Find power series of $\frac{1}{\cosh z}$

Question

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

Answer 1

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 $$\dfrac{1}{1-(-(\frac12z^2+\frac{1}{4!}z^4+...))}=1+()+()^2+()^3+()^4+...=\\ 1+(-(\frac12z^2+\frac{1}{4!}z^4+...))+(-(\frac12z^2+\frac{1}{4!}z^4+...))^2+(-(\frac12z^2+\frac{1}{4!}z^4+...))^3+(-(\frac12z^2+\frac{1}{4!}z^4+...))^4+...=\\$$