Laurent series of $f(z)=\frac{1}{z^{2}\sinh(z)}$

Question

I would really appreciate if you could help me understand this. So, I’m at this point

$\sum_{m=0}^{\infty}\frac{z^{2m+3}}{(2m+1)!}\sum_{k=-2}^{\infty}c_{2k+1}z^{2k+1}=1$

But I don’t really know how to continue. How do I multiply these two power series? How do I get the coefficients of the Laurent series?

Thanks in advance.

Answer 1

If $z\ne0$, let$$g(z)=\frac{\sinh(z)}z=1+\frac1{3!}z^2+\frac1{5!}z^4+\cdots$$and note that $g$ is an even function. Let$$h(z)=\frac1{g(z)}=\frac z{\sinh(z)}.$$Then $h$ is an even function too. Let $1+a_2z^2+a_4z^4+\cdots$ be its Taylor series centered at $0$. Then\begin{align}1&=g(z)h(z)\\&=\left(1+\frac1{3!}z^2+\frac1{5!}z^4+\cdots\right)\left(1+a_2z^2+a_4z^4+\cdots\right)\\&=1+\left(a_2+\frac16\right)z^2+\cdots\end{align}and therefore $a_2=-\frac16$. Therefore$$\frac1{z^2\sinh(z)}=\frac{h(z)}{z^3}=\frac1{z^3}-\frac1{6z}+\cdots$$

Answer 2

Hint:

Use the first three terms of the expansion in power series of $\sinh z$: $$\sinh z=z+\frac{z^3}{3!}+\frac{z^5}{5!}+o(z^5),$$ use it in the function: $$f(z)=\frac1{z^2\sinh z}=\frac1{z^3}\frac1{1+\cfrac{z^2}{6}+\cfrac{z^4}{120}+o(z^4)}$$ and perform the division of $1$ by $1+\dfrac{z^2}{6}+\dfrac{z^4}{120}$ by increasing powers of $z$ up to degree $4$.