
> 
> **QUESTION.** Let $x>0$ be a real number or an indeterminate. Is this true?
> $$\sum\_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$
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**POSTSCRIPT.** I like to record this presentable form by Alexander Burstein:
$$\sum\_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$



Proving a binomial sum identity
\begin{align}\sum\_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}&= \int\_0^1\sum\_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}y^{n+x}}{2^{2n+2}}\,{\rm d}y\\&=\int\_0^1 y^{x-1}\big((1-y)^{-1/2}-1\big){\rm d}y\\&=B\left(x,\frac12\right)-\frac1x\end{align}
and the rest follows from the properties of [beta function](https://en.wikipedia.org/wiki/Beta_function).


