Another proof for $\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1(n=1,2,\cdots)$.

Question

Problem

Prove$$\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1$$where $n=1,2,\cdots$

Proof

Notice that $$1=1^{\frac{1}{n}}\leq n^{\frac{1}{n}}=\sqrt[n]{1\cdot 1\cdot 1\cdots\sqrt{n}\cdot \sqrt{n}}\leq \frac{n-2+2\sqrt{n}}{n}<\frac{n+2\sqrt{n}}{n}=1+\frac{2}{\sqrt{n}}.$$ Thus, $$1 \leq n^{\frac{1}{n}}\leq 1+\frac{2}{\sqrt{n}}\to 1(n \to \infty).$$ By the squeeze theorem, we may obtain $$\lim_{n \to \infty}n^{\frac{1}{n}}=1.$$

Answer 1

Assuming the limit is $L = 1$, let us prove this using the definition of a limit: $$\forall \epsilon > 0, \exists n_0 \in \mathbb{N} \text{ such that } \vert n^\frac{1}{n} - 1 \vert < \epsilon, \quad \forall n > n_0$$

Notice that the following are equivalent

1. $\mid n^\frac{1}{n} - 1 \mid < \epsilon $
2. $(1- \epsilon)^n < n < (1+\epsilon)^n$

Note that $(1- \epsilon)^n < 1 \le n$. As for $(1 + \epsilon)^n$, use Binomial theorem as follows $$ (1+\epsilon)^n = \sum_{i=0}^n \binom{n}{i}\epsilon^i > \binom{n}{2}\epsilon^{2} = \frac{n(n-1)}{2} \epsilon^2 $$ Choose $n_0 = \lceil 1 + \frac{2}{\epsilon^2} \rceil$, we have $$ n < \frac{n(n-1)}{2} \epsilon^2 <(1+\epsilon)^n $$