 International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

Home

Results
Individuals
Teams

Official IMC site

Problem 10

Problem 10. For $\displaystyle R>1$ let $\displaystyle \mathcal{D}_R = \{(a,b)\in\mathbb{Z}^2 \colon 0<a^2+b^2<R\}$. Compute

$\displaystyle \lim_{R\rightarrow \infty} \sum_{(a,b) \in \mathcal{D}_R} \frac{(-1)^{a+b}}{a^2+b^2}.$

(Proposed by Rodrigo Angelo, Princeton University and Matheus Secco, PUC, Rio de Janeiro)

Solution. Define $\displaystyle \mathcal{E}_R = \{(a,b) \in \mathbb{Z}^2 \setminus \{(0,0)\} : a^2+b^2<R\ \text{ and } a + b \text{ is even} \}$. Then

$\displaystyle \sum_{(a,b) \in \mathcal{D}_R} \frac{(-1)^{a+b}}{a^2+b^2} = 2 \sum_{(a,b) \in \mathcal{E}_R}\frac{1}{a^2+b^2} - \sum_{(a,b) \in \mathcal{D}_R} \frac{1}{a^2+b^2}. \tag5$

But $\displaystyle a+b$ is even if and only if one can write $\displaystyle (a,b) = (m-n, m+n)$, and such $\displaystyle m,n$ are unique. Notice also that $\displaystyle a^2+b^2 = (m-n)^2+(m+n)^2 = 2m^2+2n^2$, hence $\displaystyle a^2+b^2 < R$ if and only if $\displaystyle m^2+n^2<R/2$. With that we get:

$\displaystyle 2\sum_{(a,b) \in \mathcal{E}_R}\frac{1}{a^2+b^2} = 2\sum_{(m,n) \in D_{R/2}}\frac{1}{(m-n)^2+(m+n)^2} = \sum_{(m,n) \in D_{R/2}}\frac{1}{m^2+n^2}. \tag6$

Replacing (6) in (5), we obtain

$\displaystyle \sum_{(a,b) \in \mathcal{D}_R} \frac{(-1)^{a+b}}{a^2+b^2} = - \sum_{R/2 \le a^2+b^2 < R} \frac{1}{a^2+b^2},$

where the second sum is evaluated for $\displaystyle a$ and $\displaystyle b$ integers.

Denote by $\displaystyle N(r)$ the number of lattice points in the open disk $\displaystyle x^2+y^2<r^2$. Along the circle with radius $\displaystyle r$ with $\displaystyle \sqrt{R/2}\le r<\sqrt{R}$, there are $\displaystyle N(r+0)-N(r-0)$ lattice points; each of them contribute $\displaystyle \frac1{r^2}$ in the sum (7). So we can re-write the sum as a Stieltjes integral:

$\displaystyle \sum_{R/2 \le a^2+b^2 < R} \frac{1}{a^2+b^2} = \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2} \,\mathrm{d}N(r).$

It is well-known that $\displaystyle N(r)=\pi r^2+O(r)$. (Putting a unit square around each lattice point, these squares cover the disk with radius $\displaystyle r-1$ and lie inside the disk with radius $\displaystyle r+1$, so there their total area is between $\displaystyle \pi(r-1)^2$ and $\displaystyle \pi(r+1)^2$). By integrating by parts,

\displaystyle \begin{align*} \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2} \,\mathrm{d}N(r) &= \left[\frac1{r^2}N(r)\right]_{\sqrt{R/2}}^{\sqrt{R}} + \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{2}{r^3}N(r) \,\mathrm{d}r \\ &= \left[\frac{\pi r^2+O(r)}{r^2}\right]_{\sqrt{R/2}}^{\sqrt{R}} + 2\int_{\sqrt{R/2}}^{\sqrt{R}} \frac{\pi r^2+O(r)}{r^3} \,\mathrm{d}r \\ &= 2\pi \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{\mathrm{d}r}{r} + O\Big(1/\sqrt{R}\Big) = \pi \log 2 + O\Big(1/\sqrt{R}\Big). \end{align*}

Therefore,

$\displaystyle \lim_{R\rightarrow \infty} \sum_{(a,b) \in \mathcal{D}_R} \frac{(-1)^{a+b}}{a^2+b^2} = -\lim_{R\rightarrow \infty} \sum_{R/2 \le a^2+b^2 < R} \frac{1}{a^2+b^2} = -\lim_{R\rightarrow \infty} \int_{\sqrt{R/2}}^{\sqrt{R}} \frac{1}{r^2} \,\mathrm{d}N(r) = -\pi \log 2.$