infinite sum with $\Gamma$ function

Question

Prove that: $$\sum_{n=1}^{\infty}\frac{2^{n}\Gamma(n)}{(2n+1)\Gamma\left(n+\frac12\right)}=2\left(\sqrt{\pi}-\frac{2}{\sqrt{\pi}}\right)$$

I tried the generating function $$f(x)=\sum_{n=1}^{\infty}\frac{2^{n}\Gamma(n)}{(2n+1)\Gamma\left(n+\frac12\right)}x^{2n+1}$$ By differentiating we get $$f'(x)= \sum_{n=1}^{\infty}\frac{2^{n}\Gamma(n)}{\Gamma\left(n+\frac12\right)}x^{2n}$$ But this leads nowhere for me.

Answer 1

If the factor $2^n$ (clearly leading to a divergent series) is replaced by $2^{-n}$ we have: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{2^{-n} \Gamma(n)}{2\,\Gamma(n+3/2)}&=&\frac{1}{\sqrt{\pi}}\sum_{n\geq 1}2^{-n}\,B(n,3/2)\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{1}\sum_{n\geq 1}2^{-n} (1-x)^{1/2}x^{n-1}\,dx\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{1}\frac{\sqrt{1-x}}{2-x}\,dx\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{1}\frac{\sqrt{x}}{1+x}\,dx\\&=&\frac{2}{\sqrt{\pi}}\int_{0}^{1}\frac{z^2}{1+z^2}\,dz\\&=&\frac{2}{\sqrt{\pi}}\left(1-\frac{\pi}{4}\right).\end{eqnarray*} $$

Answer 2

We have $$\sum_{n\geq1}\frac{\Gamma\left(n\right)}{2^{n}\Gamma\left(n+1/2\right)\left(2n+1\right)}=\frac{1}{\sqrt{\pi}}\sum_{n\geq1}\frac{\left(n-1\right)!}{\left(2n+1\right)!!} $$ $$=\frac{1}{\sqrt{\pi}}\sum_{n\geq1}\frac{2^{n}\left(n!\right)^{2}}{n\left(2n+1\right)!}=\frac{1}{\sqrt{\pi}}\sum_{n\geq1}\frac{2^{n}}{n\left(2n+1\right)\dbinom{2n}{n}} $$ and since $$\sum_{n\geq0}\frac{4^{n}x^{2n}}{\left(2n+1\right)\dbinom{2n}{n}}=\frac{\arcsin\left(x\right)}{x\sqrt{1-x^{2}}} $$ we get $$S=\frac{1}{\sqrt{\pi}}\sum_{n\geq1}\frac{2^{n}}{n\left(2n+1\right)\dbinom{2n}{n}}=\frac{2}{\sqrt{\pi}}\int_{0}^{1/\sqrt{2}}\left(\frac{\arcsin\left(x\right)}{x^{2}\sqrt{1-x^{2}}}-\frac{1}{x}\right)dx $$ $$=\frac{2}{\sqrt{\pi}}\lim_{\epsilon\rightarrow0^{+}}\left(\int_{\arcsin\left(\epsilon\right)}^{\pi/4}\frac{u}{\sin^{2}\left(u\right)}du-\log\left(\frac{1}{\sqrt{2}}\right)+\log\left(\epsilon\right)\right) $$ and since we have, integrating by parts, $$\int\frac{u}{\sin^{2}\left(u\right)}du=\log\left(\sin\left(u\right)\right)-\frac{u}{\tan\left(u\right)}+C $$ we obtain $$S=\color{red}{\frac{2}{\sqrt{\pi}}\left(1-\frac{\pi}{4}\right)}.$$ as wanted.