# Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$

Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$

I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$

I have no idea about it. :|

Consider the function

$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$

$f(x)$ has a Maclurin expansion as follows:

$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$

Differentiating, we get

$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$

Evaluate at $x=1/2$

$$\sum_{n=0}^{\infty} \frac{1}{\displaystyle \binom{2 n}{n}} = \frac{\frac12 \arcsin{\frac12}}{3 \sqrt{3}/8} + \frac{4}{3} = \frac{2\sqrt{3} \pi+36}{27}$$

There are many derivations here of the above result for the Maclurin series for $f(x)$; I refer you to this one.

Recall the Euler Beta integral

$$\beta(a, b) = \int_0^1 x^{a-1} (1-x)^{b-1} dx$$

and Euler's formula for it in terms of the Gamma function,

$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$

In particular, since ${2n \choose n} = (2n!)/(n!)^2$, we have

$$\beta(n+1, n+1) = \frac{1}{(2n+1){2n \choose n}}.$$

Thus,

$$\sum_{n=0}^\infty \frac{1}{2n \choose n} = \int_0^1 \sum_{n=0}^\infty {(2n+1) (x(1-x))^{n}} dx$$

and we have the series $\sum_{n\geq 0} (2n+1)y^n = (y+1)/(y-1)^2$. Therefore,

$$\sum_{n=0}^\infty \frac{1}{2n \choose n} = \int_0^1 \frac{x(1-x)+1}{(x(1-x)-1)^2} dx$$

and by routine integration (partial fractions or your favorite standard method), this equals $\frac{2}{27}(18+\sqrt 3 \pi)$.

This paper is very relevant to your question. In particular, $\bf Theorems \;\;3.4-5$ and $\bf Theorem \;\;3.7$