How to show that the sequence $a^{1/n}$ converges to $1$ when $n \to \infty$? (using the epsilon-delta definition)

Question

Goal: To show that the sequence $a^{1/n}$ ($a\in \mathbb{R}$ and $a>0$ ) converges to $1$ when $n \to \infty$?

Attempt: Given any $\epsilon >0$, we have to come up with an $N\in\mathbb{N}$ such that for all $n\geq N$ we have $|a^{1/n}-1|<\epsilon.$ I tried different cases $0<a<1$ and $a>1,$ but it seems to be a fruitless exercise. Any hints/suggestions will be much appreciated.

Hint (assuming $a>1$). For large enough $n$ we have $1+n\epsilon >a$. But also, by binomial theorem (or Bernoulli) $$(1+\epsilon)^n\ge1+n\epsilon.$$ If $0<a<1$ do the same with $1/a$ in place of $a$. — Jyrki Lahtonen, Oct 15 '17 at 17:21
@JyrkiLahtonen I used that and it works. We choose $\epsilon\leq \sqrt{\frac{2}{n}}.$ — Student, Oct 15 '17 at 17:21
Have a look at https://math.stackexchange.com/q/1867269/321264. — StubbornAtom, Oct 15 '17 at 18:07

Robert Z · Answer 1 · 2017-10-15T17:30:58.013

3

We may assume that $a\geq 1$ (if $0<a<1$ replace $a$ with $1/a$). Let $x=a-1\geq 0$ and use the Bernoulli inequality, $$1\leq a=(1+x)^{1/n}\leq 1+\frac{x}{n}=1+\frac{a-1}{n}.$$ Hence $$|a^{1/n}-1|\leq \frac{a-1}{n}<\epsilon$$ as soon as $n>\frac{a-1}{\epsilon}$.

edited Oct 15 '17 at 17:30

answered Oct 15 '17 at 17:21

Robert Z

147,345

But what about $0<a<1$? – Student Oct 15 '17 at 17:24
If $0<a<1$ replace $a$ with $1/a$. Then $(1/a)^{1/n}\to 1$ implies that $a^{1/n}\to 1$. – Robert Z Oct 15 '17 at 17:24
And how do you get back to the definition from this step? I am asking this because I already did what you have suggested, but I am not sure how to go back to the definition – Student Oct 15 '17 at 17:25
@ShreyAryan See my edited answer. – Robert Z Oct 15 '17 at 17:30

score 0 · Answer 2 · answered Oct 15 '17 at 17:22

Let's look at $a>1$. Then $|a^{1/n}-1|=a^{1/n}-1$ and we have $$ a^{1/n}-1<\varepsilon\\ a^{1/n}<1+\varepsilon\\ a<(1+\varepsilon)^n $$ Since $(1+\varepsilon)^n$ is increasing and unbounded with respect to $n$, it will eventually become larger than $a$, and we have our $N$. Specifically, setting $N\geq \log_{1+\varepsilon}(a)=\frac{\ln a}{\ln(1+\varepsilon)}$ will work.

How to show that the sequence $a^{1/n}$ converges to $1$ when $n \to \infty$? (using the epsilon-delta definition)

2 Answers2