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!
 A: The reciprocal of the radius of convergence, as given by the Cauchy-Hadamard Theorem is
$$
\begin{align}
\limsup_{n\to\infty}\left(\frac{n^{\log(n)}}{3^n}\right)^{1/n}
&=\frac13\limsup_{n\to\infty}n^{\log(n)/n}\\
&=\frac13\limsup_{n\to\infty}e^{\log(n)^2/n}\\
&=\frac13e^{\limsup\limits_{n\to\infty}\log(n)^2/n}\\[3pt]
&=\frac13
\end{align}
$$
Therefore, for $\left|z+2\right|\lt3$, the series is absolutely convergent.
For $\left|z+2\right|=3$, the absolute value of the terms of the series is $n^{\log(n)}$, which does not tend to $0$.
A: 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 $|z+2|/3 = 1.$
A: To estimate $n^{\ln n}$, you can use
$$n^{\ln n} = e^{\ln(n^{\ln n})} = e^{\ln n \ln n} = e^{\ln^2 n}$$
and
$$3^n = e^{\ln(3^n)} = e^{n \ln 3}$$
Which should tell you which of these grows quicker (does $n$ grow quicker than $\ln^2 n$?)
