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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.

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    $\begingroup$ Hint: $(\log n)^2 = (3\log n^{1/3})^2 \leqslant (3 n^{1/3})^2 = 9 n^{2/3}$ $\endgroup$ – RRL Dec 11 '17 at 6:10
  • $\begingroup$ @RRL Wow, it seems abundantly clear to me now. Feel free to write up an answer with that text and I'd be glad to accept it. $\endgroup$ – Sriotchilism O'Zaic Dec 11 '17 at 6:12
  • $\begingroup$ Just a note: because L'hospital's rule was needed to be proven without itself it means you can always calculate limit without L'hospital's rule, although from time to time it will be very hard $\endgroup$ – Holo Dec 11 '17 at 6:16
  • $\begingroup$ Not necessary. $\log n$ to any power is $o(n)$ in general. $\endgroup$ – RRL Dec 11 '17 at 6:16
  • $\begingroup$ @Holo I suppose I should have asked if there was a reasonable way to do it with without L'hôpital's rule. $\endgroup$ – Sriotchilism O'Zaic Dec 11 '17 at 6:21
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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.

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