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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.

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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

$$ \sum_{n=0}^\infty\frac1{\binom{2n}{n}}=\frac43+\frac{2\pi}{27}\sqrt{3}. $$

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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.

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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.

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