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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?

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    How about you actually compute the series for $\cot z$ and $\tan (z/2)$? I don't know if those have nice formulas though.2011-11-29
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    What book is this? Personally, I'd just start from the series for $\dfrac{\sin\,z}{z}$ and do long division...2011-11-29
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    The series for $\cot z$ and $\tan(z/2)$ do have closed-form formulas for the coefficients, involving Bernoulli numbers. I don't know if you'd call that "nice". But I suspect "the book" is just assuming the first few are known. Of course, the series for $\csc(z)$ is just about as well-known.2011-11-29
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    $$\cot \left( z \right) =\sum _{j=0}^{\infty }{\frac { \left( -1 \right) ^{j}{2}^{2\,j}\ {\it bernoulli} \left( 2\,j \right) {z}^{2\,j-1 }}{ \left( 2\,j \right) !}} $$2011-11-29
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    ... and $\tan(z) = \cot(z) - \cot(2 z)$.2011-11-29
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    The book says: $tanz= cotz - 2cot 2z $2011-11-29
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    T_T………………………...2011-11-29

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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...

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    Yes, you are right. I have figured it out. Thank you.2011-11-29