[proofplan] We first justify that the defining lattice series for $G_k$ is absolutely convergent, so that the terms may be grouped by rows indexed by $m \in \mathbb{Z}$. The row $m=0$ gives the constant term $2\zeta(k)$, while the positive rows are evaluated using the classical Fourier expansion of the one-dimensional row sum. The negative rows equal the positive rows because $k$ is even. Finally, we group the resulting double series by the product $N=mr$ and divide by $2\zeta(k)$, using [Euler's formula](/theorems/2014) for $\zeta(k)$ in terms of the Bernoulli number $B_k$. [/proofplan]

[step:Justify grouping the Eisenstein lattice series by horizontal rows]Fix $z \in \mathbb{H}$, and write $z = x + iy$ with $x,y \in \mathbb{R}$ and $y > 0$. Define the real-[linear map](/page/Linear%20Map) \begin{align*} T_z: \mathbb{R}^2 &\to \mathbb{C} \\ (u,v) &\mapsto uz+v. \end{align*} Identifying $\mathbb{C}$ with $\mathbb{R}^2$, the map $T_z$ is invertible because its matrix in the standard real bases is \begin{align*} \begin{pmatrix} x & 1 \\ y & 0 \end{pmatrix}, \end{align*} whose determinant is $-y \neq 0$. Hence, by equivalence of norms on the finite-dimensional space $\mathbb{R}^2$, there exists a constant $c_z>0$ such that \begin{align*} |uz+v| \geq c_z (u^2+v^2)^{1/2} \end{align*} for all $(u,v)\in\mathbb{R}^2$. Therefore, for every $(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}$, \begin{align*} \left|\frac{1}{(mz+n)^k}\right| \leq c_z^{-k}(m^2+n^2)^{-k/2}. \end{align*} Since $k\geq4$, the comparison series \begin{align*} \sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}(m^2+n^2)^{-k/2} \end{align*} converges. Indeed, grouping lattice points in annuli \begin{align*} A_j := \{(m,n)\in\mathbb{Z}^2 : j \leq (m^2+n^2)^{1/2} < j+1\}, \end{align*} there is a constant $C_0>0$ such that $\#A_j \leq C_0 j$ for all $j\geq1$, and hence \begin{align*} \sum_{(m,n)\neq(0,0)}(m^2+n^2)^{-k/2} \leq C_0\sum_{j=1}^{\infty} j^{1-k}, \end{align*} which converges because $k>2$. Thus $G_k(z)$ is absolutely convergent, and we may group the terms according to the value of $m$.[/step]

[guided]Fix $z \in \mathbb{H}$, and write $z = x+iy$ with $x,y\in\mathbb{R}$ and $y>0$. Before separating the lattice sum into rows, we must justify that such regrouping is legitimate. Absolute convergence gives this justification. Define the real-linear map \begin{align*} T_z: \mathbb{R}^2 &\to \mathbb{C} \\ (u,v) &\mapsto uz+v. \end{align*} Under the identification $\mathbb{C}\cong\mathbb{R}^2$, this map has matrix \begin{align*} \begin{pmatrix} x & 1 \\ y & 0 \end{pmatrix}. \end{align*} Its determinant is $-y$, which is nonzero because $z\in\mathbb{H}$. Hence $T_z$ is an invertible real-linear map. Since all norms on the finite-dimensional space $\mathbb{R}^2$ are equivalent, there is a constant $c_z>0$ such that \begin{align*} |uz+v| = |T_z(u,v)| \geq c_z (u^2+v^2)^{1/2} \end{align*} for every $(u,v)\in\mathbb{R}^2$. Applying this estimate to integer pairs $(m,n)\neq(0,0)$ gives \begin{align*} \left|\frac{1}{(mz+n)^k}\right| \leq c_z^{-k}(m^2+n^2)^{-k/2}. \end{align*} It remains to know that the right-hand comparison series converges. For each $j\in\mathbb{N}$, define the lattice annulus \begin{align*} A_j := \{(m,n)\in\mathbb{Z}^2 : j \leq (m^2+n^2)^{1/2} < j+1\}. \end{align*} The number of lattice points in $A_j$ is bounded above by $C_0j$ for a constant $C_0>0$, because the annulus has radius comparable to $j$ and bounded width. Thus \begin{align*} \sum_{(m,n)\neq(0,0)}(m^2+n^2)^{-k/2} \leq C_0\sum_{j=1}^{\infty} j\cdot j^{-k} = C_0\sum_{j=1}^{\infty}j^{1-k}. \end{align*} This final series converges since $k>2$. Therefore the Eisenstein lattice series is absolutely convergent, and separating it into the row $m=0$, the rows $m>0$, and the rows $m<0$ is valid.[/guided]

[step:Evaluate the row with $m=0$] The contribution of the row $m=0$ is \begin{align*} \sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{1}{n^k} = \sum_{n=1}^{\infty}\frac{1}{n^k}+\sum_{n=1}^{\infty}\frac{1}{(-n)^k}. \end{align*} Since $k$ is even, $(-n)^k=n^k$, and hence this row contributes \begin{align*} 2\sum_{n=1}^{\infty}\frac{1}{n^k}=2\zeta(k). \end{align*} [/step]

[step:Evaluate the positive rows using the row sum Fourier expansion]For $w\in\mathbb{H}$, use the classical row sum expansion (citing a result not yet in the wiki: Row Sum Fourier Expansion) \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r w}. \end{align*} For each $m\in\mathbb{N}$, the point $mz$ lies in $\mathbb{H}$ because $\operatorname{Im}(mz)=m\operatorname{Im}(z)>0$. Therefore the row sum expansion applies with $w=mz$, giving \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(mz+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r m z}. \end{align*} Since $q=e^{2\pi iz}$, this becomes \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(mz+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*} Summing over $m\geq1$, the total contribution of the positive rows is \begin{align*} \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*}[/step]

[guided]For the positive rows, we fix $m\in\mathbb{N}$ and regard the inner sum over $n\in\mathbb{Z}$ as a one-dimensional periodic row sum. The theorem we use is the classical row sum Fourier expansion (citing a result not yet in the wiki: Row Sum Fourier Expansion): \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(w+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r w}, \end{align*} valid for $w\in\mathbb{H}$ and integer $k\geq2$. We must verify its hypothesis. Since $z\in\mathbb{H}$ and $m\in\mathbb{N}$, we have \begin{align*} \operatorname{Im}(mz)=m\operatorname{Im}(z)>0. \end{align*} Thus $mz\in\mathbb{H}$, so the row sum expansion applies with $w=mz$. We obtain \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(mz+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}e^{2\pi i r m z}. \end{align*} Now define $q:=e^{2\pi iz}$. Then \begin{align*} e^{2\pi i r m z}=q^{mr}, \end{align*} so the $m$-th positive row is \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(mz+n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*} Summing this identity over all positive integers $m$ gives the total positive-row contribution \begin{align*} \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*}[/guided]

[step:Show that the negative rows contribute the same amount] Let $a\in\mathbb{N}$. The row with $m=-a$ is \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(-az+n)^k}. \end{align*} Changing the summation index by $n=-\ell$, where $\ell\in\mathbb{Z}$, gives \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(-az+n)^k} = \sum_{\ell\in\mathbb{Z}}\frac{1}{(-az-\ell)^k} = (-1)^{-k}\sum_{\ell\in\mathbb{Z}}\frac{1}{(az+\ell)^k}. \end{align*} Since $k$ is even, $(-1)^{-k}=1$. Therefore \begin{align*} \sum_{n\in\mathbb{Z}}\frac{1}{(-az+n)^k} = \sum_{\ell\in\mathbb{Z}}\frac{1}{(az+\ell)^k}. \end{align*} Thus the negative rows have the same total contribution as the positive rows. [/step]

[step:Group the double series by the product $N=mr$]Combining the row computations gives \begin{align*} G_k(z) = 2\zeta(k) + 2\frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*} Because $z\in\mathbb{H}$, we have $|q|=e^{-2\pi\operatorname{Im}(z)}<1$. Hence \begin{align*} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}|q|^{mr} = \sum_{r=1}^{\infty}r^{k-1}\frac{|q|^r}{1-|q|^r} \leq \frac{1}{1-|q|} \sum_{r=1}^{\infty}r^{k-1}|q|^r <\infty. \end{align*} Thus the double series is absolutely convergent, and we may group terms according to $N=mr$. For each $N\in\mathbb{N}$, the pairs $(m,r)\in\mathbb{N}^2$ with $mr=N$ are exactly the pairs $(N/d,d)$ where $d$ runs over the positive divisors of $N$. Therefore \begin{align*} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr} = \sum_{N=1}^{\infty}\left(\sum_{d\mid N}d^{k-1}\right)q^N = \sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N. \end{align*} Consequently, \begin{align*} G_k(z) = 2\zeta(k) + 2\frac{(-2\pi i)^k}{(k-1)!} \sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N. \end{align*}[/step]

[guided]At this point we have computed the three pieces of the lattice sum: the row $m=0$, the positive rows, and the negative rows. Putting them together gives \begin{align*} G_k(z) = 2\zeta(k) + 2\frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr}. \end{align*} The factor $2$ appears because the negative rows equal the positive rows. We now want to rewrite the double series as a single [Fourier series](/page/Fourier%20Series) in powers of $q$. Since $z\in\mathbb{H}$, \begin{align*} |q|=|e^{2\pi iz}|=e^{-2\pi\operatorname{Im}(z)}<1. \end{align*} This exponential decay justifies rearranging the double series. Indeed, \begin{align*} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}|q|^{mr} = \sum_{r=1}^{\infty}r^{k-1}\sum_{m=1}^{\infty}|q|^{mr} = \sum_{r=1}^{\infty}r^{k-1}\frac{|q|^r}{1-|q|^r}. \end{align*} Since $1-|q|^r\geq 1-|q|$ for every $r\geq1$, we have \begin{align*} \sum_{r=1}^{\infty}r^{k-1}\frac{|q|^r}{1-|q|^r} \leq \frac{1}{1-|q|} \sum_{r=1}^{\infty}r^{k-1}|q|^r. \end{align*} The final series converges because polynomial growth is dominated by exponential decay. Hence the double series is absolutely convergent. We may now group terms by the exponent $N=mr$. For a fixed $N\in\mathbb{N}$, the condition $mr=N$ says that $r$ is a positive divisor of $N$ and $m=N/r$. Thus \begin{align*} \sum_{\substack{m,r\in\mathbb{N}\\ mr=N}}r^{k-1} = \sum_{d\mid N}d^{k-1} = \sigma_{k-1}(N). \end{align*} Therefore \begin{align*} \sum_{m=1}^{\infty}\sum_{r=1}^{\infty}r^{k-1}q^{mr} = \sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N, \end{align*} and so \begin{align*} G_k(z) = 2\zeta(k) + 2\frac{(-2\pi i)^k}{(k-1)!} \sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N. \end{align*}[/guided]

[step:Normalize and convert the scalar using Bernoulli numbers] By definition, \begin{align*} E_k(z)=\frac{G_k(z)}{2\zeta(k)}. \end{align*} Using the formula for $G_k(z)$ obtained above, \begin{align*} E_k(z) = 1+ \frac{(-2\pi i)^k}{(k-1)!\zeta(k)} \sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N. \end{align*} Since $k$ is even, write $k=2\ell$ with $\ell\in\mathbb{N}$ and $\ell\geq2$. Then \begin{align*} (-2\pi i)^k = (-2\pi i)^{2\ell} = (-1)^{\ell}(2\pi)^{2\ell} = (-1)^{k/2}(2\pi)^k. \end{align*} We use Euler's evaluation of the zeta value at an even positive integer (citing a result not yet in the wiki: Euler's Formula for Even Zeta Values) \begin{align*} \zeta(k)=(-1)^{k/2+1}\frac{B_k(2\pi)^k}{2k!}. \end{align*} Therefore \begin{align*} \frac{(-2\pi i)^k}{(k-1)!\zeta(k)} &= \frac{(-1)^{k/2}(2\pi)^k}{(k-1)!} \cdot \frac{2k!}{(-1)^{k/2+1}B_k(2\pi)^k} \\ &= -\frac{2k!}{(k-1)!B_k} \\ &= -\frac{2k}{B_k}. \end{align*} Substituting this scalar into the expression for $E_k(z)$ yields \begin{align*} E_k(z) = 1-\frac{2k}{B_k}\sum_{N=1}^{\infty}\sigma_{k-1}(N)q^N. \end{align*} This is the desired Fourier expansion. [/step]