How to show that the limit $\lim_{n \in\mathbb N}\sqrt[n]n$ exists using epsilon-delta method.

Question

Finding the limit using L'Hopital's method might seem fine for me, but how can you prove that the limit exists :

$\lim_{n \in\mathbb N}\sqrt[n]n$

Answer 1

It is clear that $(\forall n\in\mathbb{N}):\sqrt[n]n>1$. Let, for each $n\in\mathbb N$, $\varepsilon_n=\sqrt[n]n-1$. Then, if $n>1$,\begin{align*}n&=\bigl(\sqrt[n]n\bigr)^n\\&=(1+\varepsilon_n)^n\\&=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots\\&>\frac{n(n-1)}2{\varepsilon_n}^2.\end{align*}Therefore,$$(\forall n\in\mathbb{N}\setminus\{1\}):0<\varepsilon_n<\sqrt{\frac2{n-1}}$$and this proves that$$\lim_{n\in\mathbb N}\sqrt[n]n=\lim_{n\in\mathbb N}(1+\varepsilon_n)=1.$$

Answer 2

Note that it is enough to show that for every $\varepsilon > 0$ there exists $N \in \mathbf{N}$ such that $(1 + \varepsilon)^N > N$.

By the binomial theorem

$$ (1 + \epsilon)^N = \sum_k \binom{N}{k} \varepsilon^k. $$

The linear term is too small to gives us the desired inequality: $N\varepsilon > N$ won't hold when $\varepsilon < 1$. Thus we use the quadratic term:

$$ (1 + \epsilon)^N \ge \frac{N(N - 1)}{2}\varepsilon^2. $$

We want to show that

$$ \frac{N(N - 1)}{2}\varepsilon^2 \ge N. $$

By the Archimedean property, we can always find $N \in \mathbf{N}$ such that

$$ N > \frac{2}{\varepsilon^2} + 1. $$

If $n \ge N$ then

$$ n \ge N \ge \frac{2}{\varepsilon^2} + 1$$

hence

$$ (1 + \varepsilon)^n \ge \frac{n(n - 1)}2 \varepsilon^2 > \frac{n\left(\frac{2}{\varepsilon^2} + 1 - 1\right)}2 \varepsilon^2 = n \ge 1. $$

Taking $n$-th roots gives

$$ 1 + \varepsilon > \sqrt[n]n \ge 1. $$

Answer 3

$$ \sqrt[n+1]{n+1}< \sqrt[n]{n}$$

$$ (n+1)^{n} < n^{n+1}$$

$$ n > \left(\frac{n+1}{n}\right)^{n}= \left(1 +\frac{1}{n}\right)^{n},\ \ n > n_{0}.$$

Sequence $ \left(\left(1 + \frac{1}{n}\right)^{n}\right)$ is bounded.

Sequence $(\sqrt[n]{n})$ is decresing and lower bounded number 1.

From Bolzano-Weierstrass theorem - sequence $\sqrt[2n]{2n}$ is convergence to the same limit g.

$ g^2 = g\cdot g =\lim_{n\to \infty}\sqrt[2n]{2n}\cdot \lim_{n\to \infty}\sqrt[2n]{2n}= \lim_{n\to \infty}(\sqrt[2n]{2n})^{2}= \lim_{n\to \infty}(\sqrt[n]{2}\cdot \sqrt[n]{n}) = \lim_{n\to\infty}\sqrt[n]{2}\cdot \sqrt[n]{n}=1\cdot g $

$$ g = 1. $$