Exercises: Power Series and Convergence

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Exercises relating to power series and their convergence properties.

Compute the radius of convergence of the power series below. $\sum _{n = 1}^\infty \frac {(x-3)^n}{2^n + 1}$ $R = \answer {2}$

When computing a limit like $\lim _{n \rightarrow \infty } \frac {2^{n} + 1}{2^{n+1} + 1},$ remember the effect of orders of growth: $1 <\!\!< 2^n$ , and so from the perspective of the limit, the $+1$ ’s in both the numerator and denominator are both negligible.

Compute the radius of convergence of the power series below. $\sum _{n = 1}^\infty \frac {(-3)^n (\ln n) (x-3)^n}{4^n}$ $R = \answer {\frac {4}{3}}$

When computing a limit like $\lim _{n \rightarrow \infty } (\ln n)^{1/n},$ remember that we already know $n^{1/n} \rightarrow 1$ and that $\ln n <\!\!<n$ as well, so we expect $(\ln n)^{1/n} \rightarrow 1$ as well.

Compute the radius of convergence of the power series below. $\sum _{n = 1}^\infty \frac {(-2)^n n! (x-3)^n}{3^n+n^2}$ $R = \answer {0}$

Compute the interval of convergence for the power series below. $\sum _{n=1}^\infty \frac {(-2)^n (x-3)^n}{\ln n}$ The left endpoint of the interval is $\answer {5/2}$ ; it isis not included in the interval of convergence. The right endpoint of the interval is $\answer {7/2}$ ; it is isis not included in the interval of convergence.

Reindex the series below: $\sum _{n=1}^\infty \frac {x^{2n+1}}{n + 2} = \sum _{n = 0} \frac {x^{\answer {2n+3}}}{\answer {n+3}}$

Differentiate the series below term-by-term: $\frac {d}{dx} \sum _{n=0}^\infty \frac {n}{n+1} x^n = \sum _{n=1} \answer {\frac {n^2}{n+1}} x^{\answer {n-1}}.$ (Note that the $n=0$ term goes away because the derivative of a constant is zero.)

Integrate the series below term-by-term: $\int _0^x \left [ \sum _{n=0}^\infty \frac {x^n}{(2n+3)^2} \right ] = \sum _{n=0} \answer {\frac {1}{(n+1)(2n+3)^2}} x^{\answer {n+1}}.$

For each step below, apply a term-by-term operation, a multiplication by a monomial, or a substitution to derive a new series formula from a known formula.

Use the formula $\frac {1}{1-x} = \sum _{n=0}^\infty x^n$ to derive a summation formula for $1/(1+x^2)$ , i.e., $\frac {1}{1+x^2} = \sum _{n=0}^\infty \answer {(-1)^n} x^{\answer {2n}}.$
Use the formula you derived above to develop a power series expansion for arctangent: $\arctan x = \sum _{n=0}^\infty \answer {\frac {(-1)^n}{(2n+1)}} x^{\answer {2n+1}}.$
The radius of convergence of this last series is $R = \answer {1}$ .

For each step below, apply a term-by-term operation, a multiplication by a monomial, or a substitution to derive a new series formula from a known formula.

Use the formula $\frac {1}{1-x} = \sum _{n=0}^\infty x^n$ to determine the sum of the series below for $-1 < x < 1$ : $\sum _{n = 0}^\infty \frac {x^{n+1}}{n+1} = \answer {- \ln |1-x|}.$ (Don’t forget absolute values if you need them.)
Use the formula you just derived to determine the sum of the series $\sum _{n=0}^\infty \frac {x^{2n+3}}{n+1} = \answer {- x \ln |1-x^2|.}$
Use the formula you just derived to determine the sum of the series $\sum _{n=0}^\infty \frac {2^{2n} x^{2n+3}}{n+1} = \answer {- \frac {1}{4} x \ln |1-4x^2|.}$

Sample Quiz Questions

Find the full interval of convergence for the power series $\sum _{m=2}^{\infty } \frac {(-3)^mm^2(x - 5)^m }{\ln m}.$ (Hints will not be revealed until after you choose a response.)

$\displaystyle \left (\frac {14}{3},\frac {16}{3}\right )$ $\displaystyle \left [\frac {14}{3},\frac {16}{3}\right )$ $\displaystyle \left (2,8\right ]$ $\displaystyle \left [2,8\right ]$ $\displaystyle \left (-\infty ,\infty \right )$

First observe that $\begin {aligned} \frac {1}{R} & = \lim _{m\rightarrow \infty } \left | \frac {\frac {(-3)^{(m+1)}{(m+1)}^2 }{\ln {(m+1)}} }{\frac {(-3)^mm^2 }{\ln m}} \right | \\ & = \lim _{m\rightarrow \infty } \left | \frac {-3{(m+1)}^2\ln m }{m^2\ln {(m+1)} } \right | \\ & = 3 \end {aligned}$ because $\lim _{m \rightarrow \infty } \frac {(m+1)^2 \ln m}{m^2 \ln (m+1)} = \lim _{m \rightarrow \infty } \frac {(m+1)^2}{m^2} \lim _{m \rightarrow \infty } \frac {\ln m}{\ln (m+1)} = 1$ by virtue of l’Hospital’s rule applied to both limits on the right-hand side.

This means that the radius equals $1/3$ . At the endpoint $x=14/3$ , the series equals $\sum _{m=2}^{\infty } \frac {m^2 }{\ln m },$ which diverges by the $n$ -th term divergence test because $\lim _{m \rightarrow \infty } m^{2} / \ln m = \infty \neq 0$ . At the endpoint $x=16/3$ , the series equals $\sum _{m=2}^{\infty } (-1)^m\frac {m^2 }{\ln m },$ which diverges for the same reason as the other endpoint, i.e., the terms do not go to zero.

Find the full interval of convergence for the power series $\sum _{m=2}^{\infty } \frac {(-4)^m(\ln m)(x - 1)^m }{(-2)^mm}.$ (Hints will not be revealed until after you choose a response.)

$\displaystyle \left (\frac {1}{2},\frac {3}{2}\right )$ $\displaystyle \left [\frac {1}{2},\frac {3}{2}\right )$ $\displaystyle \left (-1,3\right ]$ $\displaystyle \left [-1,3\right ]$ $\displaystyle \left (-\infty ,\infty \right )$

First observe that $\begin {aligned} \frac {1}{R} & = \lim _{m\rightarrow \infty } \left | \frac {\frac {(-4)^{(m+1)}(\ln {(m+1)}) }{(-2)^{(m+1)}{(m+1)}} }{\frac {(-4)^m(\ln m) }{(-2)^mm}} \right | \\ & = \lim _{m\rightarrow \infty } \left | \frac {-4(\ln {(m+1)})m }{-2(\ln m){(m+1)} } \right | \\ & = 2 \end {aligned}$ because $\lim _{m \rightarrow \infty } \frac {m\ln (m+1)}{(m+1) \ln m} = \lim _{m \rightarrow \infty } \frac {m}{m+1} \lim _{m \rightarrow \infty } \frac {\ln (m+1)}{\ln m} = 1$ by virtue of l’Hospital’s rule applied to both limits on the right-hand side.

This means that the radius equals $1/2$ . At the endpoint $x=3/2$ , the series equals $\sum _{m=2}^{\infty } \frac {(\ln m) }{m },$ which diverges by direct comparison to the harmonic series, i.e., the $p$ -series with $p = 1$ . At the endpoint $x=1/2$ , the series equals $\sum _{m=2}^{\infty } (-1)^m\frac {(\ln m) }{m },$ which converges by the alternating series test because the sign of the terms alternates and $\ln m / m$ decreases to zero as $m \rightarrow \infty$ .

Find the full interval of convergence for the power series $\sum _{m=2}^{\infty } \frac {(-2)^m(\ln m)(x + 4)^m }{6^mm}.$ (Hints will not be revealed until after you choose a response.)

$\displaystyle \left (-\frac {13}{3},-\frac {11}{3}\right )$ $\displaystyle \left [-\frac {13}{3},-\frac {11}{3}\right )$ $\displaystyle \left (-7,-1\right ]$ $\displaystyle \left [-7,-1\right ]$ $\displaystyle \left (-\infty ,\infty \right )$

First observe that $\begin {aligned} \frac {1}{R} & = \lim _{m\rightarrow \infty } \left | \frac {\frac {(-2)^{(m+1)}(\ln {(m+1)}) }{6^{(m+1)}{(m+1)}} }{\frac {(-2)^m(\ln m) }{6^mm}} \right | \\ & = \lim _{m\rightarrow \infty } \left | \frac {-2(\ln {(m+1)})m }{6(\ln m){(m+1)} } \right | \\ & = \frac {1}{3} \end {aligned}$ because $\lim _{m \rightarrow \infty } \frac {m\ln (m+1)}{(m+1) \ln m} = \lim _{m \rightarrow \infty } \frac {m}{m+1} \lim _{m \rightarrow \infty } \frac {\ln (m+1)}{\ln m} = 1$ by virtue of l’Hospital’s rule applied to both limits on the right-hand side.

This means that the radius equals $3$ . At the endpoint $x=-7$ , the series equals $\sum _{m=2}^{\infty } \frac {(\ln m) }{m },$ which diverges by direct comparison to the harmonic series, i.e., the $p$ -series with $p = 1$ . At the endpoint $x=-1$ , the series equals $\sum _{m=2}^{\infty } (-1)^m\frac {(\ln m) }{m },$ which converges by the alternating series test because the sign of the terms alternates and $\ln m / m$ decreases to zero as $m \rightarrow \infty$ .

Find the full interval of convergence for the power series $\sum _{m=1}^{\infty } \frac {\sqrt [3]{m}(x - 2)^m }{m!}.$

$\displaystyle \left (1,3\right )$ $\displaystyle \left [1,3\right )$ $\displaystyle \left (1,3\right ]$ $\displaystyle \left [1,3\right ]$ $\displaystyle \left (-\infty ,\infty \right )$

First observe that $\begin {aligned} \frac {1}{R} & = \lim _{m\rightarrow \infty } \left | \frac {\frac {\sqrt [3]{{(m+1)}} }{{(m+1)}!} }{\frac {\sqrt [3]{m} }{m!}} \right | \\ & = \lim _{m\rightarrow \infty } \left | \frac {\sqrt [3]{{(m+1)}} }{(m+1)\sqrt [3]{m} } \right | \\ & = 0 \end {aligned}$ because $m+1$ in the denominator tends to $\infty$ and $\lim _{m \rightarrow \infty } \frac {\sqrt [3]{m+1}}{\sqrt [3]{m}} = \lim _{m \rightarrow \infty } \left ( 1 + m^{-1} \right )^{1/3} = \left ( 1 + \lim _{m \rightarrow \infty } m^{-1} \right )^{1/3} = 1.$ This means that the radius is infinite and the interval of convergence is $(-\infty ,\infty )$ .

Sample Exam Questions

For which values of $x$ does the series $\displaystyle \sum _{n=1}^\infty \frac {(-1)^{n+1}(x-1)^n}{n 4^n}$ converge?

$-3 < x < 5$ $-3 \leq x < 5$ $-3 < x \leq 5$ $-5 < x \leq 3$ $-5 \leq x < 3$ $-5 \leq x \leq 3$

Find the interval of convergence of the power series below. $\sum _{n=1}^\infty \frac {(4x-1)^n}{n^\frac {3}{4} (n^2+2)}$

$\displaystyle \left ( 0 , \frac {1}{2} \right ]$ $\displaystyle \left [ 0 , \frac {1}{2} \right ]$ $\displaystyle \left ( 0 , \frac {1}{2} \right )$ $\displaystyle \left [ 0 , \frac {1}{2} \right )$ $\displaystyle \left ( -\frac {1}{2} , 0\right ]$ $\left ( - \infty , \infty \right )$

Find the interval of convergence of the power series $\displaystyle \sum _{n=2}^\infty \frac {2^n (x+5)^n}{\sqrt [3]{n}}$ .

$\displaystyle \left [ -\frac {11}{2}, -\frac {9}{2} \right ]$ $\displaystyle \left [ -\frac {11}{2}, -\frac {9}{2} \right )$ $\displaystyle \left ( -\frac {11}{2}, -\frac {9}{2} \right )$ $\displaystyle \left [ \frac {9}{2}, \frac {11}{2} \right )$ $\displaystyle \left ( \frac {9}{2}, \frac {11}{2} \right )$ $\displaystyle \left [ \frac {9}{2}, \frac {11}{2} \right ]$