# Convergence n^(1/n)

1. Jun 1, 2010

### nonequilibrium

Hello, I was wondering how to proof $$a_n = n^{1/n} \to 1.$$

Doing it straight from the definition got me nowhere. But I was thinking. It is obvious that $$\liminf a_n \geq 1$$ (since otherwise for big n you could get n^(1/n) < 1 <=> n < 1). And I also have already a proof of $$\limsup x_n^{1/n} \leq \limsup \frac{x_{n+1}}{x_n}$$ (which is a general result for any row x_n).

But proving $$\limsup \frac{a_{n+1}}{a_n} \leq 1$$ seemed to be harder than I thought.

So I'm completely stuck. Any ideas?

Thank you,
mr. vodka

2. Jun 1, 2010

### mrbohn1

I don't think you need to use lim sup/inf here.

n1/n=eln(n1/n)=eln(n)/n.

So you just need to show that ln(n)/n-->0. Try using l'Hopital's rule.

3. Jun 1, 2010

### nonequilibrium

Hm, I'd like to proof it without the use of the exponential function. It's namely introduced in my Analysis course before the exp function, and it's actually used in a proof about power series, which is later used to introduce the e-function. Thanks for your help!

4. Jun 1, 2010

### mathman

If you don't want to use exp function, how about using ln (natural log)?
ln(an) = ln(n)/n -> 0.

5. Jun 1, 2010

### nonequilibrium

Well, we defined that as the inverse of e. Any possibilities without e (or ln)? I appreciate the help though! I find the ln/e proofs very elegant, but I hope you understand I'm going to choose for logical consistency in my course :)

6. Jun 1, 2010

### Staff: Mentor

I'm not sure that there's a way around that doesn't use exp or ln.

7. Jun 1, 2010

### nonequilibrium

Oh... But it was left as an exercise for us in our course, so there must be. Hmmm, maybe I should contact one of the math assistants for this one then. Thank you guys for your time.

8. Jun 2, 2010

### nonequilibrium

For those that are interested: http://myyn.org/m/article/limit-of-nth-root-of-n/ [Broken]

Last edited by a moderator: May 4, 2017