In the book it is written:
By using$$\frac{1}{\sin z } = \cot z + \tan\frac{z}{2}$$ one can easily compute the Taylor series (of the holomorphical extension) of $\displaystyle\frac{z}{\sin z}$ at $z=0$
Definitions:$$\cot z := \frac{1}{\tan z}=\frac{\cos z}{\sin z} ,\qquad\tan z:= \frac{\sin z}{\cos z} $$
Taylor series at z=0 for $\sin z$ and $\cos z$:$$\sin z= \sum_{k=0}^{\infty} \frac{(-1)^{k}z^{2k+1}}{(2k+1)!};\qquad \cos z= \sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k}}{(2k)!} $$
$\Rightarrow$:$$\cot z + \tan\frac{z}{2} = \frac{\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k}}{(2k)!}}{\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k+1}}{(2k+1)!}} + \frac{\sum_{k=0}^{\infty} \frac{(-1)^{k}z^{2k+1}}{(2k+1)!}}{\sum_{k=0}^{\infty}\frac{(-1)^{k}(z/2)^{2k}}{(2k)!}}$$
I am stuck here. This approach doesn't seem to be very good. Does anybody see a better one?
Best Answer
Just guessing here. Maybe that book wants to get the student to figure out the series for $z/\sin z$ which involves Euler numbers, and in that book you already know series for $\tan z$ in terms of Bernoulli numbers, plus a relation between Euler numbers and Bernoulli numbers...