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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?

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    $\begingroup$ If memory serves, this Taylor expansion is somewhat messy if you expand the numerator and the denominator separately. The easier way to do it is to use the Taylor expansion for the whole expression at once. $\endgroup$
    – Dvij D.C.
    Aug 5 at 22:17
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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.

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  • $\begingroup$ Thank you! How did you get the Taylor expansion? By definition or by doing the Taylor expansion of denominator and numerator and then multiplying? $\endgroup$
    – Gottfried
    Aug 5 at 22:17
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    $\begingroup$ I did Taylor exp of numerator and denominator separately and then multiplied. You have keep track of a lot of terms though. $\endgroup$
    – SuperCiocia
    Aug 5 at 22:24
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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 $x:=\tfrac{1}{12}h^2$.)

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