TITLE: Is $\sum_{n=1}^\infty \frac{m}{(n+m)^2}$ bounded for all $m\in\mathbb{N}$?
QUESTION [5 upvotes]: I'm trying to figure out if there is a finite constant $C$ such that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq C$ for all $m\in\mathbb{N}$.
I can see that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq\sum_{n=1}^\infty \frac{m}{n^2}=mc$ for some finite constant $c$, but this is a weaker statement.
If I naively replace integers by reals and sums by integrals then I feel like the original series should be bounded.

REPLY [4 votes]: Use that $\frac{m}{(n+m+1)(n+m)}= \frac{m}{(n+m)}-\frac{m}{(n+m+1)}$ for all $n,m \in \mathbb{N}$. We can then see that your series is bounded by $C=1$ by comparing it with a telescoping series. More precisley:  
For every $N, m \in \mathbb{N}$ we have $$\sum_{n=1}^N \frac{m}{(n+m)^2} = \sum_{n=0}^N \frac{m}{(n+m+1)^2} \leq $$ $$ \leq \sum_{n=0}^N \frac{m}{(n+m+1)(n+m)} = \sum_{n=0}^N \frac{m}{(n+m)}-\frac{m}{(n+m+1)} = $$ $$= \frac{m}{m}-\frac{m}{(N+m+1)}=1 - \frac{m}{(N+m+1)} \leq 1 $$