## Related questions with answers

Find the radius of convergence and open disk of convergence of the power series.

$\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{(2 n+1)^{2}}(z-i)^{2 n}$

Solution

VerifiedThe series has center $i$.

Let's look at the magnitude of the ratio of successive terms to find the radius of convergence. We have

$\begin{align*} & \abs{\dfrac{(-1)^{n+1}\dfrac{1}{(2n+3)^2}(z-i)^{2n+2}}{(-1)^n\dfrac{1}{(2n+1)^2}(z-i)^{2n}}} = \abs{\dfrac{(2n+1)^2(z-i)^2}{(2n+3)^2}}\\ &=\abs{\dfrac{2n+3-2}{2n+3}}^2\abs{z-i}^2 = \qty(1-\dfrac{2}{2n+3})^2\abs{z-i}^2 \stackrel{n\to \infty}{\to} \abs{z-i}^2 \end{align*}$

By the ratio test, the series converges if

$\abs{z-i}^2 < 1 \implies \abs{z-i} < 1$

and diverges if

$\abs{z-i} > 1$

Therefore, the radius of convergence is 1 and the open disk of convergence is

$\abs{z-i}< 1$

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