# Proof that this series equals $\pi$

I was looking for series that sum to $$\pi$$, and I happened to come across this one:

$$\pi = \displaystyle\sum^{\infty}_{n=0}\frac{n!\left(2n\right)!\left(25n-3\right)}{2^{n-1}\left(3n\right)!}$$

Could anyone please tell me why this series does indeed sum to $$\pi$$? It just seems odd.

• I'm not familiar with the notation, is it a sum to infinity or? – mrnovice Feb 25 '17 at 20:41
• 1) You may have noticed that you have the inverse of $\binom{3n}{n}$. 2) Related: (math.stackexchange.com/q/1784843) – Jean Marie Feb 25 '17 at 22:08

We have to find the value of $$\begin{eqnarray*}\sum_{n\geq 1}\frac{n(50n-6)\,\Gamma(n)\,\Gamma(2n+1)}{2^n\,\Gamma(3n+1)}&=&\sum_{n\geq 1}\frac{n(50n-6)}{2^n}\int_{0}^{1}x^{n-1}(1-x)^{2n}\,dx\\&=&\int_{0}^{1}\frac{16\,(1-x)^2 \left(11+7 x-14 x^2+7 x^3\right)}{(2-x)^3 \left(1+x^2\right)^3}\,dx\end{eqnarray*}$$ that by partial fraction decomposition equals $\color{red}{\pi+6}$.
If you consider the original series over $n\geq 0$, you get $\pi$.
• What does the $n \geq 1$ on the bottom of the summation mean? – mrnovice Feb 25 '17 at 20:56
• @mrnovice: $$\sum_{n\geq 1}f(n) = \sum_{n=1}^{+\infty}f(n) = f(1)+f(2)+f(3)+\ldots$$ – Jack D'Aurizio Feb 25 '17 at 20:57
• If you start with $\color{#f00}{n = 0}$ you'll get $\color{#f00}{\pi}$. – Felix Marin Feb 25 '17 at 21:28