Is the sum of this series finite? Let $a\in\mathbb{R}_{+}$. I consider the series
$$
 f(q) = \left(\sum_{k=1}^q e^{-a(q-k)}\frac{1}{k}\right)^2
$$
and the sum
$$
   \sum_{q=1}^{\infty}f(q).
$$
My numerical simulations show that this sum is finite (at least, for $a=1$), but I do not know how to prove it analytically. 
 A: Here is the answer that I have found. 
Let us start defining $$z(q) = \sum_{k=1}^q e^{-aT(q-k)}\frac{1}{k}.$$ Then $f(q) = \left(z(q)\right)^2$, and the sum that we are studying is $S(p) = \sum_{q=1}^p\left(z(q)\right)^2$. The goal is to show that this sum converges as $p\to \infty$. 
Ok, the strategy is as follows. We know that for all $C>0$ and $\gamma>\frac{1}{2}$ the sum $$\sum_{q=1}^\infty\left(\frac{C}{q^\gamma}\right)^2$$ converges. Thus to show that $S$ converges we intend to show that there exists a series (having the converging sum) upper bounding the one we study. I.e., we want to show that there exist  $C>0$ and $\gamma>0$ such that for all $q\ge 1$ $$z(q) \le \frac{C}{q^\gamma}.$$ 
It can be seen that for $z(q)$ it holds $$z(q+1) = \frac{1}{e^{aT}}z(q) + \frac{1}{q+1}.$$ Suppose that for some $q$ it holds $z(q) \le \frac{C}{q^\gamma}$, and let us show that then $z(q+1)\le \frac{C}{(q+1)^\gamma}$. To this end, note that
$$z(q) \le \frac{C}{q^\gamma} \Rightarrow z(q+1) \le \frac{1}{e^{aT}}\frac{C}{q^\gamma} + \frac{1}{q+1}.$$
To simplify, choose $\gamma = 1$ and $C=2e^{aT}$. Then for $aT>0$ and $q\ge 1$
$$\begin{gathered}\left(2e^{aT}-3\right)q-2 >0 \\
\Rightarrow \frac{2}{q} + \frac{1}{q+1} \le \frac{2e^{aT}}{q+1} \\
\Rightarrow \frac{1}{e^{aT}}\frac{C}{q^\gamma} + \frac{1}{q+1} \le \frac{C}{\left(q+1\right)^\gamma} \\
\Rightarrow z(q+1)\le \frac{C}{\left(q+1\right)^\gamma},\end{gathered}$$
and $z(1)=1\le 2e^{aT}$. Thus, $f(q)$ is upperbounded by $\frac{4e^{2aT}}{q^2}$, and $S(p)$ converges.
A: $$S=\left(\sum_{k=1}^{n} e^{-an} \frac{(e^{a})^k}{k}\right)^2=\left( -e^{an} \ln(1-e^a)\right)^2=e^{-2an} \ln^2(1-e^a), ~if~ e^{a}<1$$
we have used $$\ln(1-x)=-\sum_{k=1}^n \frac{x^k}{k}, ~if ~ |x|<1$$
