I am trying to attempt a homework problem but this summation pops up:
$$\sum_{m=0}^n \left(\frac{(-1)^m}{m!(m+1/2)} \times \frac{1}{(n-m)!}\right)$$
According to Mathematica,
$$\sum_{m=0}^n \left(\frac{(-1)^m}{m!(m+1/2)} \times \frac{1}{(n-m)!}\right) = \frac{\sqrt{\pi}}{(\Gamma(n+1/2))\times (n+1/2)}$$
Is there a simple way (ie by the use of some identities etc.) for me to obtain the expression on the right? I have tried to view the summation as some sort of a binomial expansion but to no avail. My guess that that is this identity would not be "standard" in the sense that the right-hand side involves some sort of $\Gamma(n+1/2)$ and $\pi$. Thanks in advance!
Your expression is :
$$ \frac{1}{n!}\sum_0^n \binom {n} {m}\int_0^1 (-1)^m t^{m-1/2}dt=\frac{1}{n!}\int_0^1(1-t)^n t^{-1/2}dt$$
Now use
$$B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$