Is $\{(x_n)^{1/n}\}$ convergent when $\{x_n\}$ is a positive convergent sequence?

Question

Consider $\{x_n\}$ to be a positive sequence and convergent. Is it true that the sequence $\{x_n^{1/n}\}$ is convergent?

Answer 1

A simple counterexample:

$$x_n=\frac{n\bmod 2}{2^n}$$ so that

$$\sqrt[n]{x_n}=\frac{n\bmod 2}2.$$

Answer 2

If $x := \lim_{n \rightarrow \infty} x_n$ is positive, it is true. because there exists an $N \in \mathbb{N}$ such that $1/M<a_n<M$ for all $n \geq N$ for some fixed $M$. Thus $$1=\liminf_{n\rightarrow \infty} M^{-1/n}\leq\liminf_{n \rightarrow \infty} a_n^{1/n} \leq \limsup_{n \rightarrow \infty} a_n^{1/n} \leq \limsup_{n\rightarrow \infty} M^{1/n} =1.$$ Otherwise not, as the above examples show!