A doubt on limit supremum

Question

In a book, I see the following :

$\limsup_{n \to\infty} |a_n| \geq 1$ implies $\limsup_{n \to \infty} |a_n|^{\frac{1}{n}} \geq 1$. Why ?

Answer 1

Hint: $$\limsup_{n \to \infty} |a_n| \geq 1 \quad \Longrightarrow \quad \forall \varepsilon>0,\exists \{n_k\} \subset \mathbb{N}, n_k < n_{k+1} \colon\ |a_{n_k}| \ge 1-\varepsilon \quad \Longrightarrow \quad |a_{n_k}|^{1/n_k} \ge 1-\varepsilon.$$

Answer 2

If $\limsup_n |a_n|^{1/n}\lt 1$ and $n_k\uparrow \infty$, then there is $k_0$ and $c\gt 0$ such that $|a_{n_k}|^{1/n_k}\lt 1-c$ for each $k\geqslant k_0$. This gives $|a_{n_k}|\leqslant (1-c)^{n_k}$ hence $\lim_k|a_{n_k}|=0$.