TITLE: Find the sum, if exists $\sum\limits_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2(n+1)}$
QUESTION [6 upvotes]: $$
\sum\limits_{n=1}^{\infty}\dfrac{(2n)!}{2^{2n}(n!)^2(n+1)}
$$
By comparison test this series converges. Any nice way to work the sum?

I see that this can be written as:
$$
\sum\limits_{n=1}^{\infty}\dfrac{\binom{2n}{n}}{2^{2n}(n+1)}
$$

REPLY [10 votes]: Consider the following expansion:
$$\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty} {2n \choose n}x^n$$
Integrate both the sides under the limit $0$ to $x$ i.e
$$-\frac{1}{2}\left(\sqrt{1-4x}-1\right)=\sum_{n=0}^{\infty} {2n \choose n}\frac{x^{n+1}}{n+1} \Rightarrow -\frac{1}{2x}\left(\sqrt{1-4x}-1\right)=\sum_{n=0}^{\infty} {2n \choose n}\frac{x^{n}}{n+1}$$
Substitute $x=1/4$ to get:
$$\sum_{n=0}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=2 \Rightarrow \sum_{n=1}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=1$$