# Why does $\sum\limits_{n=1}^{\infty } 2^{2n} \times \frac{(n!)^2}{n(2n+1)(2n)!} =2$ ?

In my Calculus class, we were solving an integration by parts problem two different ways. One way gave an answer pretty easily, the other led to an infinite series that stumped us. After playing around with it for a long time, I came to this fact:

$\sum\limits_{n=1}^{\infty } 2^{2n} \times \frac{(n!)^2}{n(2n+1)(2n)!} =2$

How can I prove that this is true? It would be ideal if there is a way to solve this using only math that students in AP Calculus BC would have seen. I realize that may not be possible, though. Thank you!

Observe that
$\frac{(n!)^2}{(2n+1) (2n)!}=\frac{n! n!}{(2n+1) (2n)!}=\frac{1}{(2n+1){2n \choose n}}.$
Now the idea is to use the Taylor series expansion for $\arcsin^2 x$. Indeed

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

Replace $x\mapsto\sqrt x$ to get

$$2\arcsin^2\left(\sqrt x\right) = \sum_{n=1}^\infty \frac{2^{2n} x^n}{n^2 \binom{2n}n}.$$

Differentiate both side:

$$\frac{2\arcsin\left(\sqrt x\right)}{\sqrt x\sqrt{1-x}} = \sum_{n=1}^\infty \frac{2^{2n} x^{n-1}}{n \binom{2n}n}.$$

Now multiply both sides by $x$ to obtain

$$\frac{2\sqrt x\arcsin\left(\sqrt x\right)}{\sqrt{1-x}} = \sum_{n=1}^\infty \frac{2^{2n} x^n}{n \binom{2n}n}.$$

Replace $x\mapsto x^2$ to get

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

Thus for $0<x<1$ we have
$$\sum_{n=1}^\infty \frac{2^{2n} x^{2n+1}}{n(2n+1) \binom{2n}n} = \int_0^x \frac{2s\arcsin(s)}{\sqrt{1-s^2}} \, ds.$$

Letting $x\to1^-$ we get

$$\sum_{n=1}^\infty \frac{2^{2n} (n!)^2}{n (2n+1)!} = \sum_{n=1}^\infty \frac{2^{2n}}{n(2n+1)\binom{2n}n} = \int_0^1 \frac{2x\arcsin(x)}{\sqrt{1-x^2}} \, dx$$
$=-2\arcsin (x) (\sqrt{1-x^2})|_0^1+2\int_0^1(\sqrt{1-x^2})\frac{1}{(\sqrt{1-x^2})}dx$
$=0+2=2,$
where we have applied integration by parts to compute the last integral.

• This is absolutely beautiful. Thank you so much! This was driving me insane.

• Thanks! I am glad I was able to help.