<s>[INST] Proving a binomial sum identity 
QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true?
$$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$

POSTSCRIPT. I like to record this presentable form by Alexander Burstein:
$$\sum_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$
 [/INST] Let, $f(x)=\frac{\sqrt{\pi}}{\Gamma(x)}$, The values of $f(x)$ at $x=-1,-2,...-N$ points are $y_{-n}=\frac{\binom{2n}{n}}{(-4)^n}$.
Now, for $N+1$ points $n=0,-1,-2,-3,..,-N$ we define $F_N(x):=\frac{\sqrt{\pi}}{\Gamma(x+\frac{1}{2})}N^x$ and $y_N(-n)=\frac{n!\binom{2n}{n}N^{-n}}{(-4)^n}$.
Hence, from Lagrange's interpolation:
$W(x)\sum_{n=0}^{N} \frac{1}{(n+x)(-1)^nn!(N-n)!}y_N(-n)≈F_N(x)$
[ Here, $W(x)=\prod_{n=0}^{N} (x+n)$ ]
or, $\frac{W(x)}{N!}\sum_{n=0}^{N} \frac{\binom{2n-1}{n}}{2^{2n-1}}N(N-1)..(N-n+1)N^{-n}≈F_N(x)$
[$2\binom{2n-1}{n}:=1$ as we get from the previous step]
Now, tending $N \to \infty$, and using $\lim\limits_{N \to \infty} \frac{W(x)}{N!N^x}=\frac{1}{\Gamma(x)}$
Hence, we get $\frac{1}{\Gamma(x)}\left(\frac{1}{x}+\sum_{n=0}^{\infty} \frac{\binom{2n+1}{n+1}}{2^{2n-1}(n+1+x)}\right)=\frac{\sqrt{\pi}}{\Gamma(x+\frac{1}{2})}$
The error term would go to zero as $N$ is increased when $x>-N$. This proves the identity.
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