What is the radius of convergence of this series?

Question

Suppose we have this series:

$$f(z) = \frac{1}{2z^3} + \frac{1}{12z} - \frac{z}{240}.$$

What is the radius of convergence?

Answer 1

With the additional information in the comment, the series is supposed to be the Laurent series for $$ f(z) = \frac{1}{z^2\sinh z}, $$ In fact, there is some error in your computation. The first three (non-zero) terms should be $$ \frac{1}{z^2\sinh z} = \frac{1}{z^3} - \frac{1}{6z} + \frac{7z}{360} + \cdots $$

As for convergence, the general theory shows that the Laurent series of $f$ converges on the largest annulus $0 < |z| < r$ on which $f$ is holomorphic. The only singularities of $f$ occur when $z=0$ and when $\sinh z = 0$, i.e. when $z = i\pi k$ for $k \in \mathbb{Z}$. The closest singularity to $0$ (other than $0$ itself) is $\pm i\pi$. In other words, $r = \pi$.