An interesting sum over lattice points in a large disk centered at the origin Evaluate the the limit, as $r \rightarrow \infty $, of the sum $\displaystyle \sum \limits_{(m,n) \in D_r}$ $\displaystyle (-1)^{m+n} \over \displaystyle m^2 + n^2$, where $D_r$ denotes the closed disk of radius $r$ centered at the origin.  
I expect one would need Poisson summation to turn this into an exponential sum and apply some analytic estimate, but I cannot find any relevant results. Any help and especially, references, would be welcome.
 A: More generally, when $\Re(s)\ge1$
$$\sum_{(m,n)\ne(0,0)}(-1)^{m+n}/(m^2+n^2)^s=-4(1-2^{1-s})\zeta_K(s)$$
where $K=\mathbb Q(i)$ and $\zeta_K$ is the Dedekind zeta function of $K$.
A: Assuming convergence, here goes:
let $r_2(k)$ denote the number of ways a number $k\in\mathbb{N}$ can be written
as the sum of squares of two integers. Then, we compute
$$\sum_{(0,0)\neq(m,n)\in\mathbb{Z}^2}\frac{(-1)^{m+n}}{m^2+n^2}=\sum_{k=1}^{\infty}(-1)^k\frac{r_2(k)}k.$$
It is known that
$$r_2(k)=4\sum_{d\vert k}\left(\frac{-4}k\right)=4\left(1*\left(\frac{-4}k\right)\right)(k)$$
where $\left(\frac{a}b\right)$ is the Jacobi symbol. Using $\left(\frac{-4}{2k}\right)=0$ and $\left(\frac{-4}{2k+1}\right)=(-1)^k$ it follows that
\begin{align} \sum_{k=1}^{\infty}(-1)^k\frac{r_2(k)}k
&=4\sum_{k=1}^{\infty}(-1)^k\frac{(1*\left(\frac{-4}k\right)(k)}k \\
&=4\sum_{n=1}^{\infty}\frac{(-1)^n}n\sum_{k=1}^{\infty}(-1)^k\frac{\left(\frac{-4}k\right)(k)}k \\
&=4\sum_{n=1}^{\infty}\frac{(-1)^n}n\sum_{m=0}^{\infty}\frac{(-1)^m}{2m+1} \\
&=4(-\log(2))\left(\frac{\pi}4\right) \\
&=-\pi\,\log(2),
\end{align}
which confrims Henri Cohen's guesstimate.
A: The limit equals $-\pi\log 2$, in accordance with Henri Cohen's remark above.
For the proof, we combine the formula
$$r_2(k):=\#\{(m,n)\in\mathbb{Z}^2:m^2+n^2=k\}=4\sum_{d\mid k}\chi_4(d)$$
with Dirichlet's hyperbola method. With this notation, the OP's sum equals
$$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=4\sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de}.$$
We shall now use the well-known facts that
$$\sum_{d=1}^\infty\frac{\chi_4(d)}{d}=\frac{\pi}{4}
\qquad\text{and}\qquad
\sum_{e=1}^\infty\frac{(-1)^e}{e}=-\log 2,$$
where both series are alternating. Using this observation,
\begin{align*}
\sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de}
=&\sum_{\substack{d\leq r\\e\leq r^2/d}}\frac{(-1)^{e}\chi_4(d)}{de}
+\sum_{\substack{e\leq r\\d\leq r^2/e}}\frac{(-1)^{e}\chi_4(d)}{de}-
\sum_{d,e\leq r}\frac{(-1)^{e}\chi_4(d)}{de}\\
=&\sum_{d\leq r}\frac{\chi_4(d)}{d}\left(-\log 2+O\left(\frac{d}{r^2}\right)\right)\\
&+\sum_{e\leq r}\frac{(-1)^e}{e}\left(\frac{\pi}{4}+O\left(\frac{e}{r^2}\right)\right)\\
&-\left(\sum_{d\leq r}\frac{\chi_4(d)}{d}\right)
\left(\sum_{e\leq r}\frac{(-1)^e}{e}\right)\\
=&-\log 2\sum_{d\leq r}\frac{\chi_4(d)}{d}+\frac{\pi}{4}\sum_{e\leq r}\frac{(-1)^e}{e}+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\
=&-\frac{\pi}{4}\log 2-\frac{\pi}{4}\log 2+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\
=&-\frac{\pi}{4}\log 2+O\left(r^{-1}\right).
\end{align*}
Therefore, the OP's sum is
$$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=-\pi\log 2+O\left(r^{-1}\right).$$
A: It is problem number 10 of IMC 2018, you may find the solution on the official site.
