Actually I have no idea how to begin. I tried to represent it as a product of two series or as a sum of two series but have no result with this. Any idea ?
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1$$ \sum_{2}^{\infty} \frac{n!^{2}2^{n+1}}{(2n)!} = \pi$$ – Nilotpal Sinha Oct 30 '17 at 10:11
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3$$\sum _{n=0}^{\infty } \frac{(n!)^2 x^n}{(2 n)!}=\frac{4 \left(\sqrt{4-x}+\sqrt{x} \arcsin\left(\frac{\sqrt{x}}{2}\right)\right)}{\sqrt{4-x} (4-x)}$$Good luck! – Raffaele Oct 30 '17 at 10:35
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May be the Taylor expansion of $\sin^{-1}(x)$ could give you some ideas. – Claude Leibovici Oct 30 '17 at 10:36
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1See https://mikespivey.wordpress.com/2016/07/26/generating-function-for-the-reciprocals-of-the-central-binomial-coefficients/ – Gabriel Romon Oct 30 '17 at 11:56
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$$\sum_{n\geq 0}\frac{x^n}{\binom{2n}{n}} = \sum_{n\geq 0}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\cdot (2n+1)x^n = \sum_{n\geq 0} B(n+1,n+1)\cdot (2n+1) x^n $$ can be written as $$ \int_{0}^{1} \sum_{n\geq 0} u^{n}(1-u)^n (2n+1) x^n\,du =\int_{0}^{1}\frac{1+u x-u^2 x}{\left(1-u x+u^2 x\right)^2}\,du$$ and the last integral can be computed by partial fraction decomposition, leading to $$ \sum_{n\geq 0}\frac{x^n}{\binom{2n}{n}}=\frac{4 \left(\sqrt{4-x}+\sqrt{x} \arcsin\frac{\sqrt{x}}{2}\right)}{\sqrt{4-x} (4-x)}$$ for any $x\in(0,4)$. An alternative is to start with the Taylor series of the squared arcsine $$ \arcsin^2(x) = \frac{1}{2}\sum_{n\geq 0}\frac{(4x^2)^n}{n^2\binom{2n}{n}} $$ to replace $x$ with $\frac{\sqrt{x}}{2}$, then to apply $\left(x\cdot\frac{d}{dx}\right)$ twice.
Jack D'Aurizio
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