Proving that $\sum_{n=0}^{\infty }\frac{3(n!)^2}{(2n+2)!}=\sum_{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi ^2}{6}$

Question

Proving that $$\sum_{n=0}^{\infty }\frac{3(n!)^2}{(2n+2)!}=\sum_{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi ^2}{6}$$

I know the proving of second series which is very famous series to give us $\zeta(2)$, but I dont know how to prove the first series which is faster than second series, any help

Do you mean $3(n!)^2$ in the top of the fraction? Otherwise the sum is infinite. — ASCII Advocate, Sep 28 '15 at 23:01

score 10 · Accepted Answer · edited Apr 13 '17 at 12:20

10

Using the series expansion of $\arcsin^2(x)$ (see this question) given by

$$\arcsin^2(x) = \frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

we see that your sum is

$$\sum_{n=0}^\infty \frac{3(n!)^2}{(2n+2)!} = \sum_{n=1}^\infty \frac{3}{n^2{2n\choose n}} = 6\arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{6}$$

edited Apr 13 '17 at 12:20

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answered Sep 28 '15 at 23:17

Winther

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@Clayton We have $\frac{(n!)^2}{(2n+2)!} = \frac{1}{(n+1)^2{2(n+1)\choose n+1}}$ then I change the summation variable $n+1\to n$. – Winther Sep 28 '15 at 23:39
I see; thanks.${}$ – Clayton Sep 28 '15 at 23:40

Proving that $\sum_{n=0}^{\infty }\frac{3(n!)^2}{(2n+2)!}=\sum_{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi ^2}{6}$

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