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Prove that $$n^\frac{1+2}{\sqrt{\log n}} = O(n\ \log n).$$

I want to compute the two growth rates by using L'Hôpital's rule:

$$\lim_{n\to \infty} \frac{f(n)}{g(n)}$$

so I get something like this:

$$\lim_{n\to \infty}\frac{n^\frac{1+2}{\sqrt{\log n}}}{n\ \log n}$$

However the main trouble I'm having is differentiating $$ n^\frac{1+2}{\sqrt{\log n}}$$

What would be the best way to approach $f(n)$?

Thanks in advance, guys.

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do you mean n^{1+2/sqrt{logn}} ? – Lee Dae Seok Jan 24 '13 at 4:09
ya that's the correct equation – FireStorm Jan 24 '13 at 4:11
Is it {1+2}/sqrt{n}? (=3/sqrt{n})? – Lee Dae Seok Jan 24 '13 at 4:22
its supposed to be 3/sqrt{log n} unless I'm doing something wrong? – FireStorm Jan 24 '13 at 4:34
@FireStorm: Then why was it written as $1+2$? Note that when $n$ is big enough, $3/\sqrt{\log n}<1$, which already gives $O(n)$. – Jonas Meyer Jan 24 '13 at 4:36
up vote 2 down vote accepted

It is same as to show $n^{\frac{2}{\sqrt{\log n}}}=O(\log n)$ I think it's wrong. Let $n=e^{x^2}$, then it becomes $e^{2x}=O(x^2)$, But it's false..

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The OP asked if $n^{3/\sqrt{\log n}} = O(n \log n)$ and this is in fact true. We even have $n^{3/\sqrt{\log n}} = o(n \log n)$. – Fabian Jan 24 '13 at 8:01
Since when $ O(n\ \log n)$ is the same as $ O(\ \log n)$? How did this get 3 upvotes? – kuhaku Jan 23 at 14:47

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