The following sums are the ones I'm interested in:
$\sum_{i=m}^{\Omega}{\binom{i}{m}i^{-k}}$
$\lim_{\Omega\rightarrow\infty}{\sum_{i=m}^{\Omega}{\binom{i}{m}i^{-k}}}$
I already know that $\lim_{\Omega\rightarrow\infty}{\sum_{i=m}^{\Omega}{i^{-k}}}$ is Hurwitz Zeta function $\zeta(k,m)$ and converges for $k\geq 2$.
Please list the steps taken to arrive at a closed form. If it diverges, show why. If these results are known already, please provide a link to them.
This was originally at https://mathoverflow.net/questions/206189/does-this-have-summation-involving-binomial-have-a-closed-form-if-so-what-is, and I received the following comment from there (not sure if I agree with it, or maybe I don't completely understand it):
"The binomial coefficient in your sum is a polynomial of $i$ of degree $m$, hence your series is a linear combination of the shifted Hurwitz zeta functions $\zeta(k−j,m)$ for $0≤j≤m$. It is is also clear that the series converges if and only if $k>m+1$."
Consider the fixed case $m=2$ for simplicity. Then ${i\choose 2}$ is just $\frac{i(i-1)}{2}=\frac12(i^2-i)$, so your sum can be written as $\sum_{i=2}^\infty\frac12(i^2-i)i^{-s} = \frac12\sum_{i=2}^\infty \left(i^{-(s-2)}-i^{-(s-1)}\right)$, and by easy shifts both of these are clearly Hurwitz zeta functions. (In fact, using the fact that the binomial coefficient $i\choose m$ is $0$ if $1\leq i\lt m$, we can rewrite the sum as starting from $i=1$ without changing the value, and make each term an un-shifted 'proper' zeta function; in the $m=2$ case this comes to $\frac12\sum_{i=1}^\infty\left(i^{-(s-2)}-i^{-(s-1)}\right)=\frac12\left(\zeta(s-2)-\zeta(s-1)\right)$. The same can be done for any value of $m$, using the expansion of the binomial in terms of Stirling numbers.