Show that whenever $r \gt 2$ , the series $$\sum_{(n,m)\neq (0,0),n,m\in Z} |n+m\tau|^{-r}$$ converges uniformly in every half-plane Im$(\tau)\ge \delta \gt 0$ .
Note that $$\sum_{(n,m) \neq (0,0), n,m\in Z}(|n|+|m|)^{-r}$$ converges whenever $r \gt 2$ . So it suffice to prove $$|n+m\tau| \ge c(|m|+|n|)$$
For all $\tau\in C$ with Im$(\tau)\ge \delta$ .
If $\tau =s+it$ , then we have $$|n+m\tau|=[(n+ms)^2+(mt)^2]^{\frac12}\ge \frac{|n+ms|+|n\delta|}{2}$$ $$\ge \frac{\min \{\delta , 1 \}}{2}(|n+ms|+|m|)$$
The hint in Stein's complex analysis Page$_{269}$ was to consider two cases when $|n|\le 2|m||s|$ and $|n| \ge 2|m||s|$ to show that $$|n+ms|+|m|\ge c(|n|+|m|)$$
When $|n|\ge 2|m||s|$ , then we have $$|n+ms|+|m|\ge |n|-|ms|+|m|\ge |n|-|n|/2+|m|\ge \frac{|n|+|m|}{2}$$
When $|n|\le 2|m||s|$ , then we have $$|n|+|m|\le 2|m||s|+|m|\le (2|s|+1)(|n+ms|+|m|)$$ However , this estimate depends on $|s|$ .
$$f(\tau)=\sum_{(n,m)\neq (0,0),n,m\in Z} |n+m\tau|^{-r}$$ doesn't converge uniformly on $\Im(\tau)\ge\delta$, it only converges absolutely.
This is because all the $m\ne 0$ terms of $f(\tau+k)$ converge to $0$ as $k\to \infty$, if the convergence was uniform it would imply $\lim_{k \to \infty} f(\tau+k)=2\zeta(2)$ which isn't the case because $f(\tau+k)=f(\tau)>2\zeta(2)$.
From the $1$-periodicity it suffices to look at $|\Re(\tau)|\le 1/2$ in which case it does converge uniformly:
From $$\pmatrix{n\\ m} =\pmatrix{1 & \Re(\tau)\\ 0&\Im(\tau)}^{-1}\pmatrix{n+m\Re(\tau)\\ m\Im(\tau)}$$ show that $|n+m\tau|^2 \ge c_\delta (m^2+n^2)$ and look at $\sum_{n,m} (m^2+n^2)^{-r/2}\le A+\iint_{x^2+y^2>1} (x^2+y^2)^{-r/2}dxdy=A+\int_1^\infty v^{-r} 2\pi vdv$.