Expand $\frac{z-\sin z}{z^3}$ Around $z=0$

Question

Expand $\frac{z-\sin z}{z^3}$ Around $z=0$

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

So $$\frac{z-\sin z}{z^3}=\frac{z-\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}}{z^3}=\frac{1}{z^2}-\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n-2}}{(2n+1)!}=z^{-2}-\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n-2}}{(2n+1)!}$$

Can we say that the singularity will be a pole of order $3$ (independent on the point around we expand?)

$R=lim_{n\to \infty}|\frac{a_n}{a_{n+1}}|=lim_{n\to \infty}\frac{(2n+3)!}{(2n+1)!}=lim_{n\to \infty}4n+5=\infty$

So the Laurent series converges for all $z$

Is it correct?

Answer 1

$$\frac{z-\sin z}{z^3}=\frac{\displaystyle z-\sum_{n\ge 0}\dfrac{(-1)^n z^{2n+1}}{(2n+1)!}}{z^3}=\frac{\displaystyle\sum_{n\ge1}\dfrac{(-1)^{n+1} z^{2n+1}}{(2n+1)!}}{z^3}=\sum_{n\ge1}\dfrac{(-1)^{n+1} z^{2(n-1)}}{(2n+1)!},$$ which can be simplified to $$\sum_{n\ge0}\dfrac{(-1)^{n} z^{2n}}{(2n+3)!}.$$

Answer 2

The sine is an odd function and $\sin'(0)=1$ so that the terms $z$ cancel each other and the next term is in $z^3$. Then after division by $z^3$, there is no pole.

The series is certainly convergent because except for the first term and a factor $z^3$, it is the series for the sine, which is known to converge everywhere.