Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $

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 $\dfrac{2\pi}{3\sqrt3}$.

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 $$S=\sum_{k=0}^{\infty} \frac{1}{2k+1}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} [x(1-x)]^{k}= \int_{0}^{1} \frac{1}{x^2-x+1} dx $$ $$ \implies S = \frac{2}{\sqrt{3}}\tan^{-1} \frac{2x-1}{\sqrt{3}}\Bigg|_{0}^{1}=\frac{2\pi}{3\sqrt{3}}.$$