2
$\begingroup$

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.

  • 0
    Something is wrong: $\frac{\Gamma(n)}{\Gamma(n+1/2)}$ behaves like $\frac{1}{\sqrt{n}}$ for large $n$s, but due to the presence of the $2^n$ factor the given series is divergent. – 2017-02-02
  • 0
    The sequence behaves as $\frac{2^n}{n^\frac32}$ which goes to $0$ for large $n$-s. – 2017-02-02
  • 0
    No, $2^n / n^{3/2}$ diverges as $n\to +\infty$. There should probably be a $\large{2^{-n}}$ factor. – 2017-02-02
  • 0
    @aimtech1 No, that sequence does not approach $0$. – 2017-02-02
  • 0
    The identity can be rewritten as $~\displaystyle\sum_{n=0}^\infty\dfrac{(2x)^{2n}}{\displaystyle(2n+1)~{2n\choose n}}~=~\dfrac{\arcsin x}{x\sqrt{1-x^2}},~$ which is no doubt related to the more famous $~\displaystyle\sum_{n=1}^\infty\dfrac{(2x)^{2n}}{\displaystyle{2n\choose n}~n^2}~=~2\arcsin^2x.~$ – 2017-02-02
  • 0
    @Lucian ... which does not exist for $x=1$ (or $|x|\ge 1$). – 2017-02-02
  • 0
    @Dr.MV: Did I say that it does ? :-$)$ – 2017-02-02
  • 0
    @Lucian No. But the OP's series has $x=1$. – 2017-02-02

2 Answers 2

5

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

  • 0
    I was just about to write a solution using exactly that way forward. (+1) for your speed. – 2017-02-02
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.