The nth root of n tends to 1

Example

For each integer $n \ge 1$, let $a_n = n^{1/n}$ be the positive $n$-th root of $n$. Then $a_n \to 1$ as $n \to \infty$. For the proof, write $a_n = 1 + h_n$ with $h_n \ge 0$. For $n \ge 2$, the binomial theorem gives $n = (1 + h_n)^n \ge \binom{n}{2} h_n^2 = \frac{n(n-1)}{2} h_n^2$. Thus $0 \le h_n \le \sqrt{2/(n-1)}$, and the squeeze theorem gives $h_n \to 0$.

Remarks

The first terms are $1, \sqrt{2} \approx 1.414, 3^{1/3} \approx 1.442, 4^{1/4} \approx 1.414$. The largest term is at $n = 3$. In the same way, $c^{1/n} \to 1$ for each real $c > 0$. Mathlib proves the version for a real variable, $x^{1/x} \to 1$ as $x \to \infty$, in the lemma tendsto_rpow_div.

Explanations

L1 human

The number $n$ grows, but the root index $n$ grows too, and the root wins. For example, $1000^{1/1000}$ is about $1.0069$, very close to $1$.