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Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP.

The identity holds for $x>0$ and also for $x<0$, unequal to a negative integer. It fails for $x=0$. The left-hand-side then evaluates to $2\ln 2$ while the limit $x\rightarrow 0$ of the right-hand-side is $-\gamma_{\text{Euler}}$.

Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP.

The identity holds for all real $x$ unequal to a negative integer. It holds in particular for $x=0$, when the sum equals $2\ln 2$.

Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP. This also holds for $x<0$, unequal to a negative integer.

Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP.

The identity holds for $x>0$ and also for $x<0$, unequal to a negative integer. It fails for $x=0$. The left-hand-side then evaluates to $2\ln 2$ while the limit $x\rightarrow 0$ of the right-hand-side is $-\gamma_{\text{Euler}}$.

Mathematica evaluates this as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP. This also holds for $x<0$, unequal to a negative integer.

Q: Is this identity true?

A: Yes, Mathematica evaluates it as $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{\sqrt{\pi }\, \Gamma (x)}{\Gamma \left(x+\frac{1}{2}\right)}-\frac{1}{x},$$ which is another way to write the answer in the OP. This also holds for $x<0$, unequal to a negative integer.