3

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 $(\sin^{-1} x)^2$.

Rene Schipperus
  • 39,526
  • May be by considering a generating series for the inverse of central binomial coefficients like here. – Jean Marie Nov 07 '21 at 19:46
  • 1
    @Rene Schipperus, Could you please elaborate what markoff transformation mean or post a reference to it? – RAHUL Nov 07 '21 at 19:49
  • 1
    @RAHUL Basically if you have a rectangular array you can sum it in two ways, columns first or rows first. This is dealt with in ch VIII of Knopp's Infinite series, where an application to this problem is given. – Rene Schipperus Nov 07 '21 at 19:54

1 Answers1

3

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.

TheSilverDoe
  • 29,720