# Show that $\lim_{n\to\infty}\frac{2^n}{n^{\ln(n)}}=\infty$

Could anyone please give a hint for showing the following? $$\lim_{n\to\infty}\frac{2^n}{n^{\ln(n)}}=\infty$$

• $\log \frac{a}{b} = \log a - \log b$ – Daniel Fischer May 18 '15 at 10:31

HINTS: $$\lim\exp(\ldots)=\exp(\lim\ldots)$$ $$\dfrac{2^n}{n^{\ln{n}}}=\exp\left(n\ln{2}-\ln^2{n}\right)$$

• @ Demosthene: Why is it true that $\lim\exp(\ldots)=\exp(\lim\ldots)$? – Caleb Owusu-Yianoma May 18 '15 at 10:44
• @CKKOY Because $\exp(x)$ is continuous. This is basically a way of saying that we first evaluate the limit of the argument, then exponentiate it. – Demosthene May 18 '15 at 10:53

HINT: Taking the logarithm you have $$\lim_n \log(2^n)-\log(n^{\log n}) = \lim_n (\log 2)n-\log^{2}n = \infty$$

A fancy way: put

$$a_n=\frac{n^{\log n}}{2^n}\implies \sqrt[n]{a_n}=\frac{n^{\log n/n}}2\xrightarrow[n\to\infty]{}\frac12<1$$

Thus, the series $\;\sum\limits_{n=1}^\infty a_n\;$ converges, so

$$\lim_{n\to\infty} a_n=0\implies \lim_{n\to\infty}\frac1{a_n}=\infty$$

• Should the second equality read $\sqrt[n]{a_n}=\frac{n^{\log n^{\frac{1}{n}}}}2$? – Caleb Owusu-Yianoma May 18 '15 at 10:50
• @CKKOY No: $$\sqrt[n] a=a^{1/n}\implies \sqrt[n]{n^{\log n}}=\left(n^{\log n}\right)^{1/n}=n^{\frac1n\log n}=n^{\log n/n}$$ Of course, if you meant $\;n^{\log \left(n^{1/n}\right)}\;$ then yes, as $\;\log x^a=a\log x\;$ – Timbuc May 18 '15 at 10:57
• Ah - I think that I misread your notation. By $n^{\log n/n}$, do you mean the same as $n^{(\log n)/n}$? – Caleb Owusu-Yianoma May 18 '15 at 11:03
• Yes, I did mean $n^{\log(n^{1/n})}$. – Caleb Owusu-Yianoma May 18 '15 at 11:04
• The $\;n$-th root test or root test...google it. – Timbuc May 21 '15 at 17:30

First of all, I would observe that showing that $lim_{n \to \infty} \frac{(n+1)^{ln(n+1)}}{n^{ln(n)}} < 2$ is an equivalent proof.

And to do that, rather than using logarithm's algebraic properties, I would use the fact that it increases very slowly, for example:

$lim_{n \to \infty} \frac{(n+1)^{ln(n+1)}}{n^{ln(n)}} < lim_{n \to \infty} \frac{(n+1)^{ln(n+1)}}{n^{ln(n+1)}} = lim_{n \to \infty}\left(\frac{n+1}{n}\right)^{ln(n+1)}$

And now, for $n > 3$ we have: $ln(n+1) < n/2$.

We're almost done now, can you see why?

• I don't yet see how you arrived at the first sentence of your answer. – Caleb Owusu-Yianoma May 18 '15 at 11:00
• Be $a$ the growing factor for $n \to \infty$ to the formula in my first sentence. If we show that $a < 2$ then the growing factor of our original limit is $2/a > 1$ and therefore diverges like geometric series do. – Klest Dedja May 26 '15 at 23:13