Does this summation require advanced techniques?

We have, $S = \sum_{n=1}^{\infty}\dfrac{1}{\binom{2n}{n}}$ $=\sum_{n=1}^{\infty}\dfrac{n!n!}{(2n)!}$ $=\sum_{n=1}^{\infty}\dfrac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+1)}$ $=\sum_{n=1}^{\infty}\dfrac{(2n+1)\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}$ $=\sum_{n=1}^{\infty}(2n+1)\beta(n+1,n+1)$ $=\sum_{n=1}^{\infty}(2n+1)\displaystyle \int_{0}^{1}x^{n}(1-x)^{n}\text{d}x$ As ,all terms are finite and continuous independently, we can switch the summation and the integral to give: $S = \displaystyle \int_{0}^{1}\left[\sum_{n=1}^{\infty}(2n+1)x^{n}(1-x)^{n}\right]\text{d}x$ Let $x(1-x) = t$ Consider the sum (second order GP) .Here $|t|<1$ $S'= \sum_{n=1}^{\infty}(2n+1)t^{n}$ $S'=\dfrac{t(3-t)}{(t-1)^{2}}$ The above was solved using basic method of solving a second order GP. Finally we have, $S = \displaystyle \int_{0}^{1} \dfrac{(x(1-x))[3-x(1-x)]}{[1-x(1-x)]^{2}} \text{d}x$ Evaluating the above integral is left as an exercise to the reader( using partial fraction decomposition), we arrive at the answer: $S= \dfrac{9+2\sqrt{3}\pi}{27}$ Giving the answer to be: $a+b+c+d =9+2+3+27 =\boxed{41}$

Starting with the known power series expansion $f(x) \; = \; \frac{\ln\big(x + \sqrt{1 + x^2}\big)}{\sqrt{1+x^2}} \; = \; \sum_{k=0}^\infty (-1)^k \frac{2^{2k}(k!)^2}{(2k+1)!} x^{2k+1}$ valid for all $|x| < 1$ , we have $f'(x) \; = \; \frac{1}{1+x^2} - \frac{x\ln\big(x + \sqrt{1 + x^2}\big)}{(1+x^2)^{\frac32}} \; = \; \sum_{k=0}^\infty (-1)^k \frac{2^{2k}}{\binom{2k}{k}} x^{2k}$ and hence the sum is $S \; = \; f'\big(\tfrac12i\big) - 1 \; = \; \frac43 - \frac{4i}{3\sqrt{3}}\ln\big(\tfrac12\sqrt{3} + \tfrac12i\big) \; = \; \frac{9 + 2\pi\sqrt{3}}{27}$ giving the answer $\boxed{41}$ .