- The problem statement, all variables and given/known data
Show that $\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2})) =1$
where $\psi(z)=\frac{1}{z^2}+\sum\limits_{w \in \Omega}' \frac{1}{(z-w)^2}-\frac{1}{w^2}$
where $\Omega$ are the periods of $\psi(z)$
Relevant equations
The attempt at a solution
$\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2}))= 1 + \sum\limits_{w\in \Omega}'\frac{z^2}{(z-w)^2}-\frac{z^2}{w^2}$
The last time clearly vanishes.
For the second term I can write this as $\frac{1}{(1-\frac{w}{z})^2} \to 1 $ as $z \to 0 $
So I get
$\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2})) =1+ \sum_{w \in \Omega}' 1$
This is probably a stupid question but I don't really understand the summation second term here.
$\sum_{n=1}^{n=n} 1 = n$ right?
So there's an infintie number of $ w \in \Omega $ so obviously I dont want to do this, are you in affect looking at the limit '$mod \Omega $', so where the first term corresponds to $w=0 $ ?
Many thanks