[proofplan] We prove a uniform lower bound for the lattice quantity $|mz+n|$ when $z$ ranges over a fixed compact set $K \subset \mathbb{H}$. This reduces absolute convergence of the Eisenstein series to convergence of the numerical series $\sum_{(m,n)\neq(0,0)} (|m|+|n|)^{-k}$. The latter converges because the number of integer pairs with $|m|+|n|=r$ grows linearly in $r$, while $k \geq 4$. The tail estimate obtained from this summable majorant proves [uniform convergence](/page/Uniform%20Convergence) on $K$. [/proofplan] [step:Define the summands and exclude zeros in the denominator] For each $(m,n) \in \mathbb{Z}^2 \setminus \{(0,0)\}$, define the function \begin{align*} f_{m,n}: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto \frac{1}{(mz+n)^k}. \end{align*} If $z \in \mathbb{H}$ and $mz+n=0$, then either $m=0$, which forces $n=0$, or $m \neq 0$, which gives $z=-n/m \in \mathbb{R}$. Both alternatives contradict $(m,n)\neq(0,0)$ and $z \in \mathbb{H}$. Hence every $f_{m,n}$ is well-defined on $\mathbb{H}$. [/step] [step:Obtain a compact-uniform lower bound for $|mz+n|$] Fix a compact set $K \subset \mathbb{H}$. Define \begin{align*} a &:= \inf_{z \in K} \operatorname{Im}(z), \\ b &:= \sup_{z \in K} |\operatorname{Re}(z)|. \end{align*} Since $K$ is compact and $K \subset \mathbb{H}$, we have $a>0$ and $b<\infty$. Write $z=x+iy$ with $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$. Then $y \geq a$ and $|x| \leq b$ for every $z \in K$. Define \begin{align*} C_K := \min\left\{\frac{1}{3}, \frac{a}{2b+3}\right\}. \end{align*} We claim that for every $z \in K$ and every $(m,n)\in \mathbb{Z}^2 \setminus \{(0,0)\}$, \begin{align*} |mz+n| \geq C_K(|m|+|n|). \end{align*} If $|n| \geq 2(b+1)|m|$, then \begin{align*} |mx+n| \geq |n|-|m||x| \geq |n|-b|m| \geq \frac{|n|}{2}. \end{align*} Also $|m|\leq |n|/(2(b+1)) \leq |n|/2$, so \begin{align*} |m|+|n| \leq \frac{3}{2}|n|. \end{align*} Therefore \begin{align*} |mz+n| \geq |mx+n| \geq \frac{|n|}{2} \geq \frac{1}{3}(|m|+|n|). \end{align*} If $|n| < 2(b+1)|m|$, then $m \neq 0$, and \begin{align*} |mz+n|^2 = |mx+n+imy|^2 = (mx+n)^2 + m^2y^2 \geq m^2y^2. \end{align*} Hence \begin{align*} |mz+n| \geq |m|y \geq a|m|. \end{align*} Moreover, \begin{align*} |m|+|n| < (2b+3)|m|, \end{align*} and so \begin{align*} |mz+n| \geq \frac{a}{2b+3}(|m|+|n|). \end{align*} Combining the two cases gives the claimed bound. [guided] The compact set $K$ stays a positive distance above the real axis and has bounded real part. We encode these two facts by defining \begin{align*} a &:= \inf_{z \in K} \operatorname{Im}(z), \\ b &:= \sup_{z \in K} |\operatorname{Re}(z)|. \end{align*} Compactness of $K$ and the inclusion $K \subset \mathbb{H}$ give $a>0$ and $b<\infty$. For $z=x+iy \in K$, this means $y \geq a$ and $|x| \leq b$. We need a lower bound for $|mz+n|$ in terms of the integer size $|m|+|n|$, uniformly for all $z \in K$. Define \begin{align*} C_K := \min\left\{\frac{1}{3}, \frac{a}{2b+3}\right\}. \end{align*} We prove \begin{align*} |mz+n| \geq C_K(|m|+|n|) \end{align*} for every $z \in K$ and every nonzero integer pair $(m,n)$. There are two possible regimes. First suppose $|n| \geq 2(b+1)|m|$. The real part $mx+n$ cannot be too small, because $n$ dominates the possible cancellation from $mx$: \begin{align*} |mx+n| \geq |n|-|m||x| \geq |n|-b|m| \geq \frac{|n|}{2}. \end{align*} Since $|m|\leq |n|/(2(b+1))\leq |n|/2$, we also have \begin{align*} |m|+|n| \leq \frac{3}{2}|n|. \end{align*} Thus \begin{align*} |mz+n| \geq |mx+n| \geq \frac{|n|}{2} \geq \frac{1}{3}(|m|+|n|). \end{align*} Now suppose $|n| < 2(b+1)|m|$. Then $m\neq 0$, and the imaginary part of $mz+n$ controls the size: \begin{align*} |mz+n|^2 = |mx+n+imy|^2 = (mx+n)^2 + m^2y^2 \geq m^2y^2. \end{align*} Since $y \geq a$, \begin{align*} |mz+n| \geq |m|y \geq a|m|. \end{align*} The assumption in this case also gives \begin{align*} |m|+|n| < (2b+3)|m|. \end{align*} Therefore \begin{align*} |mz+n| \geq \frac{a}{2b+3}(|m|+|n|). \end{align*} The constant $C_K$ is the smaller of the two constants obtained in the two cases, so it works uniformly for all nonzero pairs $(m,n)$ and all $z \in K$. [/guided] [/step] [step:Show that the lattice majorant is summable] For each integer $r \geq 1$, define the finite set \begin{align*} A_r := \{(m,n) \in \mathbb{Z}^2 : |m|+|n|=r\}. \end{align*} The set $A_r$ has exactly $4r$ elements: for each $j \in \{0,\dots,r\}$ with $|m|=j$ and $|n|=r-j$, there are four sign choices except at the two endpoints, and the endpoint adjustment gives the same total count $4r$. Hence \begin{align*} \sum_{(m,n)\in \mathbb{Z}^2\setminus\{(0,0)\}} \frac{1}{(|m|+|n|)^k} &= \sum_{r=1}^{\infty} \sum_{(m,n)\in A_r} \frac{1}{r^k} \\ &= \sum_{r=1}^{\infty} \frac{4r}{r^k} \\ &= 4\sum_{r=1}^{\infty} \frac{1}{r^{k-1}}. \end{align*} Since $k \geq 4$, we have $k-1\geq 3>1$, so the final numerical series converges. [/step] [step:Dominate the absolute values uniformly on the compact set] For every $z \in K$ and every $(m,n)\neq(0,0)$, the compact-uniform lower bound gives \begin{align*} |f_{m,n}(z)| = \frac{1}{|mz+n|^k} \leq \frac{1}{C_K^k(|m|+|n|)^k}. \end{align*} Therefore \begin{align*} \sum_{(m,n)\in \mathbb{Z}^2\setminus\{(0,0)\}} \sup_{z\in K}|f_{m,n}(z)| \leq \frac{1}{C_K^k} \sum_{(m,n)\in \mathbb{Z}^2\setminus\{(0,0)\}} \frac{1}{(|m|+|n|)^k} <\infty. \end{align*} Thus the series of absolute values is uniformly dominated on $K$ by a convergent numerical series. [/step] [step:Pass from the majorant to absolute and locally uniform convergence] Let $F$ be a finite subset of $\mathbb{Z}^2\setminus\{(0,0)\}$, and define the finite partial sum \begin{align*} S_F: K &\to \mathbb{C} \\ z &\mapsto \sum_{(m,n)\in F} f_{m,n}(z). \end{align*} If $F \subset E$ are finite subsets of $\mathbb{Z}^2\setminus\{(0,0)\}$, then \begin{align*} \sup_{z\in K}|S_E(z)-S_F(z)| &\leq \sum_{(m,n)\in E\setminus F}\sup_{z\in K}|f_{m,n}(z)|. \end{align*} Because the series of suprema is finite, its tails tend to $0$. Hence the net of finite partial sums is uniformly Cauchy on $K$, and therefore converges uniformly on $K$. The same majorant also gives \begin{align*} \sum_{(m,n)\in \mathbb{Z}^2\setminus\{(0,0)\}} |f_{m,n}(z)| < \infty \end{align*} for every fixed $z \in K$, so the convergence is absolute on $K$. Since $K \subset \mathbb{H}$ was an arbitrary compact subset, the Eisenstein series defining $G_k$ converges absolutely at every point of $\mathbb{H}$ and uniformly on every compact subset of $\mathbb{H}$. This is precisely local uniform convergence on $\mathbb{H}$. [/step]