I know from numerical evaluation that for all $n \ge 0$ and for any $\alpha$ which is positive real value
$\sum\limits_{k = 0}^n {\frac{{{{\left( { - 1} \right)}^k}}}{{k + \alpha }}} \left( {\begin{array}{*{20}{c}}n\\k \end{array}} \right) = \frac{{\Gamma \left( \alpha \right)!n!}}{{\Gamma \left( {n + \alpha + 1} \right)}}$ .
See here for a proof (by induction) in the special case of $\alpha$ being an integer, but addressing the general case eludes me. Does anyone have any ideas of how to proceed?
Also, it is not hard to prove that
$\sum\limits_{k = 0}^n {\frac{1}{{k + 1}}\left( {\begin{array}{*{20}{c}}n\\k \end{array}} \right)} = \frac{{{2^{n + 1}} - 1}}{{n + 1}}$ ,
which look remarkably similar for the case $\alpha = 1$. Does anyone know if the non-alternating sum has a similar closed form for a more general case, either integer or real?
For the first one, note that $$\sum_{k=0}^n\frac{(-1)^k}{k+\alpha}{\binom nk}=\sum_{k=0}^n \int_0^1 (-1)^k\binom nkx^{k+\alpha-1}\,dx=\int_0^1 x^{\alpha-1}(1-x)^n \,dx.$$ This is a beta integral, and equals $$B(\alpha,n+1)=\frac{\Gamma(\alpha)\Gamma(n+1)}{\Gamma(n+\alpha+1)} =\frac{\Gamma(\alpha)n!}{\Gamma(n+\alpha+1)}.$$