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How can one show that $n \log n \subseteq O(n^{1+\epsilon})$ where $0 < \epsilon < 1$ without using limits? This question arise from a homework where I used limits to prove the relation. I'd like to know if there is another way.

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A bit nitpicky: I suggest either saying $\epsilon$ is fixed, or writing the above as $n\log n=O_\epsilon \left(n^{1+\epsilon}\right)$ to indicate the constant depends on $\epsilon$. (Otherwise it is simply not true) – Eric Naslund Sep 17 '11 at 16:54
up vote 4 down vote accepted

Let $0<\epsilon<1$ be fixed. We need only show that $\log n=O(n^\epsilon)$. Suppose $n\geq 1$ so that $\log n\geq 0$. Since $$n^\epsilon=e^{\epsilon \log n}=1+\epsilon \log n+\frac{\epsilon^2 \log^2 n}{2}\cdots \geq \epsilon \log n$$ we see that $$\log n \leq \frac{1}{\epsilon}n^\epsilon $$ which proves the desired result.

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Why couldn't you stop the expansion at $\epsilon \log n$? – Li-thi Sep 17 '11 at 17:27
@Pho: Huh. You could! Great idea. I edited accordingly. – Eric Naslund Sep 17 '11 at 17:28
Instead of expanding into a series you could use the inequality $e^x \geq 1+x$. – t.b. Sep 17 '11 at 17:36
Thanks for the answer! – Li-thi Sep 17 '11 at 17:51

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