An interesting infinite sum involving factorial

Question

This infinite sum is not like anything else I have seen. I wonder if it can be computed by Taylor series of a function evaluated at some point.

$$ \sum_{n=1}^{\infty } \frac{2^{2n}(n!)^2}{n(2n+1)!}. $$ I expect it to be equal to 2.

Answer 1

$\textbf{Hint:}$

Given $ n\in\mathbb{N} $, we can prove that : $$ \frac{2^{2n}\left(n!\right)^{2}}{\left(2n+1\right)!}=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x} $$

Now exchanging the sum and the integral signs, then using taylor series expansion for $ \ln $, gives the following : \begin{aligned} \sum_{n=1}^{+\infty}{\frac{2^{2n}\left(n!\right)^{2}}{n\left(2n+1\right)!}}&=-2\int_{0}^{1}{\ln{x}\,\mathrm{d}x}\end{aligned}

Answer 2

Starting with

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

Substitute $x\mapsto\sqrt x$.

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

Differentiate.

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

Multiply both sides by $x$.

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

Substitute $x\mapsto x^2$.

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

Now integrate. By the fundamental theorem of calculus, noting that both sides are $0$ when $x=0$, we have

$$\sum_{n=1}^\infty \frac{2^{2n} x^{2n+1}}{n(2n+1) \binom{2n}n} = \int_0^x \frac{2t\arcsin(t)}{\sqrt{1-t^2}} \, dt$$

Let $x\to1$ from below. Then the given sum is equivalent to the integral (can be done by parts)

$$\sum_{n=1}^\infty \frac{2^{2n} (n!)^2}{n (2n+1)!} = \sum_{n=1}^\infty \frac{2^{2n}}{n(2n+1)\binom{2n}n} = \int_0^1 \frac{2x\arcsin(x)}{\sqrt{1-x^2}} \, dx = 2$$

as you suspected.