1
$\begingroup$

I'm reading through a calculation for $\displaystyle z\pi\cot(\pi z)=\pi iz+\frac{2\pi iz}{e^{2\pi iz}-1}$ which confuses me. It states $ \begin{align*} z\pi\cot(\pi z) &= \pi iz+\frac{2\pi iz}{\sum_{k=1}^\infty\frac{(2\pi iz)^k}{k!}}\\ &= \pi iz+\frac{1}{1-(-\sum_{k=1}^\infty\frac{(2\pi iz)^k}{(k+1)!})}\\ &=\pi iz+\sum_{n=0}^\infty\left(-\sum_{k=1}^\infty\frac{(2\pi iz)^k}{(k+1)!}\right)^n\\ &=\pi iz+\sum_{k=0}^\infty\frac{B_k}{k!}(2\pi iz)^k \end{align*} $ where $B_k$ is the $k$th Bernoulli number. I don't follow the last equality from the third to fourth line. How does it follow? Thanks.

  • 0
    Oh, that makes it much easier to see. Thanks!2012-04-16

0 Answers 0