Convergence of Eisenstein series

Question

Let $\tau \in \mathbf{H} = \{z \in \mathbf C \;|\; \text{Im}(z) >0 \}$ and $s>1 \in \mathbf{R}$.

I would like to show that the series $\sum_{(c,d) \in \mathbf{Z}^2-(0,0)} \frac{1}{|c\tau+d|^{2s}}$ is convergent. I have read that it follows from the comparison with the integral (which is easily seen to be convergent):

$$ \int_{\mathbf R^2} \frac{dxdy}{(x^2+y^2+1)^{2s}} $$

How does one do this comparison ?

Answer 1

See Theorem 4.3 and Appendix B of https://ctnt-summer.math.uconn.edu/wp-content/uploads/sites/1632/2016/02/CTNTmodularforms.pdf.

Answer 2

Without doing this comparison, we can divide the proof into two parts.

Lemma. For each $z\in\mathcal{H}$, there exists a $\delta\in(0,1)$ such that $$|mz+n|\geq\delta|mi+n|$$ for all $m$, $n\in\mathbb{Z}$.

Proof. In the case $m=0$ it is trivial. If $m\neq 0$, then $|mz+n|\geq\delta|mi+n|$ is equivalent to $$\left|\frac{z+n/m}{i+n/m}\right|\geq\delta.$$ Set a continuous function as follows $$\begin{aligned} f:\mathbb{R}&\longrightarrow\mathbb{R},\\ x&\longmapsto\left|\frac{z-x}{i-x}\right|. \end{aligned}$$ One can verify that $f(x)>0$ for all $x\in\mathbb{R}$, and $f(x)\rightarrow 1$ as $x\rightarrow\pm\infty$. Hence, there exists a sufficiently large positive number $R$ depending on $z$, such that $f(x)\geq\frac{1}{2}$ for $|x|>R$. For $x\in[-R,R]$, compactness of $[-R,R]$ that there is some $c>0$ such that $f(x)\geq c$ can be derived from the fact $f(x)>0$. Choose $\delta=\min\left(\frac{1}{2},c\right)$, and thus $f(x)\geq\delta$ for all $x\in\mathbb{R}$.

Theorem. Eisenstein series $E_k(z)$ is absolutely convergent.

Proof. Choose $\delta$ as in the above lemma. Then $$\frac{1}{|mz+n|^k}\leq\frac{1}{\delta^k|mi+n|^k}=\frac{1}{\delta^k(m^2+n^2)^{\frac{k}{2}}}.$$ Since $\frac{k}{2}>1$, the absolute convergency of $E_k(z)$ follows from that $$\sum_{(m,n)\neq(0,0)}\frac{1}{(m^2+n^2)^{\frac{k}{2}}}$$ is absolutely convergent for $k\geq 4$. Hence $E_k(z)$ is absolutely convergent.