# Evaluating:$\sum_{n=0}^\infty\frac1{\binom{2n}{n}}$ [closed]

How to evaluate:
$$\sum_{n=0}^\infty\frac1{\binom{2n}{n}}$$
$\binom{n}{r}$ is the binomial coefficient.

If possible, present different methods as well.

Hint. One may observe that, by integrating by parts (see here), one has $$\frac1{\binom{2n}{n}}=\frac{(2n+1)}{2^{2n}}\int_0^1(1-x^2)^ndx$$ giving $$\sum_{n=0}^\infty\frac1{\binom{2n}{n}}=4\int_0^1\frac{\left(5-x^2\right)}{\left(x^2+3\right)^2}dx$$ then the integral may classically be evaluated by partial fraction decomposition giving
$$\sum_{n=0}^\infty\frac1{\binom{2n}{n}}=\frac43+\frac{2\pi}{27}\sqrt{3}.$$
Hint: In general, $$~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2} ~=~ 2\arcsin^2x.~$$ By twice differentiating-and-then- multiplying with regard to x, we arrive at the desired result.
We have $$1+\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+\frac{1}{70}+\frac{1}{252}+\cdots=\frac{4}{3}+\frac{2\pi}{9\sqrt{3}}.$$ A proof of this can be found at Sprugnoli.