- In contrast with the result in Lemma 1.5, prove that the series
\[ \sum_ {n + m \tau \in \Lambda^ {*}} \frac {1}{| n + m \tau | ^ {2}} \quad \mathrm{where} \tau \in \mathbb {H} \]
does not converge. In fact, show that
\[ \sum_ {1 \leq n ^ {2} + m ^ {2} \leq R ^ {2}} 1 / (n ^ {2} + m ^ {2}) = 2 \pi \log R + O (1) \quad \text { as } R \to \infty . \]
Solution
First consider the square lattice. For \(p=(m,n)\in\mathbb Z^2\setminus\{0\}\), let \(Q_p=p+[-1/2,1/2]^2\). For \(|p|\ge2\) and \(x\in Q_p\), the mean-value theorem applied to \(r\mapsto r^{-2}\) gives \[ \left|\frac1{|x|^2}-\frac1{|p|^2}\right| \le \frac{C}{|p|^3}, \] with an absolute constant \(C\). Hence \[ \int_{Q_p}\frac{dx}{|x|^2} =\frac1{|p|^2}+O(|p|^{-3}). \tag{1} \] The total error from summing (1) over \(1\le|p|\le R\) is bounded, because \[ \sum_{p\ne0}|p|^{-3}<\infty. \] The union of the corresponding unit squares differs from the annulus \(1\le|x|\le R\) only by a fixed inner region and an outer annulus of width \(O(1)\). The integral of \(|x|^{-2}\) over that outer annulus is \(O(1/R)\), hence in particular \(O(1)\). Consequently \[ \sum_{1\le m^2+n^2\le R^2}\frac1{m^2+n^2} =\int_{1\le|x|\le R}\frac{dx}{|x|^2}+O(1). \] In polar coordinates, \[ \int_{1\le|x|\le R}\frac{dx}{|x|^2} =\int_0^{2\pi}\int_1^R\frac1{r^2}r\,dr\,d\theta =2\pi\log R. \] Thus \[ \sum_{1\le m^2+n^2\le R^2}\frac1{m^2+n^2} =2\pi\log R+O(1). \]
Now let \(\tau\in\mathbb H\). Since the real-linear map \[ (m,n)\longmapsto n+m\tau \] is an isomorphism \(\mathbb R^2\to\mathbb C\), there is \(C>0\) such that \[ |n+m\tau|\le C\sqrt{m^2+n^2} \] for all integers \(m,n\). Hence \[ \frac1{|n+m\tau|^2}\ge \frac1{C^2(m^2+n^2)}. \] The square-lattice sum diverges logarithmically, so \[ \sum_{n+m\tau\in\Lambda^*}\frac1{|n+m\tau|^2} \] diverges as well.