# Show that $\sum_{d\in\mathbb{Z}}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}=-{2\pi i\over \tau}$?

It is an exercise about Eisenstein Series $G_2(\tau)$, to prove that $$(G_2[\gamma]_2)(\tau)=G_2(\tau)-{2\pi i\over\tau}$$ where $\gamma=\begin{pmatrix} 0&1\\ -1 &0 \end{pmatrix}$

I just cannot show that $$\sum_{d\in\mathbb{Z}}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}=-{2\pi i\over \tau}$$ I tried $$\lim_{N\to\infty}\sum_{d=-N}^{N-1}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}=\lim_{N\to\infty}\sum_{c\ne0}({1\over c\tau-N}-{1\over c\tau+N})$$ and to use the equality $$\pi\cot\pi N/\tau={\tau\over N}-\tau\sum_{c=1}^{\infty}({1\over c\tau-N}-{1\over c\tau+N})$$ but $\displaystyle\lim_{N\to\infty}\pi\cot\pi N/\tau$ doesn't exist.

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What is $\tau$ supposed to be? The infinite series $\sum_{d\in\mathbb{Z}}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}$ does not converge if $\tau \in \{0,1\}$. And if it were to converge for some $\tau \in \mathbb{R}$, the answer could not be a complex number. – Hans Engler Apr 3 '13 at 22:29
@HansEngler The book says $\tau\in\mathbb{c}\cup\{\infty\}$ – Ziqian Xie Apr 3 '13 at 22:39
$\tau$ is almost always meant to represent two complex numbers, say $\frac{\omega_1}{\omega_2}$, whose ratio is not real and is contained in the upper half plane – Brent J Apr 4 '13 at 0:52
@ Brent: So the right hand side of this identity is analytic. How about the left hand side? Is the double series expected to converge in an open set of the complex plane? If so, it should be possible to continue it analytically. (This is just a reality check. I don't actually know anything about Eisenstein series.) – Hans Engler Apr 4 '13 at 16:39
@HansEngler Not sure why you're asking me because I didn't ask the question, although I may know a solution. Anyways, an Eisenstein series, like any modular form, is holomorphic at the cusp. In fact, it is absolutely convergent to a holomorphic function of $\tau$. I believe it has an analytic continuation to the whole complex plane with a simple pole at $\pm\frac{1}{2}$ – Brent J Apr 4 '13 at 17:10