Find this sum $$\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n)!}(2x)^{2n}.\qquad (-1\le x\le 1)$$

My idea: let $$f(x)=\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n)!}(2x)^{2n},$$ then we have $$f'(x)=2\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n-1)!}(2x)^{2n-1}$$ $$f''(x)=4\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n-2)!}(2x)^{2n-2}$$ and so on. Then I can't solve it. Thank you for your help.

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    $\begingroup$ This is the Taylor series for $2\arcsin^2x$. $\endgroup$ – Lucian Mar 13 '14 at 13:01
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    $\begingroup$ How would one know that? $\endgroup$ – abnry Mar 13 '14 at 13:03
  • $\begingroup$ Personally, I've been able to recognize the central binomial coefficient from my own dabbling into Newton's binomial series, as an alternative way of expressing $\displaystyle{\pm\frac12\choose n}$, which naturally appears in the expansion of $(1\pm x)^{^{\pm\tfrac12}}$, which immediately reminded me of the derivative of the $\arcsin$ and $\arccos$ functions. So I guess the answer to your question is practice. $\endgroup$ – Lucian Mar 13 '14 at 13:20
  • $\begingroup$ hello,@Lucian,How find it by hand? $\endgroup$ – user94270 Mar 13 '14 at 13:56

Here's a derivation of sorts. Let $f(x)$ be the sum of interest. Then

$$x \frac{d}{dx} \left [x \frac{d}{dx} f(x) \right ] = 4 \sum_{n=1}^{\infty} \frac{(2 x)^{2 n}}{\displaystyle \binom{2 n}{n}}$$

Referencing this solution, we see that

$$x \frac{d}{dx} \left [x \frac{d}{dx} f(x) \right ] = 4 \frac{x \arcsin{x}}{(1-x^2)^{3/2}} + \frac{4}{1-x^2}-4 $$

Divide by $x$ and integrate to get

$$x \frac{d}{dx} f(x) = 4 \int dx \frac{\arcsin{x}}{(1-x^2)^{3/2}} + 4 \int dx \left [\frac1{x (1-x^2)} - \frac1{x} \right ]$$

To evaluate the integrals, we may use a trig sub in the form of $x=\sin{\theta}$; the integrals become

$$4 \int d\theta (\theta \, \sec^2{\theta} + \csc{\theta} \sec{\theta}-\cot{\theta})$$

Integrate the first piece by parts, and this is equal to

$$4 \theta \tan{\theta} + 4 \underbrace{\int d\theta (\csc{\theta}\sec{\theta}-\cot{\theta}-\tan{\theta})}_{\text{Nice how this vanishes, huh?}}= 4 \theta \tan{\theta} +C $$

Setting $C=0$ from $f'(0)=0$, I get

$$f'(x) = 4 \frac{\arcsin{x}}{\sqrt{1-x^2}} $$

and therefore, noting that $f(0)=0$,

$$f(x) = 2 (\arcsin{x})^2$$


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