How can we choose $z$ to make this series absolute convergent? I have the following series:
$\sum_{n=0}^\infty \frac{n^{\ln (n)}}{3^n} (z+2)^n$
The question is, how can we choose $z \in \mathbb{C}$ to make this absolute convergent?
For the first sight it seemed impossible to have such $z$, since $n^{ln n}$ should be way way bigger than $3^n$ as $n \rightarrow \infty$. Am I wrong? Any hints, ideas?
Thanks!

Take the $n$th root of the $n$th term to get $n^{(\ln n)/n}(|z+2|/3).$ Show $n^{(\ln n)/n} \to 1.$ Thus the root test gives convergence if $|z+2|/3 < 1,$ divergence if $|z+2|/3 > 1.$ You can also check that the series diverges when $|z+2|/3 = 1.$
