# Laurent Series and Taylor Expansion of $1 / (e^z - 1)$ Could someone please assist me with the second part of the second paragraph, from "By expanding $f_1$..."?

I am not convinced that my expansion for $f_1$ is right - I used the standard binomial, but with $$(z^2 + 4 \pi^2 r^2)^{-1} = z^{-2}(1 + ({2\pi r \over z})^2)^{-1}$$ then normal binomial on this, but only need the first term. I could then find the Taylor series without using my previous Laurent series for $f$, which I assume that I'm not supposed to do since I am then finding the Laurent series for $f$, and my previous was only valid for $0 < |z| < 2\pi$.

Thanks in advance! :)

• I get the following when I expand $f_1$: $f_1(z) = \frac{1}{z} + \sum_{r = 1}^n \frac{2z}{z^2\left(1 - (-\frac{4\pi^2r^2}{z^2}\right)} = \frac{1}{z} + \frac{2}{z}\sum_{r = 1}^n \left(\sum_{i = 0}^\infty (-1)^i\left(\frac{2\pi r}{z}\right)^{2i}\right)$ – Jared Apr 7 '14 at 22:28
• Thanks :). That was indeed what I got also. The next bits were the issues... – Sam T Apr 8 '14 at 7:07