# How can I prove this double sum equation?

I'm trying to prove that: $$\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{(m^2+n^2x^2)^2}=-\frac{\pi}{2x}-\frac{1}{x^2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{\left(m^2+n^2/x^2\right)^2}$$ I'm missing $$-\frac{\pi}{2x}$$ in my process, just by swapping $$n\to m,\, m\to n$$ in sum above on LHS as follows: \begin{align*}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{(m^2+n^2x^2)^2}&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2x^2-n^2}{(n^2+m^2x^2)^2}\\ &=-\frac1{x^2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2/x^2-m^2}{(m^2+n^2~/x^2)^2}\end{align*} Here I see that the exchange of summation operators probably cannot be applied, but I can't see that. Any help will be welcome!

• I'm not too familiar with this style of problem. It looks to me that $\sum_{n=1} \sum_{m=1} \frac{n^2}{(m^2+n^2)^2}$ weakly diverges by power-counting. Is it a-priori obvious that $\sum_{n=1} \sum_{m=1} \frac{x^2 n^2-m^2}{(m^2+x^2 n^2)^2}$ converges? Also, how should I interpret the equality you give when $x=1$: it seems to say $0 = \frac{\pi}{2}$? Commented Jan 29, 2023 at 19:17
• @user196574 For $x=1$ $\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2-m^2}{(m^2+n^2)^2}=-\frac\pi 4$
– lpb
Commented Jan 30, 2023 at 1:29
• Whoops, I was thinking that sum would be zero, but that was my mistake. I'm gaining appreciation for these types of sums. Commented Jan 30, 2023 at 2:52

The sum doesn't converge absolutely, so you can't simply rearrange it.

The equality itself holds for $$x>0$$, and is the particular case (at $$z=ix$$) of the following fact (known as a property of the so-called non-holomorphic Eisenstein series $$G_2$$):

For $$z\in\mathbb{C}$$ with $$\Im z>0$$, and for $$m,n\in\mathbb{Z}$$, define $$f_{m,n}(z)=\begin{cases}\hfill 0,\hfill&(m,n)=(0,0)\\(m+nz)^{-2},&\hfill\text{otherwise}\hfill\end{cases},\\F(z)=\sum_{n\in\mathbb{Z}}\sum_{m\in\mathbb{Z}}f_{m,n}(z),\qquad\overline{F}(z)=\sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}f_{m,n}(z).$$ Then $$F(z)-\overline{F}(z)=2\pi i/z$$.

Here is the most elementary proof I know. Define, a similar way, $$g_{m,n}(z)=\begin{cases}\hfill 0,\hfill&(m,n)\in\big\{(0,0),(1,0)\big\}\\(m-1+nz)^{-1}-(m+nz)^{-1},&\hfill\text{otherwise}\hfill\end{cases},\\G(z)=\sum_{n\in\mathbb{Z}}\sum_{m\in\mathbb{Z}}g_{m,n}(z),\qquad\overline{G}(z)=\sum_{m\in\mathbb{Z}}\sum_{n\in\mathbb{Z}}g_{m,n}(z).$$

Then $$F(z)-G(z)=\overline{F}(z)-\overline{G}(z)$$ since this time we have absolute convergence.

Further, it's easy to see that $$G(z)=2$$ identically; indeed $$\sum_{m\in\mathbb{Z}}g_{m,n}(z)$$ is a telescoping sum, which vanishes if $$n\neq 0$$, and equals $$2$$ if $$n=0$$.

The sum $$\overline{G}(z)$$ is also a telescoping one, but a bit harder (for me). Using the known $$\sum_{n=1}^\infty\left(\frac1{z+n}+\frac1{z-n}\right)=\pi\cot\pi z-\frac1z\qquad(z\in\mathbb{C}\setminus\mathbb{Z})$$ we find $$\sum_{n\in\mathbb{Z}}g_{m,n}(z)=\begin{cases}\hfill 1-\frac{\pi}{z}\cot\frac{\pi}{z},\hfill&m\in\{0,1\}\\\frac{\pi}{z}\left(\cot\pi\frac{m-1}{z}-\cot\pi\frac{m}{z}\right),&\hfill\text{otherwise}\hfill\end{cases}.$$

Since $$\Im z>0$$, we have $$\lim\limits_{m\to\pm\infty}\cot(\pi m/z)=\pm i$$. Thus, summing the above over $$m\in\mathbb{Z}$$, $$\overline{G}(z)=\frac{\pi}z\left(-i-\cot\frac{-\pi}z\right)+2\left(1-\frac{\pi}z\cot\frac{\pi}z\right)+\frac{\pi}z\left(\cot\frac{\pi}z-i\right)=2-\frac{2\pi i}z,$$ and the desired result follows.

• Very interesting, although I didn't understand the proof essentially. I was more relaxed with the phrase: "The sum doesn't converge absolutely, so you can't simply rearrange it." Thank you!
– lpb
Commented Jan 31, 2023 at 23:26
• @lpb: Too concise perhaps. I just didn't want to make the proof unnecessarily long and "polluted" with obvious details. The key idea is to "approximate" $f_{m,n}$ by $g_{m,n}$, so that $\sum_{m,n}(f_{m,n}-g_{m,n})$ converges absolutely (therefore admits any rearrangement) and $\sum_{m}\sum_{n}g_{m,n}$ as well as $\sum_{n}\sum_{m}g_{m,n}$ are relatively easy to compute explicitly. (A "telescoping sum" is something like $\sum_{n\in\mathbb{Z}}(a_{n-1}-a_n)$. What does it evaluate to?..) Anyway, feel free to make me clarify anything specific. Commented Feb 1, 2023 at 3:37
• +1 there. I never tried the double sum thing for $G_2$ because of the non-absolute convergence. Commented Feb 12, 2023 at 4:51