<s>[INST] Showing series expansion How would I show $$\frac{\sinh z}{z^2}=\frac{1}{z}+\sum_0^\infty \frac{z^{2n+1}}{(2n+3)!}$$ 
I have $\sinh z=\sum_0^\infty\frac{z^{2n+1}}{(2n+1)!}$ and if I multiply by $\frac{1}{z^2}$ then $\frac{1}{z^2}\sinh z=\sum_0^\infty \frac{1}{z^2}\frac{z^{2n+1}}{(2n+1)!}$ but I how would I would factor out $\frac{1}{z}$? 
 [/INST] So far you have
$$\frac{\sinh z}{z^2} = \sum_{n = 0}^\infty \frac{1}{z^2} \frac{z^{2n+1}}{(2n+1)!} = \sum_{n = 0}^\infty \frac{z^{2n-1}}{(2n+1)!} = \frac{1}{z} + \sum_{n = 1}^\infty \frac{z^{2n-1}}{(2n+1)!}.$$
If we introduce a new index $k = n - 1$, we have
$$\frac{\sinh z}{z^2} = \frac{1}{z} + \sum_{k = 0}^\infty \frac{z^{2k+1}}{(2k+3)!}$$
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