$$G_k (z)=\sum_{\displaystyle(m,n) \neq (0,0)} \frac{1}{(mz+n)^k}$$
For $k \geq 3$ how to show that $G_k (z)=0$ for odd $k$ and
$G_k (\tau z)=(cz+d)^k G_k (z)$ for $\tau \in SL_2(Z), \tau = \left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right) $
$(m,n)$ are integers
For odd values of $k$, you can see that if you expand the sum, you'll get cancellations of terms because $m$ and $n$ run through $\mathbb{Z}$ and you have negative values to an odd power and also positive values to an odd power. For the second part:
$$G_k (\tau z)=\sum_{\displaystyle(m,n) \neq (0,0)} \frac{1}{(m(\tau z)+n)^k}$$ $$=\sum_{\displaystyle(m,n) \neq (0,0)} \frac{1}{(m(\frac{az+b}{cz+d})+n)^k}$$ $$=\sum_{\displaystyle(m,n) \neq (0,0)} \frac{1}{(\frac{1}{cz+d})^{-k}(m(az+b)+n(cz+d))^k}$$ $$=(cz+d)^k\sum_{\displaystyle(m,n) \neq (0,0)} \frac{1}{((ma+nc)z+(mb+nd))^k}$$
But $m$ and $n$ $\in\mathbb{Z}$ run through all of $\mathbb{Z}$ so : $$G_k (\tau z)=(cz+d)^k\sum_{\displaystyle(m',n') \neq (0,0)} \frac{1}{(m'z+n')^k}=(cz+d)^k G_k (z)$$ Where $m'=ma+nc$ and $n'=mb+nd$.