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A friend posed the following question to me:

Evaluate the the limit, as $r rightarrow infty $, of the sum $displaystyle sum limits_{(m,n) in C_r}$ $displaystyle (-1)^{m+n} over displaystyle m^2 + n^2$ where the $(m,n)$ are the lattice points in the circle radius $r$ centered at 0.

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.

It is problem number 10 of IMC 2018, you may find the solution on the official site.

Assuming convergence, here goes:

let $r_2(k)$ denote the number of ways a number $kinmathbb{N}$ can be written
as the sum of squares of two integers. Then, we compute
$$sum_{(0,0)neq(m,n)inmathbb{Z}^2}frac{(-1)^{m+n}}{m^2+n^2}=sum_{k=1}^{infty}(-1)^kfrac{r_2(k)}k.$$
It is known that
$$r_2(k)=4sum_{dvert k}left(frac{-4}kright)=4left(1*left(frac{-4}kright)right)(k)$$
where $left(frac{a}bright)$ 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)^kfrac{r_2(k)}k
&=4sum_{k=1}^{infty}(-1)^kfrac{(1*left(frac{-4}kright)(k)}k \
&=4sum_{n=1}^{infty}frac{(-1)^n}nsum_{k=1}^{infty}(-1)^kfrac{left(frac{-4}kright)(k)}k \
&=4sum_{n=1}^{infty}frac{(-1)^n}nsum_{m=0}^{infty}frac{(-1)^m}{2m+1} \
&=4(-log(2))left(frac{pi}4right) \
&=-pi,log(2),
end{align}
which confrims Henri Cohen's guesstimate.

I think the OP meant that $C_r$ is the disk ${(x,y)inmathbb{R}^2:x^2+y^2leq r^2}$. If this is the case, then the limit equals $-pilog(2)$, in accordance with Henri Cohen's remark above.

For the proof we combine the formula
$$r_2(k):=#{(m,n)inmathbb{Z}^2:m^2+n^2=k}=4sum_{dmid k}chi_4(d)$$
with Dirichlet's hyperbola method. Here, $chi_4$ denotes the nontrivial Dirichlet character modulo $4$. With this notation and the above interpretation for $C_r$, the OP's sum equals
$$sum_{kleq r^2}frac{(-1)^kr_2(k)}{k}=4sum_{deleq r^2}frac{(-1)^{e}chi_4(d)}{de}.$$
We shall now use the well-known facts that
$$sum_{d=1}^inftyfrac{chi_4(d)}{d}=frac{pi}{4}
qquadtext{and}qquad
sum_{e=1}^inftyfrac{(-1)^e}{e}=-log 2,$$
where both series are alternating. Using this observation,
begin{align*}
sum_{deleq r^2}frac{(-1)^{e}chi_4(d)}{de}
=&sum_{substack{dleq r\eleq r^2/d}}frac{(-1)^{e}chi_4(d)}{de}
+sum_{substack{eleq r\dleq r^2/e}}frac{(-1)^{e}chi_4(d)}{de}-
sum_{d,eleq r}frac{(-1)^{e}chi_4(d)}{de}\
=&sum_{dleq r}frac{chi_4(d)}{d}left(-log 2+Oleft(frac{d}{r^2}right)right)\
&+sum_{eleq r}frac{(-1)^e}{e}left(frac{pi}{4}+Oleft(frac{e}{r^2}right)right)\
&-left(sum_{dleq r}frac{chi_4(d)}{d}right)
left(sum_{eleq r}frac{(-1)^e}{e}right)\
=&-log 2sum_{dleq r}frac{chi_4(d)}{d}+frac{pi}{4}sum_{eleq r}frac{(-1)^e}{e}+frac{pi}{4}log 2+Oleft(r^{-1}right)\
=&-frac{pi}{4}log 2-frac{pi}{4}log 2+frac{pi}{4}log 2+Oleft(r^{-1}right)\
=&-frac{pi}{4}log 2+Oleft(r^{-1}right).
end{align*}
Therefore, the OP's sum is
$$sum_{kleq r^2}frac{(-1)^kr_2(k)}{k}=-pilog 2+Oleft(r^{-1}right).$$