In the following exercises, state whether each statement is true, or give an example to show that it is false.
1. 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].
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True. If a series converges then its terms tend to zero.
2. [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].
3. 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].
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False. It would imply that [latex]{a}_{n}{x}^{n}\to 0[/latex] for [latex]|x|
4. 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 n, then the radius of convergence of [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}{x}^{n}[/latex] is greater than or equal to R.
5. 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, then it converges at any point closer to c than x.
[latex]x=1[/latex]
[latex]x=2[/latex]
[latex]x=3[/latex]
[latex]x=0[/latex]
[latex]x=5.99[/latex]
[latex]x=0.000001[/latex]
Show Solution
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].
6. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{\left(x+1\right)}^{n}[/latex] converges at [latex]x=-2[/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, then it converges at any point closer to c than x.
[latex]x=2[/latex]
[latex]x=-1[/latex]
[latex]x=-3[/latex]
[latex]x=0[/latex]
[latex]x=0.99[/latex]
[latex]x=0.000001[/latex]
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.
[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]
In the following exercises, find the radius of convergence R and interval of convergence for [latex]\displaystyle\sum {a}_{n}{x}^{n}[/latex] with the given coefficients [latex]{a}_{n}[/latex].
[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].
[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].
[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].
[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].
[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].
[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]
[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]
27. [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]
Show Solution
[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]
[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]
[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]
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, and identify its interval of convergence.
Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.
43. 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].
Show Solution
[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.
[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]
[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]
48. 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 is odd. Explain why [latex]p\left(x\right)=-p\left(\text{-}x\right)[/latex].
49. 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 is even. Explain why [latex]p\left(x\right)=p\left(\text{-}x\right)[/latex].
Show Solution
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].
50. 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(Ax\right)[/latex].
51. 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].
Show Solution
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.
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. 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.
55. [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{{n}^{2}}{x}^{{n}^{2}}[/latex] (Hint: Let [latex]{b}_{k}={a}_{k}[/latex] if [latex]k={n}^{2}[/latex] for some n, otherwise [latex]{b}_{k}=0.[/latex])
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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.
56. Suppose that [latex]p\left(x\right)[/latex] is a polynomial of degree N. Find the radius and interval of convergence of [latex]\displaystyle\sum _{n=1}^{\infty }p\left(n\right){x}^{n}[/latex].
57. [T] 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 increases.
Show Solution
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.
58. [T] 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 increases.
59. [T] 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 increases.
Show Solution
The approximation appears to stabilize quickly near both [latex]x=\pm 1[/latex].
60. [T] 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 increases.
61. [T] 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 increases.
Show Solution
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].
62. [T] 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 increases.
