Factorization of $\Delta$-function I am reading Serre's 'A course in Arithmetic' and got stuck in his Theorem 6 (Jacobi) on page 95. 
Specifically, he defines
$$ H_1(z) = \sum_n \sum_m' \frac{1}{(m-1+nz)(m+nz)}, \hspace{5mm}
H(z) = \sum_m \sum_n' \frac{1}{(m-1+nz)(m+nz)} $$
where the prime means the sum runs through the pairs $(m,n)\neq (0,0),(1,0)$. He writes: $\textit{The series $H_1$ and $H$ are easy to calculate explicitly because of the formula}$
$$ \frac{1}{(m-1+nz)(m+nz)} = \frac{1}{m-1+nz} - \frac{1}{m+nz} $$
$\textit{One finds that they converge, and that}$
$$ H_1 = 2, \hspace{5mm} H = 2-2\pi i/z $$
I understand the fraction identity, and I see how you obtain $H_1 = 2$, but I can't seem to get the right result for $H$. 
Any help is appreciated. 
 A: Let's write $A_{m,n}=1/(m+nz)$. Then
\begin{align}
H(z)&=\sum_{m\ne0,1}\sum_n(A_{m-1,n}-A_{m,n})
+\sum_{n\ne0}(A_{-1,n}-A_{0,n})+\sum_{n\ne0}(A_{0,n}-A_{1,n})\\
&=\sum_{m=2}^\infty\sum_n(A_{m-1,n}-A_{m,n})+
\sum_{m=-\infty}^{-1}\sum_n(A_{m-1,n}-A_{m,n})
+\sum_{n\ne0}(A_{-1,n}-A_{1,n})\\
&=\lim_{M\to\infty}\sum_{n}(A_{1,n}-A_{M,n})
+\lim_{M\to\infty}\sum_{n}(A_{-M,n}-A_{-1,n})
+\sum_{n\ne0}(A_{-1,n}-A_{1,n})\\
&=\lim_{M\to\infty}\sum_{n\ne0}(A_{1,n}-A_{M,n})
+\lim_{M\to\infty}\sum_{n\ne0}(A_{-M,n}-A_{-1,n})
+\sum_{n\ne0}(A_{-1,n}-A_{1,n})\\
&\quad+\lim_{N\to\infty}(A_{1,0}-A_{M,0})
+\lim_{N\to\infty}(A_{-M,0}-A_{-1,0})\\
&=\lim_{M\to\infty}\sum_{n\ne0}(A_{-M,n}-A_{M,n})+2\\
&=2+\lim_{M\to\infty}\sum_{n\ne0}\frac{2M}{n^2z^2-M^2}\\
&=2+\lim_{M\to\infty}\frac1{z}\sum_{n\ne0}\frac{2M/z}{n^2-(M/z)^2}\\
&=2+\frac2z\lim_{M\to\infty}\left(\frac zM-\pi\cot\frac{\pi M}{z}\right).\end{align}
As $z$ is in the upper half plane, $1/z$ is in the lower half plane.
Now, $\cot w\to i$ when the imaginary part of $i$ tends to $-\infty$.
We conclude that
$$H(z)=2-\frac{2\pi i}z.$$
