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$\lim_{n\to\infty} \frac{n!}{n^{log(log(n))}}$

I am struggling calculating this because I want to know $n!$ and $n^{log(log(n))}$, which function grows faster eventually.

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  • $\begingroup$ by L'HOspital? $\frac{d}{dx} \frac{n!}{n^{log(log(n))}}$ $\endgroup$ – Darío A. Gutiérrez Sep 22 '16 at 6:33
  • $\begingroup$ @DarioGutierrez It doesn't really make any differences when I calculate the derivatives of two $\endgroup$ – good2know Sep 22 '16 at 6:42
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Here is my train of thought when I see this limit....

  • Right off the bat, I expect probably the limit is infinity, because $n!$ grows famously fast, whereas $\log(n)$ grows very slow, and $\log(\log(n))$ even more slowly, so probably $n^{\log(\log(n))}$ should grow slower than $n!$.
  • Since I want to use that the top is growing fast, I should try to get a simple underestimate of it that still grows quite fast. Note that of the $n$ factors in $n! = 1 \cdot 2 \cdots n$, half of them are $\geq n/2$. So we have $$n! \geq (n/2)^{n/2} = \frac{n^{n/2}}{2^{n/2}}.$$
  • Now let's see where this estimate gets us $$ \frac{n!}{n^{\log(\log(n))}} \geq \frac{n^{n/2}}{2^{n/2} n^{\log(\log(n))}}. $$
  • At this point it seems pretty clear that the top grows much faster than the bottom, but we should still check precisely that our estimate goes to infinity. One strategy is to take the logarithm of the sequence and prove that goes to infinity. Then, the original sequence goes to infinite too, by exponentiation. We get $$\log\left( \frac{n^{n/2}}{2^{n/2} n^{\log(\log(n))}}\right) = \frac{n}{2} \log(n) - \frac{n}{2} \log(2) - \log(\log(n)) \log(n)$$

Can you continue from here?

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Let's look at the log.

$\begin{array}\\ \ln\frac{n!}{n^{\log(\ln(n))}} &=\ln (n!)-\ln(n)\ln(\ln(n))\\ &>n(\ln(n)-1)-\ln(n)\ln(\ln(n))\\ &=\ln(n)(n-n/\ln(n)-\ln(\ln(n)))\\ &\to \infty \qquad\text{since }n-n/\ln(n)-\ln(\ln(n)) \to \infty\\ \end{array} $

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