Proof of limit of n-th root of n using error principle

I was trying to prove the limit of the n-th root of n

$\lim \limits_{n \to \infty} n^{1/n} = 1$

Using the error principle for which states that for any sequence $a_n$

$a_n = L + e_n $

$\lim \limits_{n \to \infty} a_n = L \implies \lim \limits_{n \to \infty} e_n = 0 $

I start with the proof then by saying $ n^{1/n} = e_n +1 $

$ n = (e_n +1)^n $

$ n = (e_n^n +ne_n^{n-1}+...+ne_n+1) $

$ n \gt ne_n^{n-1} + ne_n+1 $

$ n-1 \gt ne_n^{n-1} + ne_n $

$ 1-\frac{1}{n} \gt e_n^{n-1} + e_n $

here technically I haven't proven that $e_n > 0$ and I would like to know how to justify it since I would like to then say

$ 1-\frac{1}{n} \gt e_n^{n-1} $

$ 1-\frac{1}{n} \gt e_n^{n-1} + e_n $

$ ({1-\frac{1}{n}})^{\frac{1}{n-1}} \gt e_n $

$ 1 \gt e_n $

But this still doesn't prove my point, I've gotten nowhere. Anyone knows how I can go about this?

Solution 1:

The idea is to first set without any assumption $$ e_n=n^{1/n}-1. $$ Then notice that $e_n>0$ for $n>1$ and, as you did, via binomial expansion and discarding most terms $$ n=(1+e_n)^n\ge 1+\binom{n}{2}e_n^2 $$ The situation is much simplified by only retaining one term with a power of $e_n$. You can try to only use the linear term, there the coefficient does not grow fast enough. The quadratic term as the next variant works well, $$ n-1\ge\frac{n(n-1)}{2}e_n^2 \\ e_n\le\sqrt{\frac{2}{n}} $$ Now one has a known null sequence as upper bound.

Solution 2:

It is seen by induction (I will clarify this point if you wish) that if $h\gt0,n\in\Bbb N$, then:

$$(1+h)^n\ge1+n\cdot h$$

Define the sequences, for $n\in\Bbb N_1$:

$$a_n=\sqrt[n]{n},\quad b_n=\sqrt{a_n}$$

We have that $b_n\ge1$, and therefore there exists a sequence of $e_n\ge0$ such that $b_n=1+e_n$. Therefore:

$$\sqrt{n}=b_n^n\ge1+n\cdot e_n,\quad e_n\le\frac{\sqrt{n}-1}{n}\lt\frac{1}{\sqrt{n}}$$

Then, as $a_n$ must be $\ge1$ for all $n$,

$$1\le a_n=b_n^2=1+2\cdot e_n+e_n^2\lt1+\frac{2}{\sqrt{n}}+\frac{1}{n}\\1\le\sqrt[n]{n}\lt1+\frac{2}{\sqrt{n}}+\frac{1}{n}$$

And the limit as $n\to\infty$ becomes clear: $a_n=\sqrt[n]{n}\to1$