Is ln(n) Less Than n^c for All c>0 and n>N?

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Homework Statement



How to rigorously (real analysis) prove that for all real c>0
Exists N such that for all n>N
ln(n)<n^c

Homework Equations





The Attempt at a Solution


The fact can be shown using graphical calculator
 
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What about taking derivatives and comparing them?
 
grossgermany said:

Homework Statement



How to rigorously (real analysis) prove that for all real c>0
Exists N such that for all n>N
ln(n)<n^c

Homework Equations





The Attempt at a Solution


The fact can be shown using graphical calculator

I think that you can do it like this:

If you view these two function as series

e.g. [tex]\sum_{n=1}^{\infty} ln(n)[/tex] and [tex]\sum_{c=1}^{\infty} n^c[/tex] and then use the comparison test from Calculus to show that

[tex]ln(n) < n^c[/tex]
 
For the comparison test, we need to show that there exists N such that for all n>N
ln(n)<Mn^c for some constant M