I am trying to prove that for all $x \ge 1, (x^{\frac{1}{n}}) \to 1$.

Question

$\textbf{Proof:}$

Let $x \ge 1$ be arbitrary and note that $\forall n \in \mathbb{N}$, $x^{\frac{1}{n}} > 1$.

So $ \forall n \in \mathbb{N},\ x^{\frac{1}{n}} - 1 \ge 0 > -1$.

Therefore I can write $x^{\frac{1}{n}} = 1 + ( x^{\frac{1}{n}} -1 )$ and use Bernoulli's inequality.

So $ \forall n \in \mathbb{N}, \left ( 1 + ( x^{\frac{1}{n}} -1) \right )^{n} \ge 1 + n( x^{\frac{1}{n}} -1)$

$\implies \> x \ge 1 + n( x^{\frac{1}{n}} -1) $

$\implies \> x - 1 \ge n( x^{\frac{1}{n}} -1) $

$\implies \> \frac{x - 1}{n} \ge ( x^{\frac{1}{n}} -1) \ge 0$

Note that as $n \to\infty, \left ( \frac{x-1}{n} \right ) \to 0 $. So by the sandwich theorem $ ( x^{\frac{1}{n}} -1) \to 0 $.

Therefore by the sum rule for convergent sequences, $(x^{\frac{1}{n}}) \to 1$.

I was wondering if my proof is correct, as it makes sense to me but I just to need to check it.

Answer 1

For $n$ sufficiently large, $$1\le x\le n$$ $$\sqrt[n]{1}\le \sqrt[n]{x}\le \sqrt[n]{n}$$ By the squeeze theorem, as $n\to \infty$, since $\sqrt[n]{1}\to 1$ and $\sqrt[n]{n}\to 1$, also $\sqrt[n]{x}\to 1$.