TITLE: Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $
QUESTION [5 upvotes]: Mathematica gives away the interesting sum:
$$\sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac{2\pi}{3\sqrt{3}}$$
The question is: How to prove it by hand?
Remark. This question is self-answered (whence the OP provided an effort).  See the answer below.

REPLY [5 votes]: Note the integral representation of the reciprocal of the binomial co-efficient:
$${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx.$$
Then
$$S=\sum_{k=0}^{\infty}  \frac{1}{2k+1}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} [x(1-x)]^{k}= \int_{0}^{1} \frac{1}{x^2-x+1} dx $$ $$ \implies S = \frac{2}{\sqrt{3}}\tan^{-1} \frac{2x-1}{\sqrt{3}}\Bigg|_{0}^{1}=\frac{2\pi}{3\sqrt{3}}.$$