Laurent series of $\sin(-\frac{1}{z^2})$ radii of convergence

Question

I am calculating radii of convergence of series for function: $$ f(z)=\sin(-\frac{1}{z^2}) $$ I started with Taylor expansion for $\sin$ and then inserted $-\frac{1}{z^2}$. I got: $$ -\frac{1}{z^2}+\frac{1}{3!}\frac{1}{z^6}-\frac{1}{5!}\frac{1}{z^{10}}+\frac{1}{7!}\frac{1}{z^{14}}...=\sum_{n=0} ^{\infty} (-1)^n \frac{1}{(2n-1)!}z^{4n-2} $$ Is this even right? But furthermore, I have to determine the radii of convergence? I tried to use root formula ($\limsup \sqrt[n] {a_n}$), I tried to calculate it with L'Hospital's rule, but I don't know the derivative of faculteta... Any idea, how to continue?

Answer 1

You can certainly insert $\frac{-1}{z^2}$ in $\sin$ if $z\neq0$ and you did it correctly but the series is

$\sum_{n=0}^{\infty} (-1)^{n+1} \frac{z^{-4n-2}}{(2n+1)!}$.

This case is easier to use the quotient rule:

$|\frac{a_{n+1}}{a_{n}}|=\frac{(2n+1)!}{(2n+3)!}|z^{-4}|\rightarrow0$ when $n\rightarrow\infty$, Then $\sin(\frac{-1}{z^2})$ converges for any $z\in\mathbb{C}$ except $z=0$. You could also notice that $\sin$ converges for any $z\in\mathbb{C}$, then in the domain of $g(z)=\frac{-1}{z^2}$ you have that $\sin(g(z))$ converges.