The question I meant to ask on Sunday was whether $\sum_{m,n=1}^\infty1/(m^2+n^2)^2$ converges or diverges, so I’ll give my solution to that problem now. In fact, the methodology closely resembles the divergence proof I gave for the alternate sum, even though this sum converges.

Since all terms in the sum are positive, the sum either converges absolutely or diverges (the sum cannot be conditionally convergent). In either case the terms of the sum may be rearranged without affecting the convergence. For suppose the sum diverges but converges for some rearrangement: since the terms of the rearranged sum are still positive, the rearranged sum would have to converge absolutely, and therefore converge for all rearrangements, contradicting the supposed diverging arrangement.

Therefore we can rearrange the terms as we please; in particular, we can sort the terms in decreasing order, which has the effect of grouping together all terms of the form $1/k^2$ where $k$ is a sum of two squares. The term $1/k^2$ will appear in the sum once for each solution of $m^2+n^2=k$ in positive integers $m$, $n$.^{1}

For our purposes it is sufficient to note there are at most $\sqrt{k}$ solutions to $m^2+n^2=k$. This follows since we know that $1\leq m\leq\sqrt{k}$ and for each value of $m$ there is at most one solution; the only possible value of $n$ which could work is $\sqrt{k-m^2}$.

The argument proceeds as follows:

\[ \newcommand{\N}{\mathbb{N}}\begin{align*}

\sum_{m,n\in\N}\frac{1}{(m^2+n^2)^2} &= \sum_{k=2}^\infty\sum_{\substack{m,n\in\N\\m^2+n^2=k}}\frac{1}{k^2} \\

&\leq \sum_{k=2}^\infty\frac{\sqrt{k}}{k^2} \\

&= \sum_{k=2}^\infty\frac{1}{k^{3/2}} \\

&= \zeta(3/2)-1

\end{align*} \]

The final sum converges since it is a $p$-series (truncated). By the comparison test, the sum in question also converges.

- The exact number of solutions to $m^2+n^2=k$ is essentially the sum of squares function $r_2(k)$, although since we are exclusively working with solutions in
*positive*numbers the actual number of solutions will be $r_2(k)/4$ if $k$ is not a perfect square and $(r_2(k)-4)/4$ if $k$ is a perfect square.Via MathWorld, we see that if $r_2(k)$ is nonzero then it is equal to $4B(k)$, where $B(k)$ is the number of divisors of $k$ solely comprised of primes congruent to $1$ mod $4$. In any case, this shows that the maximum number of times $1/k^2$ can appear in the summation is $d(k)$, the number of divisors of $k$. In Apostol (page 296) it is shown that $d(k)=o(k^\epsilon)$ for any $\epsilon>0$, so this is a fairly slowly growing function. ↩