help proof lim n to infinity (n^2) ^ (1/n)

**kanli** · October 14th 2012, 07:20 PM

I know the limit goes to 1. I can do it by l'hopital rule. I am wondering if there are other ways to prove this.

I am thinking about geometric mean arithmetic mean inequality. but that only gives me a upper bound which approaches 1. I think I need a lower bound but I don't know what is a good lower bound.

Please advice. thanks

**Plato** · October 19th 2012, 10:55 AM

Originally Posted by kanli

I know the limit goes to 1. I can do it by l'hopital rule. I am wondering if there are other ways to prove this.

Suppose that $n\ge 2$ . Define $a_n = \sqrt[n]{n} - 1$ so it is clear that $a_n>0$ .

Thus $n=(1+a_n)^n\ge\frac{n(n-1)}{2}a_n^2$ so that $0<a_n<\sqrt{\frac{2}{n-1}}$ .

Thus $(a_n)\to 0$ giving $\sqrt[n]{n}\to 1.$

Thread: help proof lim n to infinity (n^2) ^ (1/n)

LinkBack

Thread Tools

Display

help proof lim n to infinity (n^2) ^ (1/n)

Re: help proof lim n to infinity (n^2) ^ (1/n)

Similar Math Help Forum Discussions

help proof limit n to infinity (n+1)^1/3 - n (1/3)

Proof of that a limit -> 0 as n-> infinity

Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

Epsilon-proof as x->-infinity currently defeating me

limit at infinity proof

Search Tags