Eisenstein series converge absolutely for $k\geq 2$ I am looking at Eistenstein series on modular forms: https://en.wikipedia.org/wiki/Eisenstein_series
The page claims that the series converges absolutely to a holomorphic function of τ when $k\geq 2$. I can't easily see why. It's apparent that when the exponent is odd, the series equal to $0$, so we only focus on the case of $2k$. 
But why does Eisenstein series converge absolutely iff the exponent is odd $(1,3, 5,...)$ or an even number bigger than $2$?
 A: In the notation used by Wikipedia the Eisenstein series converges absolutely for every $k\in \frac{1}{2}\mathbb{N}$, $k\geq \frac{3}{2}$. In general I prefer the notation:
\begin{equation*}
G_k(z) = \sum_{(n,m)\neq (0,0)} \frac{1}{(mz+n)^k}
\end{equation*}
In this fashion $G_k$ converges absolutely for $k\geq 3$, but as you have pointed out $G_k=0$ for odd $k$. We have also $G_1=0$, so for weight $1$ the Eisenstein series converge trivially.
The convergence of $G_k$ can be proved in several ways, but the core idea is always to compare it with the Riemann zeta function. The requirement $k\geq 3$ has exactly the role to ensure convergence of the Riemann zeta that appears.
A: In the case of the classical Eisenstein Series
$G_{2k} = \sum \frac{1}{(m+n\tau)^{2k}}$,
If $k=1$, then $G_{2}$ is up to some nonzero scalar multiple of the following
$G_{2} \equiv \frac{\eta^{'}(\tau)}{\eta(\tau)}$
Where $(2\pi)^{12} \eta^{24}(\tau)=\Delta (\tau)$ is the modular discriminant, a cusp form of weight 12. It can be proven by comparing the fourier coefficients. 
The derivatives of modular forms/functions are only modular if they are of weight 0. If it were the case the Eisenstein Series rep. for $G_{2}$ was at least a conditionally convergent sum for which rearrangement could possibly work, then it would it not be a quasimodular form.  But thats not the case. I'm not too sure if this enough to imply it is not a Modular form, so really take this with a grain of salt.
A: As an extension of a (partly mythical) "integral test", you can prove that a sum $\sum_{0\not=\lambda\in L}{1\over |\lambda|^s}$, for a lattice (a discrete copy of $\mathbb Z^n$) inside $\mathbb R^n$ converges if and only if $\Re(s)>n$. This gives perspective...
