Re: Proof - the right track?
yes, but it still needs a bit of work to be convincing.
the easiest way is to use the "ratio test": consider lim_{n -> \infty}
(\log n)^k / n. Differentiate the top and bottom (l'Hopital's rule) k
times then the limit equals 0 and so (\log n)^k = O(n).
-Andy
.
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