# Proving $\sum\limits^\infty_{k=1}{\sum\limits^\infty _{n=1}\frac{a_n}{n^2+k^2}}$ converges

Given the series $\displaystyle\sum^\infty _{n=1}\frac{a_n}{n}$, where $a_n>0$, converges. Prove that $$\sum^\infty_{k=1}{\sum ^\infty _{n=1}\frac{a_n}{n^2+k^2}}$$ converges.

My idea is $$\sum^\infty_{k=1}{\sum ^\infty _{n=1}\frac{a_n}{n^2+k^2}}=\sum^\infty_{n=1}{a_n\sum ^\infty _{k=1}\frac{1}{n^2+k^2}}.$$

I have tried to make $$\sum ^\infty _{k=1}\frac{1}{n^2+k^2} \leq \frac{\alpha}{n}$$ but I did not make it.

Help me, thank you so much!

$$\sum_{m,n\geq 1}\frac{a_n}{m^2+n^2} = \sum_{m,n\geq 1}\frac{n}{m^2+n^2}\cdot\frac{a_n}{n}=\sum_{n\geq 1}\frac{a_n}{n}\sum_{m\geq 1}\frac{n}{n^2+m^2}$$ and for any fixed value of $n\geq 1$, the function $f(m)=\frac{n}{n^2+m^2}$ is decreasing on $\mathbb{R}^+$. In particular $$\sum_{m\geq 1}\frac{n}{n^2+m^2}\leq \int_{0}^{+\infty}\frac{n}{n^2+x^2}\,dx =\frac{\pi}{2}$$ and $$\sum_{m,n\geq 1}\frac{a_n}{m^2+n^2}\leq \color{red}{\frac{\pi}{2}}\sum_{n\geq 1}\frac{a_n}{n}.$$ The constant $\frac{\pi}{2}$ is optimal: if we consider $a_n = n^{-\varepsilon}$ for some small $\varepsilon >0$ we have that $\sum_{n\geq 1}\frac{a_n}{n}=\zeta(1+\varepsilon)=\frac{1}{\varepsilon}+\gamma+O(\varepsilon)$ while $$\sum_{m,n\geq 1}\frac{1}{n^\varepsilon(n^2+m^2)}=\sum_{n\geq 1}\frac{\pi n\coth(\pi n)-1}{2n^{2+\varepsilon}}=\frac{\pi}{2}\zeta(1+\varepsilon)+\underbrace{\sum_{n\geq 1}\frac{\frac{2\pi n}{e^{2\pi n}-1}-1}{2n^{2+\varepsilon}}}_{O(1)}.$$
Because$$n^2 + k^2 \geqslant \frac{1}{2} (n + k)(n + k - 1) \Longleftrightarrow (n - k)^2 + (n + k) \geqslant 0,$$ then$$\sum_{k = 1}^\infty \frac{1}{n^2 + k^2} \leqslant \sum_{k = 1}^\infty \frac{2}{(n + k)(n + k - 1)} = \frac{2}{n}.$$