<s>[INST] 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.
 [/INST] 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})
$.
</s>