Prove that if $(z_n)_{n\geq 1}$ is a null sequence, then $(|z_n|^q)_{n\geq 1}$ with $\forall q\in \mathbb{Q}:q>0$ is a null sequence and vice versa.

Question

Let $z\in \mathbb{C}, (z_n)_{n\geq 1} \subset \mathbb{C}$ be a sequence.

Prove that:

$(z_n)_{n\geq 1}$ null sequence $\Longleftrightarrow$ $(|z_n|^q)_{n\geq 1} \quad,\forall q\in \mathbb{Q}:q>0$ is a null sequence.

A null sequence is a sequence tending to zero.

For "$\Rightarrow$"

We can rewrite $(|z_n|^q)_{n\geq 1} \quad,\forall q\in \mathbb{Q}:q>0$ to $\left(|z_n|^{\frac{n}{m}}=\sqrt[m]{|z_n|^n}\right)_{n\geq 1} \quad , \forall n\in \mathbb{Z} \land \forall m\in \mathbb{N}$.

The root merely influences the speed of decrease/increase. But how do I show that?

For "$\Leftarrow$":

Can't I just square the rewritten expression by $\uparrow \frac{n}{m}$?

Answer 1

Here is the proof:

Let $(z_n)$ be a null sequence. For arbitrary $\varepsilon>0$, there exist $N>0$ such that $\forall n\geq N$, $$|z_n|<\varepsilon. $$ Therefore, for any $0< q\in\mathbb{Q}$, we have $$|z_n|^q<\varepsilon^q$$ which shows $(|z_n|)$ is a null sequence.

Now let for all $0<q\in\mathbb{Q}$, the sequence $(|z_n|^q)$ is null. Specially for $q=1$, the sequence $(|z_n|)$ is null which implies $(z_n)$ is convergent to zero or ia a null sequence.