<s>[INST] Eisenstein and Weierstrass zeta - series identity Let $\zeta_\Lambda$ be the Weierstrass $\zeta$-function for lattice $\Lambda$ and $G_2$ the Eisenstein series of weight $2$.
The quasiperiod is defined by $\eta_\Lambda(\lambda) := \zeta_\Lambda(z + \lambda) - \zeta_\Lambda(z)$.
Then the following identity holds \begin{align*} \eta_{\Lambda_\tau}(1)=G_2(\tau) .\end{align*}
I tried to check the series definition and i am not sure on how to continue with that 
\begin{align*}
 \zeta_\Lambda(z + 1) - \zeta_\Lambda(z) &= \frac{1}{z+1} + \left( \sum_{(c,d) \neq (0,0)} \frac{1}{(z+1)-(c \tau +d)} +\frac{1}{c \tau +d}+ \frac{z+1}{(c \tau +d)^2}\right) \\ &-\frac{1}{z} + \left( \sum_{(c,d) \neq (0,0)} \frac{1}{z-(c \tau +d)} +\frac{1}{c \tau +d} + \frac{z}{(c \tau +d)^2}\right) \\ &= \left(\frac{1}{z+1} - \frac{1}{z} \right) +... 
\end{align*}
Any help is really appreciated
 [/INST] $$G_2(\tau)=\sum_c\sum_d\frac{1_{(c,d)\ne (0,0)}}{(c\tau+d)^2}$$
It converges and is holomorphic for $\Im(\tau)> 0$ because $\sum_d \frac{1}{(c\tau+d)^2} = \frac{\pi^2}{\sin^2(\pi c\tau)}= O(e^{-2 \pi |c\Im(\tau)|})$. Then $$\zeta_\tau(z) =\sum_c\sum_d\frac{1}{z-c\tau-d}+1_{(c,d)\ne (0,0)}(\frac{1}{c\tau+d}+\frac{z }{(c\tau+d)^2})$$
$$ =\sum_c\sum_d\frac{1}{z-c\tau-d-1}+1_{(c,d+1)\ne (0,0)}(\frac{1}{c\tau+d+1}+\frac{z }{(c\tau+d+1)^2})$$
From there the result is immediate
$$\zeta_\tau(z+1)-\zeta_\tau(z)-G_2(\tau)\\=\sum_c\sum_d 1_{(c,d)\ne (0,0)}(\frac{1}{c\tau+d}+\frac{z+1}{(c\tau+d)^2})-1_{(c,d+1)\ne (0,0)}(\frac{1}{c\tau+d+1}+\frac{z}{(c\tau+d+1)^2})-\frac{1_{(c,d)\ne (0,0)}}{(c\tau+d)^2}\\=0$$
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