Show that $\lim_{n\to\infty}\frac{2^n}{n^{\ln(n)}}=\infty$ Could anyone please give a hint for showing the following? $$\lim_{n\to\infty}\frac{2^n}{n^{\ln(n)}}=\infty$$
 A: HINTS: 
$$\lim\exp(\ldots)=\exp(\lim\ldots)$$
$$\dfrac{2^n}{n^{\ln{n}}}=\exp\left(n\ln{2}-\ln^2{n}\right)$$
A: 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: 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
$$\lim_{n\to\infty} a_n=0\implies \lim_{n\to\infty}\frac1{a_n}=\infty$$ 
A: 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?
