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How do I prove $f(n)=n \log{\log{n}} \notin \Theta (n^k)$ for any $k$? I have no idea where to start but I tried plotting the graph in Google and noticed that $\log{\log{n}}$ is very close to 0.

But might it be because it doesn't have a lower bound? Cos as $n \rightarrow 0$, $\log{\log{n}} \rightarrow - \infty$

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General advice: You care about $n \to \infty$, not $n \to 0$. –  ShreevatsaR Feb 4 '13 at 12:49
    
Do you know the definition of $\Theta(n^k)$ (or $\Theta(g(n))$ in general)? –  ShreevatsaR Feb 5 '13 at 7:24
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1 Answer 1

up vote 2 down vote accepted

The involved sequence are positive, so we have to see whether there exists $a,b>0$ such that for all $n\geqslant 2$, $$a\leqslant \frac{n\log\log n}{n^k}\leqslant b.$$

  • No $k<1$ can work because $n^{1-k}$ diverges to $\infty$.
  • $k=1$ neither as $\log\log n\to +\infty$.
  • $k>1$ is not good as $\frac{f(n)}{n^a}=\frac{\log\log n}{n^{a-1}}$ is not below bounded.
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Erm ... sorry I dont really get what you wanted to say. It will be nice if you could rephrase/elaborate? Sorry I am bad at math ... Or did you mean: diverge to $\infty$ (+/-) means no upper/lower bound? Then by not below bounded means? –  Jiew Meng Feb 5 '13 at 2:40
    
@JiewMeng I've added details. Don't hesitate to ask me more clarifications. –  Davide Giraudo Feb 5 '13 at 9:37
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