\displaystyle{sin(\frac{1}{z}) = \sum_{n=0}^{\infty}(-1)^n\frac{(\frac{1}{z})^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty}\frac{(-1)^n}{z^{2n+1}\cdot (2n+1)!} = \frac{1}{z\cdot 1!} - \frac{1}{z^3\cdot 3!} + \frac{1}{z^5\cdot 5!} - \ldots}