February 10th
Today I learned that the Eisenstein series are modular forms, from here . These are the series, parameterized by $k\ge3,$\[G_k(z)=\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{(mz+n)^k}.\]We get the main computation out of the way first. Because $\op{SL}_2(\ZZ)$ is generated by $90^\circ$ rotations and horizontal shears, it suffices to note that\[G_k\left(\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}z\right)=G_k\left(\frac{-1}z\right)=\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{\left(-\frac mz+n\right)^k}=\frac{G_k(z)}{z^k},\]and\[G_k\left(\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}z\right)=G_k(z+1)=\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{(mz+m+n)^k}=G_k(z).\]In particular, satisfying the modularity condition is closed under multiplication and inversion, so it is enough to satisfy the modularity condition for the above matrices.
We now show the uninteresting parts of this proof. We need $G_k$ to be holomorphic and holomorphic at $\infty$ to be a proper modular form. To show that $G_k$ is holomorphic, we need to know that it converges everywhere, and then I think manually checking the limit definition or something. Anyways, we begin by showing, for $k\ge3,$\[S_k=\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{(|m|+|n|)^k}\]converges, which looks of roughly the same growth rate as $G_k.$ All terms are positive, so we may rearrange terms into showing the convergence of\[S_k=4\Bigg(\underbrace{\sum_{n=1}^\infty\frac1{n^k}}_{m=0}+\sum_{m,n=1}^\infty\frac1{(m+n)^k}\Bigg)\]by taking the upper-right quadrant of $\ZZ^2.$ The first sum here is $\zeta(k)$ and therefore converges. To deal with the double sum, we need to bound what it looks like for a single $m,$ which is\[\sum_{n=1}^\infty\frac1{(m+n)^k}\le\frac1{(m+1)^k}+\int_1^\infty\frac1{(m+t)^k}\,dt=\frac1{(m+1)^k}+\frac1{(k-1)(m+1)^{k-1}}.\]This is bounded by $2/m^{k-1},$ so we see\[\sum_{m,n=1}^\infty\frac1{(m+n)^k}\le2\sum_{m=1}^\infty\frac1{m^{k-1}}=2\zeta(k-1).\]Note $k \gt 2$ tells us that the sum is upper-bounded, and because it only has positive terms, the sum must converge.
Now we show that $G_k$ converges; as should be expected, we show it absolutely converges. That is, we focus on\[|G|_k(z)=\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{|mz+n|^k}.\]This feels like it should have similar growth rate as $S_k.$ We codify this by exhibiting a constant $C$ for which\[|mz+n|\le C(|m|+|n|),\]which will show $|G|_k(z)$ converges by comparison. For $m=0,$ any $C\ge1$ will suffice. Else, rearranging, we see we are interested in upper-bounding\[\frac{|mz+n|}{|m|+|n|}=\frac{|z+|n/m||}{1+|n/m|}=:f(|n/m|).\]The function $f:\RR_{\ge0}\to\RR$ is certainly defined over all real inputs as well as continuous, so it suffices to check its behavior as $|n/m|\to\infty.$ Letting $z=a+bi,$ this is\[\lim_{x\to\infty}\frac{|z+x|}{1+x}=\lim_{x\to\infty}\frac{|z/x+1|}{1/x+1}=1 \lt \infty.\]So because $f(x)$ is continuous on the closed interval $[0,\infty)$ and bounded as $x\to\infty,$ we see $f$ must be bounded, giving us our $C.$ This finishes the proof of $G_k$ being holomorphic.
It remains to show that $G_k$ is holomorphic at $\infty.$ For this, we merely have to show that\[\lim_{\op{Im}z\to\infty}G_k(z) \lt \infty.\]Well, it (as usual) suffices to show this for $|G|_k,$ where the statement reads\[\lim_{\op{Im}z\to\infty}\sum_{\substack{(m,n)\in\ZZ^2\\(m,n)\ne(0,0)}}\frac1{|mz+n|^k} \lt \infty.\]The $m=0$ terms gives us $2\zeta(k) \lt \infty$; else, when $m \gt 1,$ all of our terms vanish. Thus, $G_k$ is indeed holomorphic at $\infty,$ which completes the proof that $G_k$ is a modular form.
In other news, happy birthday to Sara Modur and Alan Baade!