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 $N$ 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 vanish for large $N$. This proves the identity.