Implications of $n a_n=O((n^{\frac{1}{\alpha}}|b_n|)^\alpha)$

Question

Consider two sequences of real numbers $\{a_n\}_n$, $\{b_n\}_n$. Suppose $n a_n=O((n^{\frac{1}{\alpha}}|b_n|)^\alpha)$ where $\alpha \in \mathbb{R}$, $na_n\geq 0$ and big $O$ notation is explained here

Which of the following statements are true?

(1)$\lim_{n\rightarrow \infty} na_n=\infty \leftrightarrow \lim_{n\rightarrow \infty}n^{\frac{1}{\alpha}}|b_n|=\infty$

(2)$\lim_{n\rightarrow \infty} a_n=0\leftrightarrow \lim_{n\rightarrow \infty}n^{\frac{1}{\alpha}}|b_n|=0$

(3) $a_n$ bounded away from $0$ and $\infty$ $\leftrightarrow$ $n^{\frac{1}{\alpha}}|b_n|$ bounded away from $0$ and $\infty$

Answer 1

(1) is only true if $\alpha > 0$.

Consider (2). $\lim a_n = 0$ means $a_n = o (1)$. This means $(n^{\frac{1}{\alpha}}|b_n|)^\alpha = o (1)$. If $\alpha > 0$, we have $n^{\frac{1}{\alpha}}|b_n| = o (1)$ so the statement is true. Otherwise, $\lim n^{\frac{1}{\alpha}}|b_n| = + \infty$.

(3) Let $c = \lim a_n$, where $c > 0$ and finite. Then, $\lim n^{\frac{1}{\alpha}}|b_n| = c^{\frac {1} {\alpha}} > 0$ and finite, so (3) is true in all cases.