0
$\begingroup$

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.

Thanks in advance!

$\endgroup$
1
$\begingroup$

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 $$\Phi\left(e^{-\alpha},d,k+1\right)=\frac{1}{\Gamma\left(d\right)}\int_{0}^{\infty}\frac{t^{d-1}e^{-\left(k+1\right)t}}{1-e^{-\left(\alpha+t\right)}}dt. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your response. Could you pls suggest me further how to solve (upper bound) the above expression in in terms of $\alpha, d, k, N$. $\endgroup$ – Ram May 20 '15 at 5:25
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ram May 19 '15 at 9:49
  • $\begingroup$ Since $S$ bounded below by the first term of the sum, which is $e^{-\alpha}/(k+1)^d$, it seems difficult to do better. $\endgroup$ – Julián Aguirre May 19 '15 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.