Does $ \ (\sqrt[n]{n})^p \ \to 1 \ \text{as} \ n \to \infty \ \ \text{if} \ \ p < 1 \ $?

Question

We know that $ \ \sqrt[n]{n} \to 1 \ \text{as} \ n \to \infty $.

We can show that $ \ (\sqrt[n]{n})^p \ \to 1 \ \text{as} \ n \to \infty \ \text{if} \ p \geq 1 \ $

My question is-

Does $ \ (\sqrt[n]{n})^p \ \to 1 \ \text{as} \ n \to \infty \ \ \text{if} \ \ p < 1 \ $ ?

Answer 1

For $p=0$

$$(\sqrt[n]{n})^p=(\sqrt[n]{n})^0=1$$

For $p\neq 0$

$$(\sqrt[n]{n})^p=e^{\frac {p\log n}n}\to e^0=1$$

Answer 2

The power function $x^p$ is continuous, so that

$$\lim_{n\to\infty}\left(\sqrt[n]n\right)^p=\left(\lim_{n\to\infty}\sqrt[n]n\right)^p$$

and $1^p=1$ for all $p$.

Answer 3

Yes, for all $p \in \mathbb R$. Let $a_n = \sqrt[n]n$, then $a_n \to 1$ as you said, and by definition of continuity of $x \mapsto x^p$, $a_n^p \to 1^p=1$.