Proving $\pi^2=18\sum_{n=0}^{\infty}\frac{n!n!}{(2n+2)!}$

Question

How to prove $$\pi^2=18\sum_{n=0}^{\infty}\frac{n!n!}{(2n+2)!}$$

I saw this as an exercise in Hobson's Treatise on Plane Trigonometry, pg.297. The $\pi^2$ has me flustered, I assume that there is some power series that gives this, but I could not find any promising candidates in the book.

Update:This can be proven by applying a Markoff transformation to $\sum\frac{1}{n^2}$.

Second Update: Can also be proven from the series for $(\sin^{-1} x)^2$.

Answer 1

One has $$S=\sum_{n=0}^{\infty}\frac{n!n!}{(2n+2)!} = \sum_{n=0}^{\infty}\frac{1}{(2n+2)(2n+1)\binom{2n}{n}}$$

Now you have the following Maclaurin expansion : $$\frac{\arcsin{x}}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2 n} x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$

so $$S=2 \int_0^1 \frac{\arcsin{(\frac{x}{2})}}{\sqrt{1-(\frac{x}{2})^2}} dx = 2 \left[ \arcsin^2\left(\frac{x}{2}\right)\right]_0^1 = 2 \times\left( \frac{\pi}{6} \right)^2= \frac{\pi^2}{18}$$

and you are done.