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$$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx.$$ Someone please show that this equation is correct !?

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It is known that $$ \arcsin(x)^2=\frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}} $$ see Companion to Concrete Mathematics - Mathematical Techniques and Various Applications Z. A. Melzak p. 108

Then $$ \begin{aligned} \int_{0}^{1/2}\frac{4\arcsin(x)^2}{x} &=\int_{0}^{1/2}\frac{2}{x}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}dx\\ &=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \int_{0}^{1/2}x^{2n-1}dx\\ &=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \frac{\left(\frac{1}{2}\right)^{2n}}{2n}\\ &=\sum_{n=1}^\infty\frac{1}{n^3\binom{2n}{n}} \end{aligned} $$

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Hint: $~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\displaystyle{2n\choose n}}=2\arcsin^2x$.

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    $\begingroup$ Hintier: to derive the series representation above, express the arcsine function in terms of logarithm by $\arcsin{(x)}=\frac{1}{i}\ln{\left(iz+\sqrt{1-z^2}\right)}$. $\endgroup$ – David H Sep 28 '14 at 19:16
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    $\begingroup$ It follows from the fact that $\arcsin x=\displaystyle\int_0^x\frac{dt}{\sqrt{1-t^2}}$ by means of Newton's binomial series and Cauchy's product. $\endgroup$ – Lucian Sep 28 '14 at 19:23
  • $\begingroup$ Hrmm, that works too I suppose. :) $\endgroup$ – David H Sep 28 '14 at 19:39

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