This question is related to this one.
I was able to figure out on my own that the residue of $\displaystyle \sin \left(\frac{1}{z} \right)$ is defined at $z=0$ by finding the Laurent Series of $\displaystyle \sin \left( \frac{1}{z}\right)$ centered at $z=0$, and noting that the coefficient of the $z^{-1}$ term was $1$, and so $\displaystyle res_{0} \sin \left(\frac{1}{z} \right) = 1$.
(I honestly found this to be a bit surprising, because the way the question was posed to us was "Is the residue of $\sin(1/z)$ defined at $z=0$? If yes, find the residue; if not, explain why.", and usually when our professor asks questions like this, the answer is no!)
Now, however, I have to determine the same thing about $\displaystyle \frac{1}{\displaystyle \sin\left(\frac{1}{z} \right)} = \csc \left(\frac{1}{z} \right) = \csc (z^{-1})$. I tried to approach it the same way, by setting up the Laurent Series for $\csc(z^{-1})$, but I'm not sure how to do that. Looking up what the Laurent Series for $\csc(z)$ is supposed to be, it has a strange "$B_{2n}$" term that I don't understand:
$\displaystyle \csc(z) = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}2(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}$
Could someone please explain to me, step-by-step, how to derive the $\csc(z)$ series from the $\sin(z)$ series, and confirm for me please, that the residue of both $\displaystyle \sin \left( \frac{1}{z}\right) $ at $z=0$
$z=0$ is not an isolated singularity of $\csc(1/z)$; it is a limit point of poles, since $1/(k\pi)$ is a pole for all integer $k\ne0$. There is no Laurent series of $\csc(1/z)$ around $z=0$, and therefore the residue is not defined.
The $B_{2n}$ are Bernoulli numbers.