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- Thread starter Lisa91
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- #2

- Jan 17, 2013

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I will separate the problem in two steps :

1-For 0<n<1 this is trivial since the right hand will be always negtative while

and the left hand side is always positive , also for n=1 the inequality holds .

2-Now for n>1 :

${\alpha}\ln(n)> \ln(\ln(n)) \Rightarrow \,\, \alpha> \frac{\ln(\ln(n))}{\ln(n) } $

Now this is only true iff $\frac{\ln(\ln(n))}{\ln(n) } \leq 0$

which holds iff $0<\ln(n)\leq 1\,\, \Rightarrow \,\, 1< n \leq e $

The inequality is true for all $\alpha $ iff $0<n \leq e$

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- Mar 5, 2012

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Since the right hand side approaches zero for large n, this means that for any $\alpha>0$ there is a number N such that the inequality is true for any n > N.2-Now for n>1 :

${\alpha}\ln(n)> \ln(\ln(n)) \Rightarrow \,\, \alpha> \frac{\ln(\ln(n))}{\ln(n) } $

Hey Lisa91!

Can it be there is a condition missing from your problem?

The extra condition that it holds for any n > N for some N?

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- Jan 17, 2013

- 1,667

since $\alpha $ is an independent variable of n I can choose it as small as possible so thatSince the right hand side approaches zero for large n, this means that for any $\alpha>0$ there is a number N such that the inequality is true for any n > N.

it becomes lesser than the right-hand side .

Can you give a counter example for $\alpha$ and n that disproves my argument ?

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- #6

- Mar 5, 2012

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Your argument is flawless.since $\alpha $ is an independent variable of n I can choose it as small as possible so that

it becomes lesser than the right-hand side .

Can you give a counter example for $\alpha$ and n that disproves my argument ?

It's just that you have assumed that the inequality should hold for specific n and all $\alpha$'s.

Whereas I have assumed it's not for all n.

In other words, you have solved:

Find n such that $n^\alpha > \ln n$ for all $\alpha > 0$.

Whereas I have use the first half of your argument to follow up with:

Prove that $n^\alpha > \ln n$ for $\alpha > 0$ if n is big enough given a certain $\alpha$.

That's why we need clarification on what the actual problem is.

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