Find the Laurent series of $\sin z/z^2$ using Laurent's theorem

Question

I have the function $f(z)=\frac{\sin z}{z^2}$, which is analytic over $\Bbb C\setminus\{0\}$, I want to find the Laurent series of $f$ valid for $0<|z|<R\le\infty$. Using Laurent's theorem we have that the coefficients of the series representation are given by: $$a_k=\frac1{2\pi i}\int_{C}\frac{g(z)}{(z-z_0)^{k+1}}dz$$ Using this for $f$ we get: $$a_k=\frac1{2\pi i}\int_{C}\frac{\sin z}{z^2 z^{k+1}}dz \tag{1}$$ where $$\int_{C}\frac{\sin z}{z^2 z^{k+1}}dz=\int_{C}\frac{\sin z}{z^{2(k+1)}}dz=2\pi i f^{(2k+1)}(0)= \begin{cases}\cos(0)2\pi i=2\pi i & \text{ if $k=0,4,8,\dots$} \\-\cos(0)2\pi i=-2\pi i & \text{ if $k=2,6,10,\dots$} \\\pm \sin(0)2\pi i=0 & \text{ if $k=1,3,5,\dots$} \end{cases}$$ hence $$a_k=\begin{cases}1 & \text{ if $k=0,4,8,\dots$} \\-1 & \text{ if $k=2,6,10,\dots$} \\ 0 & \text{ if $k=1,3,5,\dots$} \end{cases}$$

From here I don't know how to get the Laurent representation of $f$. Am I doing something wrong?

Answer 1

$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots=\sum_{n=1}^\infty (-1)^{n-1}\frac{z^{2n-1}}{(2n-1)!}\implies$$

$$\frac{\sin z}{z^2}=\frac1z-\frac z{3!}+\frac{z^3}{5!}-\ldots=\frac1z-\sum_{n=1}^\infty(-1)^{n-1}\frac{z^{2n-1}}{(2n+1)!}$$