I was attempting to show that the power series
\begin{equation*} \sum_{n=1}^\infty n^{\log(n)}z^n \end{equation*}
has a radius of convergence of $1$.
In order to do this I decided to use the $\alpha$ method. This meant evaluating the limit
\begin{equation*} \limsup_{n\rightarrow\infty} n^{\frac{\log(n)}{n}} \end{equation*}
I was able to prove that
\begin{equation*} \lim_{n\rightarrow\infty} \dfrac{\log(n)}{n} = 0 \end{equation*}
but I realized that was necessary but not sufficient to show that the limsup in question is $1$.
I then was able to prove that
\begin{equation*} \lim_{n\rightarrow\infty} n^{\frac{1}{n}} = 1 \end{equation*}
However I could not figure out any way of using that fact either.
I realized I could rewrite this limit as
\begin{equation*} \exp\left(\lim_{n\rightarrow\infty} \dfrac{\log(n)^2}{n}\right) \end{equation*}
However since I have not proven L'hôpital's rule, I have no way of evaluating this limit either.
This has left me pretty stuck. I'm not sure how I could tackle this problem from here. Where should I start for this limit? Is there a way to do it without L'hôpital's rule?
I'd really rather not know the whole proof if possible.
Since $n\gg1\implies\log n<\sqrt n$, you have$$n\gg1\implies n^\frac{\log n}n<n^\frac1{\sqrt n}=\left(\sqrt n^\frac1{\sqrt n}\right)^2$$and therefore$$\lim_{n\to\infty}n^\frac{\log n}n\leqslant 1.$$And you already know that the reverse inquality holds.