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.
 A: 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 $\frac{\Gamma(m+1)\,\Gamma\left(\frac{n}{k}\right)}{k\,\Gamma\left(1+m+\frac{n}{k}\right)}.$
A: We recall the Melzak's identity $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k $$ where $f $ is an algebraic polynomial up to degree $n $. So taking $f\left(z\right)\equiv1$ and $x=n/k$ we have $$\frac{1}{k}\sum_{j=0}^{m}\dbinom{m}{j}\frac{\left(-1\right)^{j}}{j+n/k}=\color{red}{\frac{1}{n\dbinom{n/k+m}{m}}}.$$
A: Assuming $n, k, m$ are positive integers, you could write this as
$$\log \left(\dfrac{\prod_{j \text{ even}} (n+jk)^{{m \choose j}}}{\prod_{j \text{ odd}} (n+jk)^{{m \choose j}}}\right)$$
For example, if $m=4$ it is
$$\log  \left( {\frac {n \left( 2\,k+n \right) ^{6} \left( 4\,k+n
 \right) }{ \left( k+n \right) ^{4} \left( 3\,k+n \right) ^{4}}}
 \right)
$$ 
Thus this is the log of a rational function of $n$ and $k$.  Numerator and denominator of that rational function are both of total degree $2^{m-1}$.  I don't see how this could be simplified any further.
A: There is a  technique for this that has appeared  on several occasions
on MSE which I am not able to locate at this time. We introduce
$$f(z) = (-1)^m \frac{m!}{n+zk} \prod_{q=0}^m \frac{1}{z-q}.$$
We suppose that $z=-n/k$ is not an integer from the range $[0, m].$ We
then obtain
$$\mathrm{Res}_{z=j} f(z)
= (-1)^m \frac{m!}{n+jk}
\prod_{q=0}^{j-1} \frac{1}{j-q} \prod_{q=j+1}^m \frac{1}{j-q}
\\ = (-1)^m \frac{m!}{n+jk} \frac{1}{j!} \frac{(-1)^{m-j}}{(m-j)!}
\\ = (-1)^j \frac{1}{nj+k} {m\choose j}.$$
It follows that
$$S = \sum_{j=0}^m \mathrm{Res}_{z=j} f(z)$$
and since residues sum to zero this means that
$$S = -\mathrm{Res}_{z=-n/k} f(z) - \mathrm{Res}_{z=\infty} f(z).$$
Now  the residue  at infinity  is zero  since $\lim_{R\to\infty}  2\pi
R/R^{m+2} = 0$ or more formally through
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) =
- \mathrm{Res}_{z=0} \frac{1}{z^2}
(-1)^m \frac{m!}{n+k/z} \prod_{q=0}^m \frac{1}{1/z-q}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
(-1)^m \frac{z\times m!}{zn+k} \prod_{q=0}^m \frac{z}{1-qz}
\\ = - \mathrm{Res}_{z=0} z^m
(-1)^m \frac{m!}{zn+k} \prod_{q=0}^m \frac{1}{1-qz}
= 0.$$
This leaves the contribution from $z=-n/k$ and we get
$$-\mathrm{Res}_{z=-n/k} \frac{1}{k}
(-1)^{m} \frac{m!}{z+n/k} \prod_{q=0}^m \frac{1}{z-q}
\\ = (-1)^{m+1} \frac{m!}{k} \prod_{q=0}^m \frac{1}{-n/k-q}
\\ = (-1)^{m+1} \times m! \times k^{m} \prod_{q=0}^m \frac{1}{-n-qk}
\\ = m! \times k^{m} \prod_{q=0}^m \frac{1}{n+qk}.$$
