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I have been trying around different ranges of summations:

$$ \sum_{(m,n) \in \mathbb{Z}^2} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$

That's not any good. What about if we restrict to $m, n \in \mathbb{Z}$ as positive integers.

$$ \sum_{m > 0, n > 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$

Now here's an anti-symmetry. I do not like taking out the symmetry, but perhaps I can ask about:

$$ \sum_{m > n > 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} = \; ? \tag{$*$} $$

There doesn't seem to be a change of variables that can work. And we've used symmetry about as much as we can. This looks related to:

$$ \sum \frac{1}{n^2} = \frac{\pi^2}{6}$$

Perhaps this other series ($*$) also has a special value.

cactus314
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    Unless I'm mistaken the series $$\sum_{(m,n)\in\Bbb{Z}^2\setminus{(0,0)}}\frac{m^2-n^2}{(m^2+n^2)^2}$$does not converge absolutely. Therefore the sum will depend on the order of summation. – Jyrki Lahtonen Nov 19 '17 at 18:04
  • Doesn't it matter in what way $m$ and $n$ increase wrt one another? Are you expecting one sum? – imranfat Nov 19 '17 at 18:06
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    Using the integral test, since $$\int_{1 \le y \le x} \frac{x^2-y^2}{(x^2+y^2)^2} \mathrm{d} x \mathrm{d}y$$ is divergent, it seems that your sum is divergent too. – Crostul Nov 19 '17 at 18:19
  • @Cactus314 can you stop asking many random questions ? Your question doesn't make sense. If $\chi$ is an odd Dirichlet character, that is $\chi(-1) = -1$ then we look at $\sum_{n=-\infty}^\infty \chi(n) n e^{2i \pi n^2 z} \in M_{3/2}(\Gamma_0(q^2),\chi^2)$. The simplest Hecke character of $\mathbb{Z}[i]$ is $\psi(a+ib) = \frac{(a+ib)^2}{(a-ib)^2}$ which is well-defined because $\psi(a) =1$ for $a \in \mathbb{Z}[i]^\times$. Thus $L(s,\psi) = \sum_{(n,m)\in \mathbb{Z}^2 \setminus (0,0)} \psi(a+ib) |a+ib|^{-2s}$ is a Hecke L-function, coming from a modular form. – reuns Nov 19 '17 at 20:23

3 Answers3

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Actually the notation $\sum_{(m,n)\in\mathbb{Z}_+^2}$ does not make sense here, since the given series is not absolutely convergent. On the other hand, $$ \sum_{m\in\mathbb{Z}^+}\sum_{n\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=\sum_{m\geq 1}\left(-\frac{1}{2m^2}+\frac{\pi^2}{2\sinh^2(m\pi)}\right) $$ $$ \sum_{n\in\mathbb{Z}^+}\sum_{m\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=-\sum_{n\geq 1}\left(-\frac{1}{2n^2}+\frac{\pi^2}{2\sinh^2(n\pi)}\right) $$ by applying $\frac{d^2}{dx^2}\log(\cdot)$ to the Weierstrass product for the sine function.
The identity $\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n)}=\frac{1}{6}-\frac{1}{2\pi}$ is pretty well-known and it can be derived through the residue theorem and/or modular forms (see Zucker, The Summation of Series of Hyperbolic Functions) and/or the Poisson summation formula. In particular

$$ \sum_{m\in\mathbb{Z}^+}\sum_{n\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=\color{red}{-\frac{\pi}{4}}.$$

Jack D'Aurizio
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Fix $m>0$ for a moment and consider the sum $$ H(m):=\sum_{n=0}^{\lfloor m/2\rfloor}\frac{m^2-n^2}{(m^2+n^2)^2}. $$ Here the numerator of the term, call it $x(n,m)$, is at least $3m^2/4$, and the denomimator is at most $4m^4$. Therefore $$ x(n,m)\ge\frac{3m^2/4}{4m^4}=\frac{3}{16m^2} $$ in this range. There are at least $m/2$ terms in $H(m)$, so we arrive at the lower bound $$ H(m)\ge \frac{3}{32m}. $$ This shows that the double sum $$ \sum_{m>n>0}\frac{m^2-n^2}{(m^2+n^2)^2} $$ diverges.

Jyrki Lahtonen
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  • The thinking is that there are $O(m)$ terms of order $O(1/m^2)$, so the series diverges absolutely. A similar sum over a lattice is used when constructing the Weierstrass $\wp$-function. For exactly this same reason e.g. Apostol first constructs the doubly periodic function $\wp'(z)$ with poles of order three at points of a lattice $\Lambda$ on the complex plane, and then integrates it termwise to produce a doubly periodic function with poles of order two. Having $O(m)$ terms of order $O(1/m^3)$ allows for absolute convergence. – Jyrki Lahtonen Nov 19 '17 at 18:29
  • that's not quite what I had in mind... Onto 4th power then! – cactus314 Nov 19 '17 at 18:54
  • this is suspicious – Guy Fsone Nov 19 '17 at 20:37
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Consider: \begin{align} \int_{0}^{\infty} e^{-s t} \, t \, dt &= \frac{1}{s^{2}} \\ S_{1} &= \sum_{n=1}^{\infty} e^{- n^{2} t} = \frac{\theta_{3}(0, e^{- t}) - 1}{2} \\ S_{2} &= \sum_{n=1}^{\infty} n^{2} \, e^{- n^{2} t} = \frac{\theta^{'}_{3}(0, e^{- t})}{2} \end{align} then \begin{align} \sum_{n,m=1}^{\infty} \frac{m^{2} - n^{2}}{(m^{2} + n^{2})^{2}} &= \int_{0}^{\infty} t \, \left[ S_{1} \cdot S_{2} - S_{1} \cdot S_{2} \right] \, dt = 0. \end{align}

Verification, numerically, was obtained by use of Mathematica by calculating the first 10000 terms of the series and obtained $0$.

For $m \geq 1$, $1 \leq n \leq m$, the series is divergent, and is also the case for $m > n$ in other forms.

Leucippus
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  • $ m > n$. We've concluded the sum is divergent. I might ask a separate question about "regularizing" this sum in some kind of way. – cactus314 Nov 19 '17 at 19:14
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    @cactus314 "Regularizing this sum", are you kidding ? Leucippus is saying $2i\pi \sum_{n,m} (m^2-n^2) e^{2 i \pi n^2 z} = \Theta'(z)\Theta(z)-\Theta'(z)\Theta(z) = 0$. – reuns Nov 19 '17 at 20:29