2
$\begingroup$

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.

$\endgroup$
7
  • 4
    $\begingroup$ Hint: $(\log n)^2 = (3\log n^{1/3})^2 \leqslant (3 n^{1/3})^2 = 9 n^{2/3}$ $\endgroup$
    – RRL
    Dec 11, 2017 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. $\endgroup$ Dec 11, 2017 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, 2017 at 6:16
  • $\begingroup$ Not necessary. $\log n$ to any power is $o(n)$ in general. $\endgroup$
    – RRL
    Dec 11, 2017 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. $\endgroup$ Dec 11, 2017 at 6:21

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.