# Prove or disprove : The limit $\lim_{n\rightarrow\infty}z^3_n$ exists if and only if the limit $\lim_{n\rightarrow\infty}z_n$ exists

Let $$z_n$$ be a complex sequence. The limit $$\lim_{n\rightarrow\infty}z^3_n$$ exists if and only if the limit $$\lim_{n\rightarrow\infty}z_n$$ exists

I think the statement is true. From the definition, an infinite sequence of complex numbers has a limit $$z$$ if, for each positive number $$ε$$, there exists a positive integer $$n_0$$ such that $$|z_n − z|<ε$$ whenever $$n>n_0$$.

How can I use the definition to prove the statement above? Is it true in the first place?

• If $\zeta$ is a primitive third root of $1$ and $z_n= \zeta^{n \mod 3}$, then $z_n$ doesn't converge, but $z_n^3$ is constant. Apr 15, 2021 at 22:27

The statement is false. Let $$\omega=-\frac12+\frac{\sqrt3}2i=e^{2\pi i/3}$$, and let $$z_n=\omega^n$$. Then $$(\forall n\in\Bbb N):z_n^{\,3}=1$$, but the sequence $$(z_n)_{n\in\Bbb N}$$ diverges.

• Can this counter example be used to disprove the opposite statement as well? ($\lim_{n\rightarrow\infty}z^3_n = inf$ if and only if $\lim_{n\rightarrow\infty}z_n = inf$ )
– Y.I.
Apr 15, 2021 at 22:38
• What is $inf$? Do you mean $\infty$? Apr 15, 2021 at 22:38
• Yes I mean $\infty$
– Y.I.
Apr 15, 2021 at 22:39
• No. It is actually true that $\lim_{n\to\infty}z_n^{\,3}=\infty\iff\lim_{n\to\infty}z_n=\infty$. Apr 15, 2021 at 22:41
• Thank you, I will try to think of the proof for this.
– Y.I.
Apr 15, 2021 at 22:42