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Question

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

n=0(1)n1(2n+1)2(zi)2n\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{(2 n+1)^{2}}(z-i)^{2 n}

Solution

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Answered 2 years ago
Answered 2 years ago
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The series has center ii.

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

(1)n+11(2n+3)2(zi)2n+2(1)n1(2n+1)2(zi)2n=(2n+1)2(zi)2(2n+3)2=2n+322n+32zi2=(122n+3)2zi2nzi2\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

zi2<1    zi<1\abs{z-i}^2 < 1 \implies \abs{z-i} < 1

and diverges if

zi>1\abs{z-i} > 1

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

zi<1\abs{z-i}< 1

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