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Exercises relating to power series and their convergence properties.

Compute the radius of convergence of the power series below.
When computing a limit like 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.
When computing a limit like remember that we already know $n^{1/n} \rightarrow 1$ and that $\ln n <\!\! as well, so we expect $(\ln n)^{1/n} \rightarrow 1$ as well.
Compute the radius of convergence of the power series below.
Compute the interval of convergence for the power series below. 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:
Differentiate the series below term-by-term: (Note that the $n=0$ term goes away because the derivative of a constant is zero.)
Integrate the series below term-by-term:
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 to derive a summation formula for $1/(1+x^2)$, i.e.,
• Use the formula you derived above to develop a power series expansion for arctangent:
• 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 to determine the sum of the series below for $-1 < x < 1$: (Don’t forget absolute values if you need them.)
• Use the formula you just derived to determine the sum of the series
• Use the formula you just derived to determine the sum of the series

### Sample Quiz Questions

Find the full interval of convergence for the power series (Hints will not be revealed until after you choose a response.)

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

Find the full interval of convergence for the power series (Hints will not be revealed until after you choose a response.)

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

Find the full interval of convergence for the power series (Hints will not be revealed until after you choose a response.)

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

Find the full interval of convergence for the power series

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

### 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.

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

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

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