My question is, generally, is there any relationship between convergence and analyticity of a complex-valued function (namely, does one property imply the other etc?)
For example, a function defined by the power-series $f(z) = \sum_{n=0}^{\infty}n!z^{n}$ diverges for all $z \neq 0$. So, does the derivative of $f$ at the origin, defined as follows, \begin{eqnarray} f'(z)|_{z=0} = \lim_{z \rightarrow 0} \frac{f(z) - 1}{z} \end{eqnarray} not exist? I.e. $f(z)$ is not analytic at the origin?
Considering this example, does divergence imply non-analyticity? (Equivalently, in other words, analyticity implies convergence?) I've used an example above where a function is defined by power-series but I'd like to know the general answer, as I also deal with functions defined by integrals.
Note on definition: I use the definition that a function is analytic at a point if its derivative exists at that point and in a neighbourhood of that point (equivalent term used is "holomorphic").