# Power Series With Bernoulli Numbers

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

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

$f(z)+f(-z)=\frac{z(e^{t/2}+e^{-t/2})}{(e^{t/2}-e^{-t/2})}$

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

$\cot(z)-\cot(2z)=\csc(2z)$

So subtract the power series to get your result.

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