Why does $ \sum\limits_{n=1}^{\infty } 2^{2n} \times \frac{(n!)^2}{n(2n+1)(2n)!} =2 $ ?
$ \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!
Answer
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:
$$\frac{2\arcsin\left(\sqrt x\right)}{\sqrt x\sqrt{1x}} = \sum_{n=1}^\infty \frac{2^{2n} x^{n1}}{n \binom{2n}n}.$$
Now multiply both sides by $x$ to obtain
$$\frac{2\sqrt x\arcsin\left(\sqrt x\right)}{\sqrt{1x}} = \sum_{n=1}^\infty \frac{2^{2n} x^n}{n \binom{2n}n}.$$
Replace $x\mapsto x^2$ to get
$$\frac{2x\arcsin(x)}{\sqrt{1x^2}} = \sum_{n=1}^\infty \frac{2^{2n} x^{2n}}{n \binom{2n}n}$$
Thus for $0<x<1$ we have
$$\sum_{n=1}^\infty \frac{2^{2n} x^{2n+1}}{n(2n+1) \binom{2n}n} = \int_0^x \frac{2s\arcsin(s)}{\sqrt{1s^2}} \, ds.$$
Letting $x\to1^$ we get
$$\sum_{n=1}^\infty \frac{2^{2n} (n!)^2}{n (2n+1)!} = \sum_{n=1}^\infty \frac{2^{2n}}{n(2n+1)\binom{2n}n} = \int_0^1 \frac{2x\arcsin(x)}{\sqrt{1x^2}} \, dx $$
\[=2\arcsin (x) (\sqrt{1x^2})_0^1+2\int_0^1(\sqrt{1x^2})\frac{1}{(\sqrt{1x^2})}dx\]
\[=0+2=2,\]
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.
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