11.13: Additional Exercises
- Page ID
- 149554
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)These problems require the techniques of this chapter, and are in no particular order. Some problems may be done in more than one way.
Exercises \(\PageIndex{}\)
Exercises \(\PageIndex{1}\) to \(\PageIndex{17}\), determine whether the series converges.
\(\displaystyle \sum_{n=0}^{\infty}\dfrac{n}{n^2+4}\)
- Answer
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Diverges
\(\displaystyle \dfrac{1}{1\cdot 2}+ \dfrac{1}{3\cdot 4}+ \dfrac{1}{5\cdot 6}+ \dfrac{1}{7\cdot 8}+ \cdots\)
- Answer
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Converges
\(\displaystyle \sum_{n=0}^{\infty}\dfrac{n}{(n^2+4)^2}\)
- Answer
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Converges
\(\displaystyle\sum_{n=0}^\infty \dfrac{n!}{8^n}\)
- Answer
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Diverges
\(\displaystyle 1- \dfrac{3}{4}+ \dfrac{5}{8}- \dfrac{7}{12}+ \dfrac{9}{16}+ \cdots\)
- Answer
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Diverges
\(\displaystyle\sum_{n=0}^\infty \dfrac{1}{\sqrt{n^2+4}}\)
- Answer
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Diverges
\(\displaystyle\sum_{n=0}^\infty \dfrac{\sin^3(n)}{n^2}\)
- Answer
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Converges
\(\displaystyle\sum_{n=0}^{\infty} \dfrac{n}{e^n}\)
- Answer
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Converges
\(\displaystyle\sum_{n=0}^\infty \dfrac{n!}{1\cdot 3\cdot 5\cdots(2n-1)}\)
- Answer
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Converges
\(\displaystyle\sum_{n=1}^\infty \dfrac{1}{n\sqrt{n}}\)
- Answer
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Converges
\(\displaystyle \dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{2}{3\cdot 4\cdot 5}+ \dfrac{3}{4\cdot 5\cdot 6}+ \dfrac{4}{5\cdot 6 \cdot 7}+ \cdots\)
- Answer
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Converges
\(\displaystyle\sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots(2n-1)}{(2n)!}\)
- Answer
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Converges
\(\displaystyle\sum_{n=0}^\infty \dfrac{6^n}{n!}\)
- Answer
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Converges
\(\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{\sqrt{n}}\)
- Answer
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Converges
\(\displaystyle\sum_{n=1}^\infty \dfrac{2^n 3^{n-1}}{n!}\)
- Answer
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Converges
\(\displaystyle 1+ \dfrac{5^2}{2^2}+ \dfrac{5^4}{(2\cdot 4)^2}+ \dfrac{5^6}{(2\cdot 4\cdot 6)^2}+ \dfrac{5^8}{(2\cdot 4\cdot 6\cdot 8)^2}+ \cdots\)
- Answer
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Converges
\(\displaystyle\sum_{n=1}^\infty \sin(1/n)\)
- Answer
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Diverges
Exercises \(\PageIndex{18}\) to \(\PageIndex{24}\), find the interval and radius of convergence. You need not check the endpoints of the intervals.
\(\displaystyle\sum_{n=0}^\infty \dfrac{2^n}{n!}x^n\)
- Answer
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\((-\infty,\infty)\)
\(\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{1+3^n}\)
- Answer
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\((-3,3)\)
\(\displaystyle\sum_{n=1}^\infty \dfrac{x^n}{n3^n}\)
- Answer
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\((-3,3)\)
\(\displaystyle x+ \dfrac{1}{2} \dfrac{x^3}{3} + \dfrac{1\cdot 3}{2\cdot 4} \dfrac{x^5}{5}+ \dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \dfrac{x^7}{7}+\cdots\)
- Answer
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\((-1,1)\)
\(\displaystyle\sum_{n=1}^\infty \dfrac{n!}{n^2} x^n\)
- Answer
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Radius is \(0\) - it converges only when \(x=0\).
\(\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^n}{n^2 3^n} x^{2n}\)
- Answer
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\((-\sqrt{3},\sqrt{3})\)
\(\displaystyle\sum_{n=0}^\infty \dfrac{(x-1)^n}{n!}\)
- Answer
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\((-\infty,\infty)\)
Exercises \(\PageIndex{25}\) to \(\PageIndex{30}\), find a series for each function, using the formula for Maclaurin series and algebraic manipulation as appropriate.
\(2^x\)
- Answer
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\(\displaystyle \sum_{n=0}^{\infty}\dfrac{\big(\ln(2)\big)^n}{n!}x^n\)
\(\ln(1+x)\)
- Answer
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\(\displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{n+1}x^{n+1}\)
\(\ln\left(\dfrac{1+x}{1-x}\right)\)
- Answer
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\(\displaystyle \sum_{n=0}^{\infty}\dfrac{2}{2n+1}x^{2n+1}\)
\(\sqrt{1+x}\)
- Answer
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\(\displaystyle 1+\dfrac{x}{2}+\sum_{n=2}^{\infty}(-1)^{n+1}\dfrac{1\cdot 3\cdot 5\cdots(2n-3)}{2^{n}n!}x^n\)
\(\dfrac{1}{1+x^2}\)
- Answer
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\(\displaystyle \sum_{n=0}^{\infty}(-1)^{n}x^{2n}\)
\(\arctan(x)\)
- Answer
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\(\displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}x^{2n+1}\)
Use the answer to the previous problem to discover a series for a well-known mathematical constant.
- Answer
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\(\displaystyle \pi=\sum_{n=0}^{\infty}(-1)^n\dfrac{4}{2n+1}\)