Summing Up Reciprocal of Central Binomial Coefficient

We differentiate the special case of Kishlaya's Identity to get $\frac{d}{dx}\left(\sum_{n=1}^\infty \frac{x^n}{n{2n \choose n}}\right) = \frac{d}{dx}\left(\sqrt{\frac{4x}{4-x}}\tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right)\right)$ $\Rightarrow \sum_{n=1}^\infty \frac{x^{n-1}}{{2n \choose n}} = \frac{4\tan^{-1}\sqrt{\frac{x}{4-x}}}{(4-x)^2\sqrt{\frac{x}{4-x}}} + \frac{1}{4-x}$

Setting $x=1$ and simplifying, yields our desired result

$\boxed{\sum_{n=1}^\infty \frac{1}{{2n \choose n}} = \frac{9+2\sqrt{3}\pi}{27}}$

My method is nowhere near as general as Kishlaya's as his wonderful identity is a far more powerful/general result. After reading through the wiki for Kishlaya's identity, solving this question using the identity seems like 'nuking a mosquito' to me :D. Which is why, I'd like to outline another approach:

Consider, $I(m,n)=\int_{0}^{1}x^m(1-x)^ndx$ Integrate it by parts once to get: $I(m,n)=\frac{n}{m+1}I(m+1,n-1)$ This step can be repeated for $I(m+1,n-1)$ . Again and again till we reach $I(m+n,0)$ which makes for trivial integration. Finally, what we get is: $I(m,n)=\int_{0}^{1}x^m(1-x)^ndx=\frac{1}{(m+n+1)\binom{m+n}{m}}$ . For this question, setting $m=n$ we get: $\int_{0}^{1}(x(1-x))^ndx=\frac{1}{(2n+1)\binom{2n}{n}}$ . Let, $\alpha=x(1-x)$ . Then, $\int_{0}^{1}(2n+1)\alpha^ndx=\frac{1}{\binom{2n}{n}}$ .

Summing from n=1 to $\infty$ will give us the required expression. Note that the L.H.S is the sum of a GP and an AP-GP in $\alpha$ .

Completing the proof isn't hard, but it is rather lengthy.