What is the convergence rate of the tail of the series $$ a_n=\sum_{k>n}\exp(-\rho k)k(\log k)^2, $$ where $\rho>0$?
The series $a_n$ is some sort of an arithmetic geometric series with an additional logarithmic factor. If this was just a geometric series (i.e. $\sum_{k>n}\exp(-\rho k)$ with $\rho>0$), we would have that $$ \sum_{k>n}\exp(-\rho k)=\frac{\exp(-\rho(n+1))}{1-\exp(-\rho)}=O(\exp(-\rho n)) $$ as $n\to\infty$. Hence, I would guess that $$ a_n =O(\exp(-\rho n)n(\log n)^2) $$ as $n\to\infty$, but I am not sure if this is true and how to show that this is true.
One possible approach would be to bound the series with an integral from above in the following way $$ \sum_{k>n}\exp(-\rho k)k(\log k)^2\le\int_n^\infty\exp(-\rho x)x(\log x)^2\ dx $$ but this integral seems to be complicated and I am not sure how to proceed.
Any help is much appreciated!
The function $$ f(t)=\frac{t\log^2 t}{e^{\rho t}}, \quad t>0, $$ is eventually positive and monotonically decreasing, and $$ a_n-a_{n-1}=f(n), $$ hence, eventually $$ \int_{n}^{n+1} f(t)\,dt \le a_n-a_{n-1} \le \int_{n-1}^n f(t)\,dt $$ and hence, eventually. $$ \int_{n+1}^\infty f(t)\,dt\le a_n \le \int_n^\infty f(t)\,dt. $$ Now $$ \int_{n+1}^\infty f(t)\,dt=\int_1^\infty f(t+n)\,dt =\int_1^\infty\frac{(t+n)\log^2(t+n)}{e^{\rho(t+n)}}=\frac{n\log^2 n}{e^{\rho n}}\int_1^\infty\frac{(1+t/n)\log^2(t+n)}{\log^2 n\,e^{\rho t}} $$ Use Lebesgue Dominated Convergence Theorem to show that $$ \int_1^\infty\frac{(1+t/n)\log^2(t+n)}{\log^2 n\,e^{\rho t}} \to \int_1^\infty\frac{1}{e^{\rho t}}=\frac{1}{\rho e^\rho}. $$ Therefore, eventually $$ \frac{1}{\rho e^\rho} \frac{(n+1)\log^2(n+1)}{e^{-\rho(n+1)}}\le a_n \le \frac{1}{\rho e^\rho} \frac{n\log^2 n}{e^{-\rho n}} $$ and finally, $$ a_n=\mathcal O(n\log^2 n\, e^{-\rho n} ). $$