Limit of the nth root of (n ln n)

[txy]

Homework Statement

Find lim_{n \rightarrow \infty} \sqrt[n]{n ln(n)}.
Do not use L'Hospital's Rule or Taylor Series.

Homework Equations

The Attempt at a Solution

I suspect I need to set up some inequality for this and then apply Squeeze Theorem. But I can't find any inequality that can help me do this. Any help would be appreciated.

 

[Dick]

Start by taking the log of the expression and trying find the limit of that.

 

[dirk_mec1]

You should find something like:

n \ln(L) = ln( n \cdot ln (n) ) = ln(n) + \ln(ln(n))

 

[lurflurf]

does
n<n log n<n^(1+epsilon)
help?

 

[txy]

That's great. In either case, I need to use the fact that ln n < n for all n > 1. How do I prove this?

 

[Dick]

That's great. In either case, I need to use the fact that ln n < n for all n > 1. How do I prove this?

If you need to prove it I think the easiest way is to define f(x)=x-ln(x). Notice f(1)=1. Now can you show f'(x)>0 for x>1? That would show f(x) is positive for x>1.

 

[txy]

If you need to prove it I think the easiest way is to define f(x)=x-ln(x). Notice f(1)=1. Now can you show f'(x)>0 for x>1? That would show f(x) is positive for x>1.

But this question that I posed is in the Limits chapter of my textbook, which comes before the chapter on Differentiation. I wonder if there's another way to prove that inequality. Perhaps I could change it to the form x < e^x , but I still can't make any progress.

 

[Dick]

Ok, I think you might be fussing a little to much over this proof, but how about using the definition of e^x. e^x=limit (1+x/n)^n as n->infinity? Can you use that, together with the binomial theorem to prove e^x>x for x>0?

 

[txy]

Ok, I think you might be fussing a little to much over this proof, but how about using the definition of e^x. e^x=limit (1+x/n)^n as n->infinity? Can you use that, together with the binomial theorem to prove e^x>x for x>0?

Yes now I can prove it. Thanks!