Convergence of $\sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2}$ Assume $r \in (0,1)$, I'm looking at the following sum, conjecture that it converges.
$$  \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} $$
This is very similar to the result of one earlier question, but I'm not sure how to cleanly deal with the $(k-u)^2$ denominator.
Edit:
Through playing around with Wolfram Mathematica, I'm getting
$$
\begin{align*}
 \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} &=  \sum_{k=1}^{\infty}\sum_{u=1}^{k}\frac{r^{k-u}}{u^2}\\ &= 
\sum_{k=1}^{\infty}  \left(\sum_{u=1}^{k} \int_{0}^{r} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \sum_{k=1}^{\infty} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \frac{x^{-u} (u x-u+1)}{u^2 (x-1)^2}  dx  \right) \\
   &=   \left( \int_{0}^{r} \frac{\text{Li}_2\left(\frac{1}{x}\right)-x \log
    \left(\frac{x-1}{x}\right)+\log
    \left(\frac{x-1}{x}\right)}{(x-1)^2} dx   \right) \\
   &=   \left( -\frac{\text{Li}_2\left(\frac{1}{r}\right)}{r-1}+\text{Li}_2\left(\frac{r-1}{r}\right)+\log \left(\frac{1}{r}\right) \log
   \left(\frac{r-1}{r}\right)-\frac{\pi ^2}{6}   \right) \\
\end{align*}
$$
However, I can't validate whether all of the steps are valid and not misspecified by the software, especially the complex logarithms..
 A: Using the Iverson bracket:
\begin{align}
\sum_{k=1}^\infty\sum_{u=0}^{k-1}\frac{r^u}{(k-u)^2} 
& = \sum_{k=1}^\infty\sum_{u=0}^{\infty}\frac{r^u[u<k]}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=1}^{\infty}\frac{[u<k]}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=u+1}^{\infty}\frac{1}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=1}^{\infty}\frac{1}{k^2}
= \frac{1}{1-r}\frac{\pi^2}{6}
\end{align}
A: $$\sum_{k=1}^\infty\sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2}=\sum_{k=1}^\infty\sum_{n=1}^k\frac{r^{k-n}}{n^2}=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{r^{k-n}}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}\frac{1}{1-r}=\frac{\pi^2}{6(1-r)}$$ (similarly to your prior question, rearranging is valid because of nonnegativity of the terms).
A: Intuition:
Consider the innermost sum. For large values of $k$, "most" of the terms of this sum will be negligibly small because of the exponential factor $r^u$. However, the "first few" terms of the inner sum will behave like $1/k^2$. It turns out that we can get away with considering only the "first few" terms of this inner sum (by "first few", I mean $\log(k)$ asymptotically, as we shall see momentarily) and noting that the contribution of the rest of the sum is insignificant compared to these first few values.
Rigor:
Define
$$s_k = \sum_{u=0}^{k-1} \frac{r^u}{(k-u)^2}$$
Let's break off the first $k'=\lfloor \log_r (1/k^3)\rfloor$ terms of this sum:
$$s_k = \color{red}{\sum_{u=0}^{k'} \frac{r^u}{(k-u)^2}} + \color{blue}{\sum_{u=k'+1}^{k-1} \frac{r^u}{(k-u)^2}}$$
For each term in the red sum, we have that
$$\frac{r^u}{(k-u)^2} \le \frac{1}{(k-k')^2}$$
meaning that the red sum satisfies
$$\color{red}{\sum_{u=0}^{k'} \frac{r^u}{(k-u)^2}} \le \frac{k'}{(k-k')^2}$$
For each term in the blue sum, we have that $r^u \le 1/k^3$, and hence
$$\frac{r^u}{(k-u)^2} \le \frac{1}{k^3}$$
Thus, the blue sum satisfies
$$\color{blue}{\sum_{u=k'+1}^{k-1} \frac{r^u}{(k-u)^2}} \le \frac{k-k'-2}{k^3} \le \frac{1}{k^2}$$
Thus, by combining our inequalities for the red and blue sums, we have that
$$s_k \le \frac{k'}{(k-k')^2} + \frac{1}{k^2}$$
Now, recalling that $k'=\lfloor \log_r (1/k^3)\rfloor$, it is easy to show that the $s_k$ is $\mathcal{O}(k^{-2}\log k)$, which is enough to show that it converges.
