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Given the following series: $$\frac{2^2}{3.4}x^4 + \frac{2^2.4^2}{3.4.5.6}x^6+\cdots = \sum a_n,$$ where $$a_n = \frac{2^2.4^2\cdots(2n)^2}{3.4.5.6\cdots (2n+2)} x^{2n+2}$$

Now we try Ratio test first: $$\frac{a_{n+1}}{a_n} = \frac{2n^2+4n+2}{n+2}x^2$$

but then I cant conclude from that, how to proceed next?

User8976
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1 Answers1

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Apply Ratio Test $$ \begin{align} \frac{2^2}{3\cdot4}x^4 + \frac{2^2\cdot4^2}{3\cdot4\cdot5\cdot6}x^6+\cdots &=\sum_{k=2}^\infty\frac{(k-1)!^22^{2k-2}}{(2k)!/2!}x^{2k}\\ &=\frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}}\\ \end{align} $$ The Ratio Test says $$ \begin{align} \lim_{k\to\infty}\frac{a_{k+1}}{a_k} &=\lim_{k\to\infty}\frac{k^2}{(k+1)^2}\frac{4x^2}{\frac{(2k+2)(2k+1)}{(k+1)^2}}\\ &=\lim_{k\to\infty}\frac{4k^2}{(2k+2)(2k+1)}x^2\\[9pt] &=x^2 \end{align} $$ Therefore, the series converges for $|x|\lt1$.

The Case $\boldsymbol{|x|=1}$

Using the estimate $(10)$ given in this answer, we have $$ \begin{align} \frac12\frac{4^k}{k^2\binom{2k}{k}} &\le\frac12\frac{4^k}{k^2\frac{4^k}{\sqrt{\pi(k+1/3)}}}\\ &=\frac{\sqrt{\pi(k+1/3)}}{2k^2}\\[9pt] &\le\frac{\sqrt{\pi/3}}{k^{3/2}} \end{align} $$ So the series converges for $|x|=1$.

Applying Raabe's Test

Another approach for the case $|x|=1$ is to use Raabe's Test: $$ \begin{align} \lim_{k\to\infty}k\left(\frac{a_k}{a_{k+1}}-1\right) &=\lim_{k\to\infty}k\left(\frac{(2k+2)(2k+1)}{4k^2}-1\right)\\ &=\lim_{k\to\infty}k\left(\frac{6k+2}{4k^2}\right)\\[3pt] &=\frac32 \end{align} $$ which is greater than $1$, so the series converges for $|x|=1$.

The Actual Sum

Applying the substitution $x\mapsto4x^2$ to this answer gives $$ \sum_{k=0}^\infty\frac{(2x)^{2k}}{\binom{2k}{k}} =\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right] $$ Integrating once gives $$ \frac12\sum_{k=0}^\infty\frac{(2x)^{2k+1}}{(2k+1)\binom{2k}{k}} =\frac{\sin^{-1}(x)}{\sqrt{1-x^2}} $$ Integrating again gives $$ \frac14\sum_{k=0}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}} =\frac{\sin^{-1}(x)^2}{2} $$ Therefore, $$ \begin{align} \frac12\sum_{k=2}^\infty\frac{(2x)^{2k}}{k^2\binom{2k}{k}} &=\frac12\sum_{k=1}^\infty\frac{(2x)^{2k+2}}{(2k+2)(2k+1)\binom{2k}{k}}\\[6pt] &=\sin^{-1}(x)^2-x^2 \end{align} $$

robjohn
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