I am trying to prove the following for a very, very long time:
$\sum_{k=0}^j \frac{ (-1)^k}{k! (j-k)!} \frac{1}{2k+1} = \frac{1}{2} \frac{\sqrt{\pi}}{\Gamma(3/2 + j)}$
or, equivalently
$\sum_{k=0}^j \frac{ (-1)^k}{k! (j-k)!} \frac{1}{2k+1} = \frac{(j+1)! 4^{j+1}}{2\cdot (2(j+1))!}$
I would be extremely happy if somebody could help me with this!