What's the behavior of $\displaystyle\sum_{n=1}^\infty (z+\sqrt{5}+2i)^{n!}$ outside its radius of convergence?

Question

I want to check the behavior of $$\displaystyle\sum_{n=1}^\infty (z+\sqrt{5}+2i)^{n!}$$ outside its radius of convergence.

I've tried to use the ratio test as follows: $$\left|\frac{(z-\sqrt{5}+2i)^{(n+1)!}}{(z-\sqrt{5}+2i)^{n!}}\right|=|(z-\sqrt{5}+2i)^{nn!}|$$ This will converge to zero, if $|z-\sqrt{5}+2i|<1$.

However, what's about the cases $|z-\sqrt{5}+2i|=1$. How can I show convergence or divergence in this case?

EDIT: I've updated my question to make things more clear. Why I was down-voted for that?

Answer 1

Evidently, this is a power series around $-\sqrt 5 -2i$, so you really should make a change of variable and look at the power series $$\sum z^{n!}.$$

The ratio test for series will indeed give you the correct radius of convergence 1, as you have written in your question.

However, the ratio test never helps you for boundary behaviour.

On the boundary, you usually use one of the following approaches:

1. terms do not even converge to zero, therefore divergent

2. terms are positive for $z=R$ and converge (because of known series, integral test, etc), therefore absolute convergence on boundary

3. terms converge for $z\not=R$ but on boundary due to Leibniz test/Abel criterion/Raabe test etc.

Answer 2

If we wish to known the radius of convergence of $$\sum_{n=0}^\infty z^{n!}=z+z+z^2+z^6+z^{24}+\cdots, $$ then we calculate, $$\frac{1}{R} = \limsup \sqrt[n]{|\text{$n^{th}$ coefficient}|} = \limsup \sqrt[n]{\{0,2,1,0,0,0,1,0,0,0,\ldots\}}=1. $$ Rudin Principles of Mathematical Analysis - page 69 - Theorem 3.39