{ln}\\left(2\\right)=2\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}[\/latex]. We have [latex]2\\displaystyle\\sum _{n=1}^{3}\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}=0.69300\\ldots[\/latex], while [latex]2\\displaystyle\\sum _{n=1}^{4}\\frac{1}{{3}^{2n - 1}\\left(2n - 1\\right)}=0.69313\\ldots[\/latex] and [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=4[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.<\/p>\n

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42.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=0}^{\\infty }\\left({x}^{k}-{x}^{2k+1}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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43.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{3k}}{6k}[\/latex]<\/p>\n<\/div>\n

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[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{k}}{k}=\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] so [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{x}^{3k}}{6k}=-\\frac{1}{6}\\text{ln}\\left(1-{x}^{3}\\right)[\/latex]. The radius of convergence is equal to 1 by the ratio test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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44.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{\\left(1+{x}^{2}\\right)}^{\\text{-}k}[\/latex] using [latex]y=\\frac{1}{1+{x}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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45.\u00a0<\/strong>[latex]\\displaystyle\\sum _{k=1}^{\\infty }{2}^{\\text{-}kx}[\/latex] using [latex]y={2}^{\\text{-}x}[\/latex]<\/p>\n<\/div>\n

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If [latex]y={2}^{\\text{-}x}[\/latex], then [latex]\\displaystyle\\sum _{k=1}^{\\infty }{y}^{k}=\\frac{y}{1-y}=\\frac{{2}^{\\text{-}x}}{1-{2}^{\\text{-}x}}=\\frac{1}{{2}^{x}-1}[\/latex]. If [latex]{a}_{k}={2}^{\\text{-}kx}[\/latex], then [latex]\\frac{{a}_{k+1}}{{a}_{k}}={2}^{\\text{-}x}<1[\/latex] when [latex]x>0[\/latex]. So the series converges for all [latex]x>0[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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46.\u00a0<\/strong>Show that, up to powers x<\/em>3<\/sup> and y<\/em>3<\/sup>, [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] satisfies [latex]E\\left(x+y\\right)=E\\left(x\\right)E\\left(y\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n
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47.\u00a0<\/strong>Differentiate the series [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex] term-by-term to show that [latex]E\\left(x\\right)[\/latex] is equal to its derivative.<\/p>\n<\/div>\n

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Answers will vary.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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48.\u00a0<\/strong>Show that if [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] is a sum of even powers, that is, [latex]{a}_{n}=0[\/latex] if n<\/em> is odd, then [latex]F={\\displaystyle\\int }_{0}^{x}f\\left(t\\right)dt[\/latex] is a sum of odd powers, while if f<\/em> is a sum of odd powers, then F<\/em> is a sum of even powers.<\/div>\n<\/div>\n<\/div>\n
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49. [T]<\/strong> Suppose that the coefficients an<\/sub><\/em> of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}_{n}=\\frac{{a}_{n - 1}}{n}+\\frac{{a}_{n - 2}}{n\\left(n - 1\\right)}[\/latex]. For [latex]{a}_{0}=0[\/latex] and [latex]{a}_{1}=1[\/latex], compute and plot the sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{a}_{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/p>\n<\/div>\n

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

The solid curve is [latex]{S}_{5}[\/latex]. The dashed curve is [latex]{S}_{2}[\/latex], dotted is [latex]{S}_{3}[\/latex], and dash-dotted is [latex]{S}_{4}[\/latex]<\/div>\n<\/div>\n

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50. [T]<\/strong> Suppose that the coefficients an<\/sub><\/em> of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}_{n}=\\frac{{a}_{n - 1}}{\\sqrt{n}}-\\frac{{a}_{n - 2}}{\\sqrt{n\\left(n - 1\\right)}}[\/latex]. For [latex]{a}_{0}=1[\/latex] and [latex]{a}_{1}=0[\/latex], compute and plot the sums [latex]{S}_{N}=\\displaystyle\\sum _{n=0}^{N}{a}_{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/div>\n<\/div>\n<\/div>\n
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51. [T]<\/strong> Given the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex], determine how many terms N<\/em> of the sum evaluated at [latex]x=-\\frac{1}{2}[\/latex] are needed to approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]\\frac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\frac{{x}^{n}}{n}[\/latex].<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
When [latex]x=-\\frac{1}{2},\\text{-}\\text{ln}\\left(2\\right)=\\text{ln}\\left(\\frac{1}{2}\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n{2}^{n}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=11}^{\\infty }\\frac{1}{n{2}^{n}}<\\displaystyle\\sum _{n=11}^{\\infty }\\frac{1}{{2}^{n}}=\\frac{1}{{2}^{10}}[\/latex], one has [latex]\\displaystyle\\sum _{n=1}^{10}\\frac{1}{n{2}^{n}}=0.69306\\ldots[\/latex] whereas [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=10[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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52. [T]<\/strong> Given the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{\\infty }{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex], use the alternating series test to determine how many terms N<\/em> of the sum evaluated at [latex]x=1[\/latex] are needed to approximate [latex]{\\tan}^{-1}\\left(1\\right)=\\frac{\\pi }{4}[\/latex] accurate to within [latex]\\frac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{k=0}^{N}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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53. [T]<\/strong> Recall that [latex]{\\tan}^{-1}\\left(\\frac{1}{\\sqrt{3}}\\right)=\\frac{\\pi }{6}[\/latex]. Assuming an exact value of [latex]\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex], estimate [latex]\\frac{\\pi }{6}[\/latex] by evaluating partial sums [latex]{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex] of the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{\\infty }{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{2k+1}[\/latex] at [latex]x=\\frac{1}{\\sqrt{3}}[\/latex]. What is the smallest number N<\/em> such that [latex]6{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)[\/latex] approximates \u03c0<\/em> accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?<\/p>\n<\/div>\n

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Show Solution<\/span><\/p>\n
[latex]6{S}_{N}\\left(\\frac{1}{\\sqrt{3}}\\right)=2\\sqrt{3}\\displaystyle\\sum _{n=0}^{N}{\\left(-1\\right)}^{n}\\frac{1}{{3}^{n}\\left(2n+1\\right)}[\/latex]. One has [latex]\\pi -6{S}_{4}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00101\\ldots[\/latex] and [latex]\\pi -6{S}_{5}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00028\\ldots[\/latex] so [latex]N=5[\/latex] is the smallest partial sum with accuracy to within 0.001. Also, [latex]\\pi -6{S}_{7}\\left(\\frac{1}{\\sqrt{3}}\\right)=0.00002\\ldots[\/latex] while [latex]\\pi -6{S}_{8}\\left(\\frac{1}{\\sqrt{3}}\\right)=-0.000007\\ldots[\/latex] so [latex]N=8[\/latex] is the smallest N to give accuracy to within 0.00001.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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