Casimir Effect and Regularization In Quauntum Field Theory in a Nutshell by A. Zee, Section 1.9, about the Casimir Effect he introduces a regulator $e^{-an/d}$ where $a$ has units of length in natural units and $d$ is the space between two conducting plates. The energy at $d$ is given by:
$$ f(d) = \frac{\pi}{2d} \sum_{n=1}^{\infty} e^{-an/d}= -\frac{\pi}{2\color{red}{d}} \frac{\partial}{\partial a} \sum_{n=1}^{\infty} e^{-an/d} = -\frac{\pi}{2\color{red}{d}} \frac{\partial}{\partial a} \frac{1}{1-e^{-a/d}} = \frac{\pi}{2d} \frac{e^{a/d}}{(e^{a/d}-1)^{2}}$$
Should not the third equality be
$$ -\frac{\pi}{2\color{red}{d}} \frac{\partial}{\partial a} \frac{e^{-a/d}}{1-e^{-a/d}} $$
Also he says that since we want $a^{-1}$ to be large, we take the limit $a$ small so that:
$$f(d) = \frac{\pi d}{2a^{2}} - \frac{\pi}{24 d} + \frac{\pi a^{2}}{480 d^{3}} +O(a^{4}/d^{5})$$
I tried Taylor expanding the last equality but did not work out. Any hints?
 A: 
Should not the third equality be

Yes, it's a typo.
They probably wanted to write:
$$ \frac{1}{\mathrm{e}^{a/d}-1}$$

I tried Taylor expanding the last equality but did not work out. Any
hints?

The Taylor expansion is
$$ \frac{\exp(h)}{(\exp(h)-1)^2} \approx \frac{1}{h^2}-\frac{1}{12}+\frac{h^2}{240} \dots... $$
substitute $h = a/d$ and use the pre-factor $\pi/(2d)$ (from which you dropped the $d$ in the later steps), and you get the required expression.
A: To supplement @SuperCiocia's answer, I'll calculate its Laurent series more efficiently than with the method suggested in their comment. Squaring$$\tfrac{e^h-1}{e^{h/2}}=e^{h/2}-e^{-h/2}=h\left(1+\tfrac{h^2}{24}+\tfrac{h^4}{1920}+o(h^4)\right)$$gives$$\tfrac{(e^h-1)^2}{e^h}=h^2\left(1+\tfrac{h^2}{12}+\tfrac{h^4}{360}+o(h^4)\right).$$The Ansatz$$\frac{e^h}{(e^h-1)^2}=h^{-2}\left(1-\tfrac{h^2}{12}+Ah^4+o(h^4)\right)$$gives$$0=\tfrac{1}{360}-\tfrac{1}{144}+A\implies A=\tfrac{1}{240}.$$(You can arguably make the arithmetic a little easier with strategic powers of $x:=\tfrac{1}{12}h^2$.)
