# Summation of a series

I am interested to get a tight upper bound on the summation of the following series $S$.

$\displaystyle S=\sum_{i=1}^N\frac{e^{-\alpha i}}{(k+i)^d}$ for integers $k,N \geq 1$, and positive reals $d$ and $\alpha$.

Pls let me know if there is any confusion about the problem.

Note that your series admit a closed form in terms of special functions. We have $$\sum_{i=1}^{N}\frac{e^{-\alpha i}}{\left(k+i\right)^{d}}=\sum_{i=1}^{\infty}\frac{e^{-\alpha i}}{\left(k+i\right)^{d}}-\sum_{i=N+1}^{\infty}\frac{e^{-\alpha i}}{\left(k+i\right)^{d}}=\sum_{i=0}^{\infty}\frac{e^{-\alpha i}}{\left(k+1+i\right)^{d}}-e^{-\alpha N}\sum_{i=0}^{\infty}\frac{e^{-\alpha i}}{\left(k+N+1+i\right)^{d}}=\Phi\left(e^{-\alpha},d,k+1\right)-e^{-\alpha N}\Phi\left(e^{-\alpha},d,k+N+1\right)$$ where $\Phi\left(z,s,a\right)$ is the Lerch Trascendent. So you can get a better approximation from the last expression. For example, in this case we have the integral rappresentation $$\Phi\left(e^{-\alpha},d,k+1\right)=\frac{1}{\Gamma\left(d\right)}\int_{0}^{\infty}\frac{t^{d-1}e^{-\left(k+1\right)t}}{1-e^{-\left(\alpha+t\right)}}dt.$$
• Thanks for your response. Could you pls suggest me further how to solve (upper bound) the above expression in in terms of $\alpha, d, k, N$. – Ram May 20 '15 at 5:25
Is $$S\le\frac{1}{(k+1)^d}\sum_{i=1}^Ne^{-\alpha i}=\frac{1}{(k+1)^d}\frac{e^{-\alpha}}{1-e^{-\alpha}}(1-e^{N\alpha})\le \frac{1}{(k+1)^d}\frac{e^{-\alpha}}{1-e^{-\alpha}}$$ tight enough?
• Thanks for your response! If we take out the largest term $\frac{1}{(k+1)^d}$ from the series, then this simply reduced to computing the sum of a geometric series. I was wondering if we could keep the term $\frac{1}{(k+i)^d}$ inside the summation and get a tighter bound on the summation. – Ram May 19 '15 at 9:49
• Since $S$ bounded below by the first term of the sum, which is $e^{-\alpha}/(k+1)^d$, it seems difficult to do better. – Julián Aguirre May 19 '15 at 10:20