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

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    $\begingroup$ $\log \frac{a}{b} = \log a - \log b$ $\endgroup$ – Daniel Fischer May 18 '15 at 10:31
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HINTS: $$\lim\exp(\ldots)=\exp(\lim\ldots)$$ $$\dfrac{2^n}{n^{\ln{n}}}=\exp\left(n\ln{2}-\ln^2{n}\right)$$

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  • $\begingroup$ @ Demosthene: Why is it true that $\lim\exp(\ldots)=\exp(\lim\ldots)$? $\endgroup$ – Caleb Owusu-Yianoma May 18 '15 at 10:44
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    $\begingroup$ @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. $\endgroup$ – Demosthene May 18 '15 at 10:53
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HINT: Taking the logarithm you have $$\lim_n \log(2^n)-\log(n^{\log n}) = \lim_n (\log 2)n-\log^{2}n = \infty$$

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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

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

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  • $\begingroup$ Should the second equality read $\sqrt[n]{a_n}=\frac{n^{\log n^{\frac{1}{n}}}}2$? $\endgroup$ – Caleb Owusu-Yianoma May 18 '15 at 10:50
  • $\begingroup$ @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 $\;\log x^a=a\log x\;$ $\endgroup$ – Timbuc May 18 '15 at 10:57
  • $\begingroup$ Ah - I think that I misread your notation. By $n^{\log n/n}$, do you mean the same as $n^{(\log n)/n}$? $\endgroup$ – Caleb Owusu-Yianoma May 18 '15 at 11:03
  • $\begingroup$ Yes, I did mean $n^{\log(n^{1/n})}$. $\endgroup$ – Caleb Owusu-Yianoma May 18 '15 at 11:04
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    $\begingroup$ The $\;n$-th root test or root test...google it. $\endgroup$ – Timbuc May 21 '15 at 17:30
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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?

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  • $\begingroup$ I don't yet see how you arrived at the first sentence of your answer. $\endgroup$ – Caleb Owusu-Yianoma May 18 '15 at 11:00
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    $\begingroup$ 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. $\endgroup$ – Klest Dedja May 26 '15 at 23:13

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