A proof for the convergence of Eisenstein series In https://ctnt-summer.math.uconn.edu/wp-content/uploads/sites/1632/2016/02/CTNTmodularforms.pdf
the convergence of Eisenstein series for $k>2$ is proved by showing that this series converges $$ \sum_{n,m \in \mathbb{Z}^2\setminus \{ \underline{0} \} } \frac{1}{(m^2+n^2)^\frac{k}{2}} $$
Can we prove that the last series converges since the integral $$\int_{0}^{2\pi} \int_{1}^{\infty}\frac{1}{r^{k-1}} dr \ d\theta  $$ converges for $k>1$?
It seems to me the that this proof is easier than the others I've seen.
 A: You can certainly prove the convergence of your series using an argument similar to what you describe. If $k>2$, then
\begin{equation}
\sum_{(m,n)\in\mathbb{Z}^2\backslash\{0\}}\frac{1}{(m^2+n^2)^{k/2}}=4\sum_{n\geq 1}\frac{1}{n^k}+4\sum_{m,n\geq 1}\frac{1}{(m^2+n^2)^{k/2}}=4\zeta(k)+4\iint_{\mathbb{R}_+^2}\frac{1}{(\left\lceil x\right\rceil^2+\left\lceil y\right\rceil^2)^{k/2}}dA
\end{equation}
Letting $D$ be the closed unit disk in $\mathbb{R}^2$, it is thus sufficient to check that the following integral converges.
\begin{equation}
\iint_{\mathbb{R}_+^2\backslash D}\frac{1}{(\left\lceil x\right\rceil^2+\left\lceil y\right\rceil^2)^{k/2}}dA\leq
\iint_{\mathbb{R}_+^2\backslash D}\frac{1}{(x^2+y^2)^{k/2}}dA=\int_0^{\pi/2}\int_1^\infty \frac{1}{r^{k-1}}drd\theta=\frac{\pi}{2(k-2)}
\end{equation}
Therefore our series must converge.
In case you're interested, here is an even shorter proof of the absolute convergence of the Eisenstein series for $k>2$ that I similarly stumbled upon a few semesters ago.
Suppose $\mathfrak{I}(\tau)>0$, then the map $||\cdot||_M:(x,y)\mapsto |x\tau+y|$ is a norm on $\mathbb{R}^2$. Note that any two norms on $\mathbb{R}^2$ are equivalent, so there exists a $C>0$ such that $C||\mathbf{z}||_M\geq ||\mathbf{z}||_\infty=\max\{|\mathbf{z}_1|,|\mathbf{z}_2|\}$ for all $\mathbf{z}\in\mathbb{R}^2$. Therefore, for $k>2$ we have that
\begin{equation}
\sum_{(m,n)\in\mathbb{Z}^2\backslash\{0\}}\frac{1}{|m\tau+n|^k}\leq
\sum_{(n,m)\in\mathbb{Z}^2\backslash\{0\}}\frac{C^k}{(\max\{|n|,|m|\})^k}
=\sum_{n\geq 1} \frac{C^k(8n)}{n^k}=8C^k\zeta(k-1)
\end{equation}
which concludes the proof.
