Limit by $\epsilon-N$ method

Question

Question is to prove that $\lim_{n\rightarrow \infty}n^{\frac{1}{n}}=1$.

This question is discussed in this site so many times and in all of the answers i have seen they have not considered the basic definition case :

Given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\left|n^{\frac{1}{n}}-1\right|<\epsilon$ for all $n\geq N$.

We want to see if $(1+\epsilon)^n>n$ I have tried the following.

Let $\epsilon\geq 1$ then $(1+\epsilon)^n=1+n\epsilon+*\epsilon^2+\cdots+\epsilon^n>1+1+\cdots+1=n$.

So, we do have $(1+\epsilon)^n>n$ for all $n$ irrespective of $\epsilon$.

Suppose $\epsilon<1$ first of all i want one $N$ such that $(1+\epsilon)^N>N$. I was looking for $N$ such that $(1+\epsilon)^N>1+N\epsilon>N$ i.e., $\dfrac{1}{1-\epsilon}>N$.. As $\epsilon<1$ it is possible to get such $N$.. But this argument does not make much sense.

Help me to solve this.

Can you use $n^{1/n}=e^{\log(n)/n}$ to simplify the limit? or are you looking for a more formal way for demonstrating the limit? — user3209698, Apr 01 '16 at 08:28
@seoanes : I know the other ways of proving which includes what you have said $n^{\frac{1}{n}}=e^{\frac{log n}{n}}$,,, I am looking for a a more formal way — , Apr 01 '16 at 08:29
Can't you proceed by induction on the inequality $(1+\varepsilon)^n >n$? — Siminore, Apr 01 '16 at 08:56
@Siminore i think i can.. this says more.. $(1+\epsilon)^n>n$ for all $n$... — , Apr 01 '16 at 09:22
I am on my phone, so cannot explain very well, but here are some duplicates of the question which I believe adopt a similar approach to yours (utilising binomial theorem): here, here and here — Inazuma, Apr 01 '16 at 09:56

score 1 · Answer 1 · answered Apr 01 '16 at 10:06

1

If you define $x_n=\sqrt[n]{n}-1$, then $$ n = (1+x_n)^n \geq \frac{n(n-1)}{2}x_n^2. $$ Hence $$ 0 \leq x_n \leq \sqrt{\frac{2}{n-1}} $$ for all $n \geq 2$. Now you can play with $\epsilon$.

Bibliography

W. Rudin. Principles of mathematical analysis.

answered Apr 01 '16 at 10:06

Siminore

35,136

This might be roundabout way of doing it.. Thanks anyways :) – Apr 01 '16 at 10:24

score 1 · Accepted Answer · 2016-04-01T10:32:22.690

1

Take

$$n>N=\frac1{\ln^2(1+\epsilon)}.$$

Then

$$\frac{\ln(n)}n<\frac1{\sqrt n}<\ln(1+\epsilon)$$ and $$|n^{1/n}-1|<\epsilon.$$

Note that for $n>4$,

$$\ln(n)<\sqrt n$$ because $$\ln(4)<\sqrt4$$and after derivation

$$\frac1n<\frac1{2\sqrt n}.$$

edited Apr 01 '16 at 10:32

answered Apr 01 '16 at 10:15

How did you come up with that $\dfrac{1}{\ln^2(1+\epsilon)}$? – Apr 01 '16 at 10:25
No.. Even if i do not like i will not downvote.. I would loose one reputation unnecessarily :D – Apr 01 '16 at 10:29
@cello: My solution comes from the inequality $\ln(n)<\sqrt n$. – Apr 01 '16 at 10:30
Shame on the anonymous downvoter. – Apr 01 '16 at 10:31
I don't understand the downvoter, your solution is the simplest so far. – Hugo Apr 01 '16 at 10:43

Hugo · Answer 3 · 2016-04-01T10:27:39.267

We have to prove that $\lim_{n\to \infty}n^{1/n}=1$. Consider $a_n = n^{1/n}-1$. Since $n\geq 1$ we have that $n^{1/n}\geq 1$ and therefore $a_n\geq 0$. On there other hand, $n = (a_n+1)^n$. Using the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) we conclude that $$(a_n+1)^n=\binom{n}{0} + \binom{n}{1}a_n+\binom{n}{2}a_n^2 + \cdots + \binom{n}{n-1}a_n^{n-1} +\binom{n}{n}a_n^{n}.$$ From that equality we obtain $$(a_n+1)^n \geq \binom{n}{2}a_n^2 = \frac{n(n-1)}{2}a_n^2.$$ So, since $n = (a_n+1)^n \geq (n(n-1)/2)a_n^2$, for every $n>1$ we have $$\frac{n(n-1)}{2}a_n^2 \leq n,$$ $$a_n^2 \leq \frac{2}{n-1},$$ $$a_n \leq \frac{\sqrt 2}{\sqrt{n-1}}.$$

Fix $\epsilon >0$ and take $N\in \mathbb N$ larger than $2/\epsilon^2+1$. If $n> N$ then $n>2/\epsilon^2+1$ and therefore $$\frac{\sqrt{2}}{\sqrt{n-1}}<\epsilon.$$ So for every $n>N$ we have $a_n < \epsilon$ and therefore $|n^{1/n}-1|=|a_n|<\epsilon$ because $a_n = |a_n|$.

How is this different from what user Siminore's answer? – Apr 01 '16 at 10:27 — , Apr 01 '16 at 10:27
That is fine. Thanks for your interest – Apr 01 '16 at 10:30 — , Apr 01 '16 at 10:30

user3209698 · Answer 4 · 2016-04-01T09:47:27.683

0

If you notice that $n^{1/n}>1$, for any $n$, you can forget about the absolute value, and $$n^{\frac{1}{n}}-1<\epsilon\to n^{\frac{1}{n}}<\epsilon+1$$ take logarithms on both sides (you can do that since the logarithm is a monotonously increasing function) $$\log(n)/n<\log(1+\epsilon)$$ This equation has solution, however I am worried it is not analytic. Then, for any positive $\epsilon$ you can find an $n$ that fulfills the convergence expression.

I hpoe this will be useful for you!

edited Apr 01 '16 at 09:47

answered Apr 01 '16 at 08:38

user3209698

1,742

Hem, $n^{1/n}\ge1$. – Apr 01 '16 at 09:45
thank you, it was a typo. All the reasoning was done supposing the correct direction of the inequality. Corrected – user3209698 Apr 01 '16 at 09:48
This is not what i wanted as i have said in the question.. It is not easy to get a solution for $\log n /n< \log(1+\epsilon)$ – Apr 01 '16 at 10:22
I have to apologize, I thought your question was: Given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\left|n^{\frac{1}{n}}-1\right|<\epsilon$ for all $n\geq N$. This solution demonstrate it exists, but I agree that it does not give an analytic and close solution to determine $N$. I would suggest you to modify then your question including this remark – user3209698 Apr 01 '16 at 10:29
It says it exists. I would like to know what it is.. That is the only difference.. I wanted to find an explicit $N$ – Apr 01 '16 at 10:34

score 0 · Answer 5 · answered Apr 01 '16 at 09:38

I think it can be done as follows :

We have $(1+\epsilon)^n=1+n\epsilon+***$

Suppose $\epsilon>1$ then we have $(1+\epsilon)^n\geq n\epsilon>n$..

I have done seperately for $\epsilon\leq 1$.

So, given $\epsilon>0$ we have $(1+\epsilon)^n>n$ for all $n$ and so we are done

Limit by $\epsilon-N$ method

5 Answers5