1
$\begingroup$

Could someone please provide a detailed derivation of the following result? $$\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+\frac 12) = \pi n!$$ Also, is it possible to generalize the result for the following sum, given a generic integer $m$? $$\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+m+\frac 12).$$

Thanks in advance for your help.

Graziano

$\endgroup$
  • $\begingroup$ Hint: beta function. In particular, $\Gamma(k+\frac12)\Gamma(n-k+\frac12) = n!\int_0^1 t^{k-\frac12}(1-t)^{n-k-\frac12} dt$ $\endgroup$ – achille hui Apr 22 '18 at 9:27
2
$\begingroup$

Since $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(1/2)=\sqrt{\pi}$, we have that for any non-negative integer $N$, $$\Gamma\left(N+\frac{1}{2}\right)=\Gamma\left(\frac{1}{2}\right)\prod_{k=0}^{N-1}\left(k+\frac{1}{2}\right)=\sqrt{\pi}\,\frac{(2N)!}{4^N N!}.$$ Hence $$\begin{align} \sum_{k=0}^n \binom{n}{k}\left(k+\frac{1}{2}\right)\Gamma\left(n-k+\frac{1}{2}\right)&=\sum_{k=0}^n \frac{n!}{k!(n-k)!}\cdot\sqrt{\pi}\,\frac{(2k)!}{4^k k!}\cdot\sqrt{\pi}\, \frac{(2(n-k))!}{4^{n-k} (n-k)!}\\ &=\frac{\pi n!}{4^n}\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=\pi n! \end{align}$$ where at the last step we used the Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ .

In a similar way you should be able to show the more general identity, $$\sum_{k=0}^n \binom{n}{k}\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(n-k+m+\frac{1}{2}\right) =\frac{\pi(n+m)!}{4^m}\binom{2m}{m}.$$

$\endgroup$
1
$\begingroup$

In an alternative way,

$$ \Gamma(k+1/2)\Gamma(n-k+1/2) = n! B(k+1/2,n-k+1/2)=n!\int_{0}^{1}(1-x)^{k-1/2}x^{n-k-1/2}\,dx $$ hence by multiplying both sides by $\binom{n}{k}$ and summing over $k=0,1,\ldots,n$ we get $$ \sum_{k=0}^{n}\binom{n}{k}\Gamma(k+1/2)\Gamma(n-k+1/2)=n!\int_{0}^{1}\frac{x^n}{\sqrt{x(1-x)}}\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1-x}{x}\right)^k\,dx $$ and the RHS equals $$ n!\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}} = n!\cdot\Gamma\left(\tfrac{1}{2}\right)^2 = \color{red}{\pi\cdot n!}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.