sequences and series - Find the sum, if exists $sumlimits_{n=1}^{infty} frac{(2n)!}{2^{2n}(n!)^2(n+1)}$

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

By comparison test this series converges. Any nice way to work the sum?

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

Answer

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: $$\sum_{n=0}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=2 \Rightarrow \sum_{n=1}^{\infty} \frac{{2n \choose n}}{n+1}\frac{1}{4^n}=1$$