2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Yes. This is the standard method. $\endgroup$
    – reuns
    May 16, 2021 at 16:42
  • $\begingroup$ @reuns In both the paper I've linked and my lecture notes it's proved in an entirely different (and more difficult) way though. So it seemed to me unlikely that I stumbled upon a much easier proof. $\endgroup$
    – Alessandro
    May 16, 2021 at 17:20

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

This is not an answer to your specific question but an extended comment in which I prove the convergence.

Defining

$$ s(k) = \sum_{n,m \in \mathbb{Z}^2\setminus \{ \underline{0} \} } \frac{1}{(m^2+n^2)^\frac{k}{2}}\tag{1}$$

Splitting the summation ranges into the four half-axes and into the four quadrants we can write the sum as

$$\begin{align}s(k) & = 4\sum_{m\ge1}\frac{1}{(m^2)^\frac{k}{2}}+4\sum_{n\ge1,m\ge1}\frac{1}{(m^2+n^2)^\frac{k}{2}}\end{align}\tag{2} $$

Now

$$\sum_{m=1}^{\infty} \frac{1}{(m^2)^\frac{k}{2}}=\zeta(k)\tag{3}$$

and from the inequality $0\le (m-n)^2 = m^2+n^2-2 n m$ or

$$m^2+n^2\ge 2 m n\tag{4}$$

we find an inequality with a decoupled double sum

$$\begin{align}s_{2}(k) & :=\sum_{n\ge1,m\ge1}\frac{1}{(m^2+n^2)^\frac{k}{2}}\\ & \le \sum_{n\ge1,m\ge1}\frac{1}{(2 m n)^\frac{k}{2}}= 2^{-\frac{k}{2}}\sum_{n\ge1,m\ge1}\frac{1}{( m n)^\frac{k}{2}}\\ & =2^{-\frac{k}{2}}\sum_{n\ge1}\frac{1}{n^\frac{k}{2}}\sum_{m\ge1}\frac{1}{m^\frac{k}{2}}=2^{-\frac{k}{2}}(\sum_{n\ge1}\frac{1}{n^\frac{k}{2}})^2\\ &=2^{-\frac{k}{2}} \zeta(\frac{k}{2})^2\end{align} \tag{5}$$

which finally gives us the following an upper bound

$$s(k)\le 4 \zeta(k) + 4\;2^{-\frac{k}{2}}\zeta(\frac{k}{2})^2\tag{6}$$

Noticing thath the sum for the $\zeta$- function is convergent if its argument is $\gt 1$ the convergence of the double sum in $(1)$ follows for $k\gt2$.

Generalization to $p$ summands

A similar idea leads to an upper bound for the sum

$$s_{p}(k) = \sum_{n_{1}\ge 1, ..., n_{p}\ge1} \frac{1}{(\sum_{i=1}^{p} n_{i}^2)^{\frac{k}{2}}}\tag{1a}$$

which at the same time simplifies the rather lengthy proof of convergence in [1], appendix B.

Generalizing $(4)$ to the inequality between arithmetic and geometric mean

$$\frac{\sum _{i=1}^p n(i)^2}{p}\ge \left(\prod _{i=1}^{p} n(i)\right)^{2/p}\tag{4a}$$

and proceeding like in $(5)$ we obtain the upper bound

$$s_1(p,k) \le \frac{1}{p^{\frac{k}{2}}}\zeta(\frac{k}{p})^p\tag{6a}$$

Which in turn means convergence of $s_1$ (by requesting the argument of the $\zeta$-function being greater than unity) if

$$k >p\tag{7}$$

How well does the integral of the OP approximate the sum?

Let us study the problem in cartesian coordinates. How are sum and integral related

$$s(q)=\sum_{n\ge1,m\ge1} \frac{1}{(n^2+m^2)^q}\sim \int_{1^2}^{\infty^2}\frac{dxdy}{(x^2+y^2)^q}=i(q)\tag{8}$$

It turns out that for $q\gt 1$ the sum is greater that the integral, a fact which could shed some doubt on the reasoning for convergence in the OP.

The integral can be calculated explicitly in terms of hypergeometric and gamma functions:

$$\begin{align}& i(q) =\frac{1}{2 (q-1) (2 q-1) \Gamma (q)} \\ & \times \left(\sqrt{\pi } \Gamma \left(q+\frac{1}{2}\right)-\left((2 q-1) \, _2F_1\left(\frac{1}{2},q;\frac{3}{2};-1\right)\\ -\, _2F_1\left(q-\frac{1}{2},q;q+\frac{1}{2};-1\right)\right) \Gamma (q)\right)\end{align}\tag{9}$$

Close to $q=1^+$ this is

$$i(q\simeq 1)\simeq \frac{\pi }{4 (q-1)}+O(1)\tag{10}$$

where $O(1) \simeq -2.00476$.

The first few term in the format $\{q,i(q)\}$ are

$$\left( \begin{array}{cc} 2 & \frac{1}{8} (-2+\pi ) \\ 3 & \frac{1}{64} (-8+3 \pi ) \\ 4 & \frac{1}{576} (-44+15 \pi ) \\ 5 & \frac{5 (-64+21 \pi )}{6144} \\ \end{array} \right)\tag{11}$$

The difference (times $2^q$) and the quotient between sum and integral for integer $q\ge 2$ are numerically

$$\left\{q,2^q (s-i),s/i\right\}|_{q=2}^{20} = \left( \begin{array}{ccc} 2 & 1.12648 & 2.97352 \\ 3 & 1.00099 & 6.62044 \\ 4 & 0.97371 & 12.2211 \\ 5 & 0.97098 & 19.8937 \\ 6 & 0.974597 & 29.6611 \\ 7 & 0.979196 & 41.5076 \\ 8 & 0.98326 & 55.4095 \\ 9 & 0.986506 & 71.3465 \\ 10 & 0.989011 & 89.3048 \\ 11 & 0.990932 & 109.275 \\ 12 & 0.992415 & 131.254 \\ 13 & 0.993572 & 155.236 \\ 14 & 0.994487 & 181.222 \\ 15 & 0.995222 & 209.21 \\ 16 & 0.99582 & 239.199 \\ 17 & 0.996313 & 271.189 \\ 18 & 0.996723 & 305.181 \\ 19 & 0.997069 & 341.173 \\ 20 & 0.997363 & 379.165 \\ \end{array} \right)\tag{12}$$

The difference has a minimum at $q=5$ and tends to 1 as $q$ becomes large.

The quotient, however gets large with increasing $q$.

Qualitatively the behaviour can be described by

$$s(q) \simeq \frac{1}{2^q}\left( 1+\frac{1}{2^q}\right), i(q) \simeq \frac{1}{2^{2q}}\tag{13a}$$

resulting in

$$2^q\left(s(q)-i(q)\right)\simeq 1, s(q)/i(q) \simeq 2^q+1\tag{13b}$$

Summarizing, we have found that the sum is greater than the integral and the integral diverges for $q\to 1$, hence the sum also diverges in this limit.

Question: is it correct to say that this does not prove that the sum converges for $q>1$?

References

[1] https://ctnt-summer.math.uconn.edu/wp-content/uploads/sites/1632/2016/02/CTNTmodularforms.pdf

$\endgroup$
1
  • $\begingroup$ @reuns I'd appreciate your opinion on my reasoning here. Thank you in advance. $\endgroup$ May 22, 2021 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .