<s>[INST] Peculiar Sum regarding the Reciprocal Binomial Coefficients Whilst playing around on Wolfram Alpha, I typed in the sum
$$\sum_{x=0}^\infty \frac{1}{\binom{2x}{x}}=\frac{2}{27}(18+\pi\sqrt 3)$$
I'm not sure how to derive the answer. My first instinct was to expand the binomial coefficient to get
$$\sum_{x=0}^\infty \frac{x!^2}{(2x)!}$$
and then to try using a Taylor Series to get the answer. I thought that if I could find a function $f(n)$ with
$$f(n)=\sum_{x=0}^\infty \frac{x!^2n^x}{(2x)!}$$
Then my sum would be equal to $f(1)$. How do I find such a function?
EDIT: I continued on this path and realized that I can use this to set up a recurrence relation for $f^{(x)}(0)$:
$$f^{(0)}(0)=1$$
$$f^{(x)}(0)=\frac{x^2}{2x(2x-1)}f^{(x-1)}(0)$$
However, I'm not sure how this helps me find $f(1)$...
Am I on the right track? Can somebody help me finish what I started, or point me towards a better method of calculating this sum?
Thanks!
 [/INST] \begin{eqnarray*}
\binom{2n}{n} ^{-1} = \frac{2n+1}{2^{2n+1}} \int_{-1}^{1} (1-x^2)^n dx  
\end{eqnarray*}
Substitute this for summand and inerchange the order of the integral and sum.
\begin{eqnarray*}
 \int_{-1}^{1} \sum_{n=0}^{ \infty} \frac{2n+1}{2^{2n+1}}  (1-x^2)^n dx  &=& \frac{1}{2} \int_{-1}^{1}   \left(2\frac{(\frac{1-x^2}{4})}{(1-(\frac{1-x^2}{4}))^2}+ \frac{1}{1-(\frac{1-x^2}{4})} \right) dx \\
 = \int_{-1}^{1}   \frac{16}{(3+x^2)^2} dx - \int_{-1}^{1}   \frac{2}{(3+x^2)} dx
\end{eqnarray*}
Now use the standard integrals
\begin{eqnarray*}
 \int_{-1}^{1} \frac{1}{(3+x^2)} dx =  \frac{ \pi}{3 \sqrt{3}} \\ \int_{-1}^{1} \frac{1}{(3+x^2)^2} dx = \frac{ 1}{12} + \frac{ \pi}{18 \sqrt{3}} \\
\end{eqnarray*}
and the result follows.
</s>