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

Question

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

Answer 1

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

Answer 2

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

Answer 3

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

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

Answer 4

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?

Answer 5

Another way to prove that

$$\lim_{n\to\infty}\frac{2^n}{n^{\ln n}}=\infty\;.$$

For any $M\!>\!0$ there exists $k_M\!\!=\!\max\!\left\{\left(1\!+\!\lfloor\ln\!M\rfloor\right)^{\,2}\!,34^2\right\}\!\!\in\!\Bbb N\,$ such that for any $\;n\in\Bbb N\;\land\;n>k_M\;$ it results that

$\dfrac{2^n}{n^{\ln n}}=\dfrac{e^{n\ln2}}{e^{\ln^2\!n}}=e^{n\ln2-\ln^2\!n}>e^{\frac12n-16\ln^2\left(\sqrt[4]n\right)}\!\!\!\!\!\!\!\!\!\!\underset{\overbrace{\text{ because }\,0\leqslant\ln x<x}\\\text{ for all }x\in[1,\infty[\,.}{>}\!\!\!\!\!\!\!\!\!e^{\frac12n-16\sqrt n}=$

$=e^{\frac12\sqrt n\left(\sqrt n-32\right)}>e^{\sqrt n}>e^{\left|1+\lfloor\ln M\rfloor\right|}\geqslant e^{1+\lfloor\ln M\rfloor}>e^{\ln M}=M\,.$

By definition of limit, it follows that $\,\lim\limits_{n\to\infty}\dfrac{2^n}{n^{\ln n}}=\infty\,.$

Answer 6

Another way, not so much effective in this case, by ratio test

$$\frac{\frac{2^{n+1}}{(n+1)^{\ln(n+1)}}}{\frac{2^n}{n^{\ln(n)}}}=2\frac{n^{\ln(n)}}{(n+1)^{\ln(n+1)}}=2\frac{n^{\ln(n)}}{(n+1)^{\ln(n)}}\frac{(n+1)^{\ln(n)}}{(n+1)^{\ln(n+1)}}=2\frac{(n+1)^{\log\left(1-\frac1{n+1}\right)}}{\left(1+\frac1n\right)^{\log n}}\to 2$$

indeed $\left(1+\frac1n\right)^{\log n}=e^{\log n\log\left(1+\frac1n\right)}\le e^{\frac{\log n}n}\to e^0=1$.