What is the Taylor series of $\frac{1}{\sin(z)}$ about $z_0 = 1$? This was a exam question so I know it cannot take too long to write out the proof. Only I cannot see an answer.
I would imagine you write $\sin(z) = \sin(1+(z-1)) = \sin(1)\cos(z-1) + \sin(z-1)\cos(1)$ and then use the everywhere-defined Taylor series for $\sin$ and $\cos$ to write $\frac{1}{\sin(z)}$ as the reciprocal of a power series. Then you manipulate it into the form $\displaystyle \frac{1}{1-P(z)}$ where $P$ is a power series and then invert using the geometric series formula. Only my $P$ looks horrible and thus the condition for convergence $|P(z)|<1$ is impossible to compute. 
Another series which I cannot do but which I imagine could be done by similar methods is $\displaystyle \frac{1}{2\cos(z) -1}$ about $z_0 = 0$.
Any tips?
 A: Using the Taylor series expansion for $\csc(x)=\frac{1}{\sin(x)}$, we may write write $$\frac{1}{\sin(z)}=\csc(z)=\sum_{k=0}^{\infty}\csc^{(k)}(1)\frac{(z-1)^{k}}{k!}.$$ This series will have radius of convergence $1$ since the poles of $\csc(x)$ are precisely at the the zeros of $\sin(x)$, all of which occur at integer multiples of $\pi$. 
Now, this might not be satisfying at all, but this is likely the best that you can do. Ideally we would want to specify the coefficients $\csc^{(k)}(1)$ exactly  (by giving a power series expansion we are doing precisely that) however they are extremely messy. The only way I can think of writing them without derivatives involves the Bernoulli numbers and powers of $\sin$ and $\cos$ evaluated at $1$. 
A: The function $z\mapsto 1/\sin(z)$ is meromorphic and has simple poles at points of $\pi\Bbb{Z}$. Thus it has a power series expansion $\sum_{n=0}^\infty a_n(z-1)^n$ 
around $z_0=1$, with radius of convergence $R=d(1,\pi\Bbb{Z})=1$.
Now to determine the coefficients we may can use the identity
$1=\sin(z)\left(\sum_{n=0}^\infty a_n(z-1)^n\right)$ 
in the neighborhood of $z_0=1$. Or, by setting $z=1+t$:
$$
\left(\sin(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}t^{2k}
+\cos(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}t^{2k+1}\right)
\left(\sum_{n=0}^\infty a_n t^n\right)=1
$$
So, the coefficients $\{a_n\}$ may be obtained inductively by the formula 
$$\eqalign{
a_0&=\frac{1}{\sin(1)},\cr
a_n&=-\sum_{k=1}^n\frac{(-1)^{\lfloor k/2\rfloor}}{k!}\delta_ka_{n-k},\
}
$$
where $\delta_k=\left\{
\matrix{\cot(1)&\hbox{if $k$ is odd,}\phantom{z}\cr
1&\hbox{if $k$ is even.}}
\right.$
