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I stumbled upon (in the literature) two identities for $\pi$, but they were not referenced as they are probably well-known. Hoping someone could point out who found them first.

Basically, the relations are: $$\sum_{p=0}^{\lceil M/2\rceil-1}\frac{2 (M!)^2}{(2p+1)^2(M-2p-1)!(M+2p+1)!}=\sum_{m=1}^M\frac{2^{2m-1}((m-1)!)^2}{(2m)!}=\frac{\pi^2}{4}$$

Mike Pierce
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science404
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    Hint: the sums are rational numbers while $\pi^2/4$ is not. – Dilip Sarwate Nov 24 '15 at 14:22
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    @DilipSarwate But in the second one, taking $ M\rightarrow\infty$ does give $ \pi^2/4 $, according to Mathematica. Probably this is what was intended? – Steven Charlton Nov 24 '15 at 14:26
  • Ok now I'm not sure if they are indeed converging to $\pi^2/4$, but I like where this is going... just looking for more information, so pointing out if I'm wrong is of course welcome – science404 Nov 24 '15 at 15:55
  • @StevenCharlton indeed, I find that it approaches $\pi^2/4$ in Mathematica, but I'm not sure how it computes this series. Matlab can only compute it to $M=80$ or so.. – science404 Nov 24 '15 at 16:18
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    The second sum can be written $\sum_{m=1}^\infty \frac{2^{2m-1}}{m^2{2m\choose m}} = \arcsin^2(1) = \frac{\pi^2}{4}$. See this answer and refs within for how to derive it. – Winther Nov 24 '15 at 18:43

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