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The exercise reads "Express the power series for $\large \frac{z}{\sin (z)} = \frac{2 i z}{e^{iz} - e^{-iz}} $ in terms of Bernoulli numbers."

I am given in a previous exercise that the Bernoulli numbers are defined by $$ \frac{z}{e^z - 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n, $$ where $ B_n $ is the $ n^{th} $ Bernoulli number. I've been looking for a clever way to write $ \;\large\frac{1}{e^{iz} - e^{-iz}} \;$ in the form of a linear combination involving terms of the form $\; \large\frac{1}{e^z - 1}\; $, but haven't had any luck. Any pointers would be greatly appreciated!

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You could clear the denominator to get $\exp(iz)\frac{2iz}{e^{2iz}-1}$ and you know the fraction's expansion in terms of bernoulli numbers so you can multiply by the expansion of $\exp(iz)$ and match coefficients of $z^n$ – Alex R. Dec 22 '12 at 4:21
@Alex I did notice this earlier. I tried it and couldn't find any sort of pattern in the expansion. I believe this would be a good solution for a computer though. – tylerc0816 Dec 22 '12 at 4:40
up vote 7 down vote accepted

If you start with $f(z):=\frac{z}{e^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}$, consider what $f(z)+f(-z)$ is. You'll get a sum of only even bernoulli numbers. Now notice that


which you can reverse engineer to verify (multiply by the right ratio which equuals 1). This essentially gives you an expansion for $\cot(z)$ in terms of the even Bernoulli numbers. However you want $\csc(z)$. This is amenable by recalling that


So subtract the power series to get your result.

The correct answer is:

$$\csc(z)=\sum_{n=0}^\infty \frac{(-1)^{n+1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}$$

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Very nice! Thanks a million! – tylerc0816 Dec 22 '12 at 14:11

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