Another cute solution

Previously I asked if the summation $\sum_{m,n=1}^\infty1/(m^2+n^2)$ converges or diverges. Actually, the I intended the denominator of the summation term to be $(m^2+n^2)^2$. But never mind, let’s solve the problem as given. To do this, we’ll employ the comparison test.

First, note that

\[ m^2+n^2 \leq m^2+2mn+n^2 = (m+n)^2 , \]

so $1/(m^2+n^2)\geq1/(m+n)^2$.

Next, assume the sum converges; since all terms are positive the sum absolutely converges and the terms may be rearranged without affecting its value. In particular, we rearrange the terms in decreasing order, by grouping all terms equal to $1/k^2$ together for each possible value of $k$.

Finally, we use the fact that there are exactly $k-1$ solutions to $m+n=k$ in positive integers $m$, $n$. Putting it all together, the argument goes as follows:

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

The final inequality just uses $k-1\geq k/2$ for $k\geq2$ and we are left with a harmonic series, which diverges. By the comparison test the original sum diverges; this contradicts the assumption that it converges, so the sum really does diverge, as required.

Next up, I’ll show the question I intended to ask: does $\sum_{m,n=1}^\infty1/(m^2+n^2)^2$ converge or diverge?