Series does not converge

Question

How would I go about showing that the series$$\sum_{n + m\tau \in \Lambda} {1\over{{|n + m\tau|}^2}}$$does not converge, where $\tau \in \mathbb{H}$?

Answer 1

Suppose that $\text{Im}\,\tau > 0$. We show that the series $\sum_{(m, n) \neq (0, 0)} {1\over{|n + m\tau|^2}}$ diverges. Suppose by contradiction that it congerges. Put $\delta = \max\{1, |\tau|\} > 0$. We have$$|n + \tau m| \le |n| + |\tau||m| \le \delta(|n| + |m|).$$Thus,$$\infty > \sum_{(m, n) \neq (0, 0)} {1\over{|n + \tau m|^2}} \ge {1\over{\delta^2}} \sum_{(m, n) \neq (0, 0)} {1\over{(|n| + |m|)^2}} \ge {1\over{\delta}} \sum_{m,n \ge 1} {1\over{(n+m)^2}}.$$Thus,$$A = \sum_{m, n \ge 1} {1\over{(n+m)^2}} < \infty.$$For each $k \ge 2$, the term ${1\over{k^2}}$ occurs exactly $k-1$ times in $A$ as$${{1\over{\left(1 + (k-1)\right)^2}}},\, {{1\over{\left(2 + (k-2)\right)^2}}},\, \dots\,,\, {{1\over{\left((k-1) + 1\right)^2}}}.$$Thus,$$A = \sum_{k=2}^\infty {{k-1}\over{k^2}} = \sum_{k=2}^\infty \left({1\over{k}} - {1\over{k^2}}\right).$$Since $A < \infty$,$$\sum_{k=2}^\infty {1\over{k}} = A + \sum_{k=2}^\infty {1\over{k^2}} < \infty.$$This is a contradiction.