Find power series of $\frac{1}{\cosh z}$ I`m not sure what is the best way to find the power series for $\frac{1}{\cosh z}$ about the point 0. 
I would appreciate it if you give me any help. 

Thank you
Let $f(z)=\displaystyle{\frac{1}{\cosh z}}$. Since $f(z)$ is analytic at $z=0$, so it has a Taylor series expansion at $0$. We can try by finding out the derivatives $f^{(n)}(z) $  of $f$ and put it in the formula 
$f(z)=f(0)+\displaystyle{\frac{f'(0)}{1}z+\frac{f''(0)}{2!}z^2 +\cdots}$.

I have found $f(0)=1$. $f'(z) =\displaystyle{-\frac{\sinh z}{\cosh^2 z}}$, $f''(z) =\displaystyle{\frac{\sinh^ z-1}{\cosh^3 z}}$ and $f'''(z) =\displaystyle{\frac{\sinh z(2-\sinh z) }{\cosh^4 z}}$ and so on.
Hope that it  help you :$$\cosh z=\dfrac{e^z+e^{-z}}{2}=\\
\dfrac{2(1+\frac12z^2+\frac{1}{4!}z^4+...)}{2}=\\1+\frac12z^2+\frac{1}{4!}z^4+...
 \to \\\dfrac{1}{\cosh z}=\dfrac{1}{1+\frac12z^2+\frac{1}{4!}z^4+...}=\\
\dfrac{1}{1-(-(\frac12z^2+\frac{1}{4!}z^4+...))}$$around $z=0\to |z|\leq 1$so 
$$\dfrac{1}{1-(-(\frac12z^2+\frac{1}{4!}z^4+...))}=1+()+()^2+()^3+()^4+...=\\
1+(-(\frac12z^2+\frac{1}{4!}z^4+...))+(-(\frac12z^2+\frac{1}{4!}z^4+...))^2+(-(\frac12z^2+\frac{1}{4!}z^4+...))^3+(-(\frac12z^2+\frac{1}{4!}z^4+...))^4+...=\\$$