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

  • Thread starter Thread starter grossgermany
  • Start date Start date
grossgermany
Messages
53
Reaction score
0

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
 
Physics news on Phys.org
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. \sum_{n=1}^{\infty} ln(n) and \sum_{c=1}^{\infty} n^c and then use the comparison test from Calculus to show that

ln(n) &lt; n^c
 
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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top