expanding $\frac 1{\log\left( \frac{z+1}{z-1} \right) }$ at infinity Just as the title says, how do we expand such a function at infinity? I am told that the first term is $z/2$. My knowledge in complex analysis is very rusty, so if anyone can help me I'd be really thankful.

The "quick and dirty" way to do this is to view this as a composition of series with well-known forms:
\begin{align*}
\left[ \ln \left( \frac{z + 1}{z - 1} \right) \right]^{-1} &= \left[ \ln \left( 1 + \frac{1}{z} \right) - \ln \left(1 - \frac{1}{z} \right) \right]^{-1} \\
&= \left[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n z^n} + \sum_{n=1}^{\infty} \frac{1}{n z^n} \right]^{-1} \\
&= \left[ \sum_{\text{odd $n$}} \frac{2}{n} z^{-n}\right]^{-1} \\
&= \left( \frac{2}{z} \right)^{-1} \left( 1 + \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right)^{-1} \\
&= \frac{z}{2} \left[ 1 - \left( \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right) + \left( \frac{1}{3} z^{-2} + \frac{1}{5} z^{-4} + \dots \right)^2 + \dots \right]
\end{align*}
At this point you can expand as many terms in the "series of series" in the last expression to get the series to the order that you need.  This is likely to be tedious for anything but the lowest few orders in the series, but in principle it can be done.
