# Proving limit of the nth root of n

1. Sep 4, 2010

### Elzair

1. The problem statement, all variables and given/known data

Prove the following limit:

$$lim_{n \rightarrow \infty } n^{ 1 / n } = 1$$

2. Relevant equations

Not sure.

3. The attempt at a solution

Given any $$\epsilon > 0$$, choose $$N \in \mdseries N$$ s.t.

$$\left| n^{ 1 / n } - 1 \right| < \epsilon$$ for all $$n > N$$

I am not sure how to proceed.

2. Sep 4, 2010

### jgens

If you're still working on this problem and need to do it with epsilons and deltas, I think that choosing N = exp(log(1+ε)-1) should suffice. I can't find a nice/elegant epsilon delta solution to this problem, but maybe someone else can.

Last edited by a moderator: Sep 5, 2010
3. Jul 11, 2012

### klondike

Prove $$\lim _{n\to \infty} \sqrt[n]{n}=1$$.
Proof: We want:
$$|\sqrt[n]{n}-1|<\epsilon$$
The abs sign can be safely dropped, it follows that
$$n<(1+\epsilon)^n$$
Using binomial theorem to expand the first 3 terms of RHS.
$$n<1+n\epsilon+\frac{1}{2}n(n-1)\epsilon^2+...$$
As long as we make n<0.5n(n-1)εε, the first inequality holds. It requires
$$n>1+\frac{2}{\epsilon^2}$$

With all that said,
For any ε>0, there exists N=[1+2/(εε)], such that if n>N, then
$$|\sqrt[n]{n}-1|<\epsilon$$

Q.E.D
P.S. I love ε-δ proof:)

Last edited: Jul 11, 2012
4. Jul 11, 2012

### micromass

Staff Emeritus