Does limit ln(n)/n^c -> 0 for any c>0?

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Homework Statement



Does limit ln(n)/n^c -> 0 for any c>0?

Homework Equations





The Attempt at a Solution


I wonder if there is an
1.Epsilon Delta Proof
2.Proof using BigO SmallO notation.

Thanks
 
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It is obviously true. There are a lot of ways to prove this. What kind of class is this for? Are you expected to be rigorous in your solutions? N-epsilon proofs and asymptotic behaviour are very different arguments in terms of the development required in an analysis setting.
 
Yes, this is for real analysis class
 
What tools do you have available? How was ln(n) derived?

There is a nifty theorem that says that if a sequence converges then any subsequence will converge to the same limit. This will enable you to look at ln(x)/x^c in R rather than in N.
 
Yes, we can use Bolzano Weierstrasse Theorem. Please feel free to proceed.
My level is on baby Rudin, the this is the first course in real analysis.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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