Is $\sum_{n=1}^\infty \frac{m}{(n+m)^2}$ bounded for all $m\in\mathbb{N}$? 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.
 A: 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 $$\sum_{n=1}^N \frac{m}{(n+m)^2} = \sum_{n=0}^N \frac{m}{(n+m+1)^2} \leq $$ $$ \leq \sum_{n=0}^N \frac{m}{(n+m+1)(n+m)} = \sum_{n=0}^N \frac{m}{(n+m)}-\frac{m}{(n+m+1)} = $$ $$= \frac{m}{m}-\frac{m}{(N+m+1)}=1 - \frac{m}{(N+m+1)} \leq 1 $$
A: By the integral comparison test,
$\sum_{n=1}^\infty \frac{1}{(n+m)^2}
=\sum_{n=m+1}^\infty \frac{1}{n^2}
\approx \int_m^{\infty} \frac{dx}{x^2}
=\frac{-1}{x}\big|_m^{\infty}
=\frac{1}{m}
$,
so
$\sum_{n=1}^\infty \frac{m}{(n+m)^2}
\approx 1
$.
To be more precise,
$\int_m^{\infty} \frac{dx}{x^2}
> \sum_{n=m+1}^\infty \frac{1}{n^2}
>\int_{m+1}^{\infty} \frac{dx}{x^2}
$,
so
$\frac{1}{m}
> \sum_{n=m+1}^\infty \frac{1}{n^2}
>\frac{1}{m+1}
$
or
$1
>\sum_{n=m+1}^\infty \frac{m}{n^2}
=\sum_{n=1}^\infty \frac{m}{(n+m)^2}
>(1-\frac1{m+1})
$.
A: For any $x>0$:
$$\sum_{n=1}^{+\infty}\frac{x}{(n+x)^2}= x\cdot \psi'(x+1) = x\cdot\frac{d^2}{dx^2}\log\Gamma(x+1)\tag{1} $$
but a way simpler bound is achieved by considering that, under the same assumption:
$$ \int_{0}^{+\infty}\frac{x}{(y+x)^2}\,dy = 1.\tag{2}$$
It is interesting to point that we may use this rather trivial bound to prove some version of Stirling's inequality: it is enough to divide the RHS of $(1)$ by $x$ and integrate twice.
