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

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SUMMARY

The limit of ln(n)/n^c approaches 0 for any c > 0, as established in the discussion. Participants suggest using Epsilon-Delta proofs and Big O/Small O notation to demonstrate this limit rigorously. The Bolzano-Weierstrass Theorem is highlighted as a useful tool for analyzing convergence in this context. The discussion is framed within a real analysis course, indicating a foundational understanding of these concepts is necessary.

PREREQUISITES
  • Understanding of Epsilon-Delta proofs
  • Familiarity with Big O and Small O notation
  • Knowledge of the Bolzano-Weierstrass Theorem
  • Basic concepts of real analysis
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  • Study Epsilon-Delta proofs in detail
  • Explore Big O and Small O notation applications
  • Research the Bolzano-Weierstrass Theorem and its implications
  • Review convergence of sequences in real analysis
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Students in real analysis courses, educators teaching mathematical rigor, and anyone interested in the convergence of logarithmic functions relative to polynomial growth.

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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.
 

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