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Concepts and Vocabulary
True or False Every power series ∞∑k=0ak(x−c)k converges for at least one number.
True
True or False Let bn denote the nth term of the power series ∞∑k=0akxk. If limn→∞|bn+1bn|<1 for every number x, then \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} is absolutely convergent on the interval ( -\infty ,\,\infty ) .
True
True or False If the radius of convergence of a power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} is 0, then the power series converges only for x=0.
True
True or False If a power series converges at one endpoint of its interval of convergence, then it must converge at its other endpoint.
False
True or False The power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} and \sum\limits_{k\,=\,0}^{\infty }a_{k}(x-3)^{k} have the same radius of convergence.
True
True or False The power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} and \sum\limits_{k\,=\,0}^{\infty }a_{k}(x-3)^{k} have the same interval of convergence.
False
True or False If the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} converges for x=8, then it converges for x=-8.
False
True or False If the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} converges for x=3, then it converges for x=1.
True
True or False If the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} converges for x=-4, then it converges for x=3.
True
True or False If the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} converges for x=3, then it diverges for x=5.
False
True or False A possible interval of convergence for the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} is [-2,4].
False
True or False If the power series \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} diverges for a number x_{1}, then it converges for all numbers x for which \vert x\vert <\vert x_{1}\vert .
False
Skill Building
In Problems 13–16, find all numbers x for which each power series converges.
\sum\limits_{k\,=\,0}^{\infty} kx^{k}
-1\,{<}\,x\,{<}\,1
\sum\limits_{k\,=\,0}^{\infty}\dfrac{kx^{k}}{3^{k}}
\sum\limits_{k\,=\,0}^{\infty }\dfrac{(x+1)^{k}}{3^{k}}
-4<x<2
\sum\limits_{k=1}^{\infty}\dfrac{(x-2) ^{k}}{k^{2}}
In Problems 17–26:
\sum\limits_{k\,=\,0}^{\infty}\dfrac{x^{k}}{2^{k}(k+1) }
(a, b) R=2; -2\le x < 2\quad (c) Answers will vary.
\sum\limits_{k\,=\,0}^{\infty}(-1) ^{k}\dfrac{x^{k}}{2^{k}(k+1) }
\sum\limits_{k\,=\,0}^{\infty}\dfrac{x^{k}}{k+5}
(a, b) R=1; -1\leq x<1\quad (c) Answers will vary.
\sum\limits_{k\,=\,0}^{\infty}\dfrac{x^{k}}{1+k^{2}}
\sum\limits_{k\,=\,0}^{\infty}\dfrac{k^{2}x^{k}}{3^{k}}
(a, b) R=3; -3< x<3\quad (c) Answers will vary.
\sum\limits_{k\,=\,0}^{\infty}\dfrac{2^{k}x^{k}}{3^{k}}
\sum\limits_{k\,=\,0}^{\infty}\dfrac{kx^{k}}{2k+1}
(a, b) R=1; -1< x<1\quad (c) Answers will vary.
\sum\limits_{k\,=\,0}^{\infty}(6x)^{k}
\sum\limits_{k\,=\,0}^{\infty}(x-3)^{k}
(a, b) R=1; 2< x<4\quad (c) Answers will vary.
\sum\limits_{k\,=\,0}^{\infty}\dfrac{k(2x)^{k}}{3^{k}}
In Problems 27–44, find the radius of convergence and the interval of convergence of each power series.
\sum\limits_{k=1}^{\infty}\dfrac{x^{k}}{k^{3}}
R=1; -1\leq x\leq 1
\sum\limits_{k=2}^{\infty}\dfrac{x^{k}}{\ln k}
\sum\limits_{k=1}^{\infty}\dfrac{(x-2)^{k}}{k^{3}}
R=1; 1\leq x\leq 3
\sum\limits_{k\,=\,0}^{\infty}\dfrac{k(x-2) ^{k}}{3^{k}}
\!\sum\limits_{k\,=\,0}^{\infty}\dfrac{(-1)^{k}}{(2k+1)!}x^{2k+1}
R=\infty; -\infty< x<\infty
\sum\limits_{k=1}^{\infty}(kx)^{k}
\sum\limits_{k=1}^{\infty}\dfrac{kx^{k}}{\ln (k+1)}
R=1; -1< x<1
\sum\limits_{k=1}^{\infty}\dfrac{x^{k}}{\ln (k+1)}
\sum\limits_{k=0}^{\infty}\dfrac{k(k+1)x^{k}}{4^{k}}
R=4; -4< x<4
610
\sum\limits_{k=1}^{\infty}\dfrac{(-1)^{k}(x-5)^{k}}{k(k+1)}
\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{(x-3)^{2k}}{9^{k}}
R=3; 0< x<6
\sum\limits_{k\,=\,0}^{\infty}\dfrac{x^{k}}{e^{k}}
\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{(2x)^{k}}{k!}
R=\infty; -\infty< x<\infty
\sum\limits_{k\,=\,0}^{\infty}\dfrac{(x+1)^{k}}{k!}
\sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{(x-1)^{4k}}{k!}
R=\infty; -\infty< x<\infty
\sum\limits_{k=1}^{\infty}\dfrac{(x+1)^{k}}{k(k+1)(k+2)}
\sum\limits_{k=1}^{\infty}\dfrac{k^{k}x^{k}}{k!}
R=\dfrac{1}{e}; -\dfrac{1}{e}< x<\dfrac{1}{e}
\sum\limits_{k\,=\,0}^{\infty}\dfrac{3^{k}(x-2)^{k}}{k!}
A function f is defined by the power series f(x) =\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{3^{k}}.
A function f is defined by the power series f(x)=\sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}\left(\dfrac{x}{2}\right)^{k}.
A function f is defined by the power series f(x)=\sum\limits_{k\,=\,0}^{\infty }\dfrac{(x-2) ^{k}}{2^{k}}.
A function f is defined by the power series f(x) =\sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}(x+3)^{k}.
If \sum\limits_{k=0}^{\infty }a_{k\,}x^{k} converges for x=3, what, if anything, can be said about the convergence at x=2? Can anything be said about the convergence at x=5?
Converges at x=2; no information about x=5
If \sum\limits_{k=0}^{\infty }a_{k}(x-2)^{k} converges for x=6, at what other numbers x must the series necessarily converge?
If the series \sum\limits_{k=0}^{\infty }\,a_{k}x^{k} converges for x=6 and diverges for x=-8, what, if anything, can be said about the truth of the following statements?
If the radius of convergence of the power series \sum\limits_{k=0}^{\infty }\,a_{k}(x-3) ^{k} is R=5, what, if anything, can be said about the truth of the following statements?
In Problems 53–58:
f(x) =\dfrac{1}{1+x^{3}}
f(x) =\dfrac{1}{1-x^{2}}
f(x) =\dfrac{1}{6-2x}
f(x) =\dfrac{4}{x+2}
f(x) =\dfrac{x}{1+x^{3}}
f(x) =\dfrac{4x^{2}}{x+2}
In Problems 59–62:
f(x) =\sum\limits_{k\,=\,0}^{\infty}\dfrac{(-1) ^{k}x^{2k+1}}{(2k+1) !}
f(x)=\sum\limits_{k\,=\,0}^{\infty}\dfrac{(-1)^{k}x^{2k}}{(2k) !}
f(x)=\sum\limits_{k\,=\,0}^{\infty}\dfrac{x^{k}}{k!}
f(x)=\sum\limits_{k\,=\,0}^{\infty}\dfrac{(-1)^{k}x^{k}}{k!}
In Problems 63–70, find a power series representation of f. Use a geometric series and properties of a power series.
f(x) =\dfrac{1}{(1+x)^{2}}
f(x)=\sum\limits_{k=1}^{\infty}{(-1)^{k-1}kx^{k-1}}, -1 < x < 1
f(x) =\dfrac{1}{(1-x)^{3}}
f(x)=\dfrac{2}{3 (1-x) ^{2}}
f(x)=\dfrac{2}{3}\sum\limits_{k=1}^{\infty}{kx^{k-1}}, -1 < x < 1
f(x) =\dfrac{1}{(1-x) ^{4}}
f(x) =\ln \left( \dfrac{1}{1+x}\right)
f(x)=\sum\limits_{k=0}^{\infty}{\dfrac{(-1)^{k+1}x^{k+1}}{k+1}}, -1 \lt x \leq 1
f(x) =\ln (1-2x)
f(x) =\ln (1-x^{2})
f(x)=-\sum\limits_{k=0}^{\infty}{\dfrac{x^{2k+2}}{k+1}}, -1 < x < 1
f(x) =\ln (1+x^{2})
611
Applications and Extensions
In Problems 71–78, find all x for which each power series converges.
\sum\limits_{k=1}^{\infty}\dfrac{x^k}{k}
-1\leq x<1
\sum\limits_{k=1}^{\infty}\dfrac{(x-4)^{k}}{k}
\sum\limits_{k=1}^{\infty}\dfrac{x^{k}}{2k+1}
-1\leq x<1
\sum\limits_{k=1}^{\infty}\dfrac{x^{k}}{k^{2}}
\sum\limits_{k\,=\,0}^{\infty}x^{k^{2}}
-1<x<1
\sum\limits_{k=1}^{\infty}\dfrac{k^{a}}{a^{k}}(x-a)^{k}, \quad a\neq 0
\sum\limits_{k\,=\,0}^{\infty}\dfrac{(k!)^{2}}{(2k)!}(x-1)^{k}
-3<x<5
\sum\limits_{k\,=\,0}^{\infty}\dfrac{\sqrt{k!}}{(2k)!}x^{k}
Use the power series found in Problem 80 to get an approximation for \ln 2 correct to three decimal places.
0.693
Use the first 1000 terms of Gregory’s series to approximate \dfrac{\pi }{4}. What is the approximation for \pi ?
If R>0 is the radius of convergence of \sum\limits_{k=1}^{\infty }a_{k}x^{k}, show that \lim\limits_{n\,\rightarrow \,\infty }\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =\dfrac{1}{R}, provided this limit exists.
See Student Solutions Manual.
If R is the radius of convergence of \sum\limits_{k=1}^{\infty }a_{k}x^{k}, show that the radius of convergence of \sum\limits_{k=1}^{\infty }a_{k}x^{2k} is \sqrt{R}.
Prove that if a power series is absolutely convergent at one endpoint of its interval of convergence, then the power series is absolutely convergent at the other endpoint.
See Student Solutions Manual.
Suppose \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} converges for \vert x\vert <R and that \lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert exists. Show that \sum\limits_{k=1}^{\infty }ka_{k}x^{k-1} and \sum\limits_{k\,=\,0}^{\infty } \dfrac{a_{k}}{k+1}x^{k+1} also converge for \vert x\vert <R.
Challenge Problems
Consider the differential equation (1+x^{2})\,y^{\prime \prime} -4xy^\prime +6y=0
Assuming there is a solution y(x)=\sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} , substitute and obtain a formula for a_{k}. Your answer should have the form y(x)=a_{0}(1-3x^{2}) +a_{1}\!\left( x-\dfrac{1}{3}x^{3}\right)\qquad a_{0},~a_{1} \hbox{ real numbers}
See Student Solutions Manual.
If the series \sum\limits_{k=0}^{\infty }a_{k}3^{k} converges, show that the series \sum\limits_{k=1}^{\infty }ka_{k}2^{k} also converges.
Find the interval of convergence of the series \sum\limits_{k=1}^{\infty }\dfrac{(x-2)^{k}}{k(3^{k})}.
-1\leq x<5
Let a power series S(x) be convergent for \vert x\vert <R. Assume that S(x)=\sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k} with partial sums S_{n}(x)=\sum\limits_{k\,=\,0}^{n}a_{k}x^{k}. Suppose for any number \varepsilon >0, there is a number N so that when n>N, \vert S(x)-S_{n}(x) \vert <\dfrac{\varepsilon }{3} for all \vert x\vert <R. Show that S(x) is continuous for all \vert x\vert <R.
Find the power series in x, denoted by f(x), for which f^{\prime \prime} (x) + f(x) =0 and f(0) =0, f^\prime (0) =1. What is the radius of convergence of the series?
\sum\limits_{k=0}^{\infty}{\dfrac{(-1)^{k}x^{2k+1}}{(2k+1)!}}; R=\infty
The Bessel function of order m of the first kind, where m is a nonnegative integer, is defined as J_{m}(x) =\sum\limits_{k\,=\,0}^{\infty }(-1) ^{k} \dfrac{1}{(k+m) !\, k!}\left( \dfrac{x}{2}\right) ^{2k+m}
Show that: