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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.

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

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  • $\begingroup$ What does the $n \geq 1$ on the bottom of the summation mean? $\endgroup$ – mrnovice Feb 25 '17 at 20:56
  • $\begingroup$ @mrnovice: $$\sum_{n\geq 1}f(n) = \sum_{n=1}^{+\infty}f(n) = f(1)+f(2)+f(3)+\ldots $$ $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 20:57
  • $\begingroup$ Ah ok, so it's what I thought thanks $\endgroup$ – mrnovice Feb 25 '17 at 20:57
  • $\begingroup$ If you start with $\color{#f00}{n = 0}$ you'll get $\color{#f00}{\pi}$. $\endgroup$ – Felix Marin Feb 25 '17 at 21:28

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