\sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{2 \pi}{9\sqrt{3}}+\frac43

Today we will evaluate the following nice infinte sum

$\sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{2 \pi}{9\sqrt{3}}+\frac43$

First recall the expansion (proved here)

$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n-1}}{2n}$ (1)

We can rewrite (1) as

$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n+1}}{\left(2n+1\right)}$ (2)

Proof:

$\begin{aligned} \frac{\arcsin(x)}{\sqrt{1-x^2}}&=\frac12\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n-1}}{n}\\ &=\frac42\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n+2\\ n+1\end{array}\right)}\frac{x^{2n+1}}{n+1} \qquad (n \to n+1 )\\ &=\frac42\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{1}{\frac{(2n+2)(2n+1)}{(n+1)^2}}\frac{x^{2n+1}}{(n+1)}\\ &=\frac42\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n+1}}{\frac{2(n+1)(2n+1)}{(n+1)}}\\ &=\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n+1}}{\left(2n+1\right)} \qquad \blacksquare\\ \end{aligned} \\$

If we differentiate both sides of (2) w.r. to x we obtain

$\sum_{n=0}^\infty\frac{ 4^{n}x^{2n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{1}{1-x^2}+\frac{x \arcsin(x)}{\left(1-x^2\right)^{3/2}}$ (3)

Proof:

$\begin{aligned} \sum_{n=0}^\infty\frac{ 4^{n}x^{2n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}&=\frac{\left(\arcsin(x) \right)^\prime\sqrt{1-x^2}-\arcsin(x)\left(\sqrt{1-x^2}\right)^\prime}{\left(\sqrt{1-x^2}\right)^2}\\ &=\frac{1}{1-x^2}+\frac{x \arcsin(x)}{\left(1-x^2\right)^{3/2}} \qquad \blacksquare \end{aligned} \\$

Letting $x=\frac12$ in (3) we obtain

$\begin{aligned} \sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)} &=\frac{1}{1-\frac14}+\frac12\frac{ \arcsin\left(\frac12\right)}{\left(1-\frac14\right)^{3/2}}\\ &=\frac43+\frac1{12}\frac{ \pi}{\frac{\sqrt{27}}{8}}\\ &=\frac{2 \pi}{9\sqrt{3}}+\frac43 \qquad \blacksquare \end{aligned} \\$

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