Thursday, 3 May 2018

calculus - Proving nepsilon>log(n) for sufficiently large n



I'd like to show that for all ϵ>0, there exists an N such that for all nN the following holds: nϵ>log(n). I'm having trouble justifying this.



My intuition says this should hold. Writing n=2k gives an inequality of the form (2ϵ)k>k. This should hold for sufficiently large k (and therefore n) because 2ϵ>1, so the function (2ϵ)k grows much faster than k. (This is also the approach described in this question.)




However, I was wondering if there is a more "rigorous" way to justify this inequality.


Answer



Provided you have proved L'Hôpital's rule (as you should have in a real analysis course), this is pretty straightforward.



Fix ϵ>0, we have
lim
Proving that for sufficiently large n, $\log n

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