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.
Best Answer
I think this might help.
$$\frac{1}{n \choose k}=(n+1)\int_0^1 x^k(1-x)^{n-k}dx.$$ This is easily obtained by integration by parts.
Putting $n=2k$ and $x=\sin^2t$, we get $$\sum_{k\geq0}\int_0^{\frac{\pi}{2}}2\sin^{2k+1}t\cos^{2k+1}t \ dt.$$
Bringing the sum inside the integral and evaluating the G.P. give $$\int_0^{\frac{\pi}{2}}\frac{\sin( 2t)}{1-\frac{\sin^2(2t)}{4}}dt.$$
Putting $\cos(2t)=u$ We get, $$2\int_{-1}^{1}\frac{du}{3+u^2},$$ which is indeed equal to $\dfrac{2\pi}{3\sqrt3}$.