Looking for Closed Form of Sum Is there a closed form of the following sum?
$$\sum^{m}_{j=0}\frac{(-1)^{j}{m \choose j}}{n+jk}$$
I figure it should but the binomial is throwing me off. Any help would be greatly appreciated.

The given sum equals
$$ \int_{0}^{1}\sum_{j=0}^{m}\binom{m}{j}(-1)^j x^{jk+n-1}\,dx =\int_{0}^{1}x^{n-1}(1-x^k)^m\,dx$$
and by the substitution $x=z^{1/k}$ and Euler's Beta function this equals $\frac{\Gamma(m+1)\,\Gamma\left(\frac{n}{k}\right)}{k\,\Gamma\left(1+m+\frac{n}{k}\right)}.$
