Prove $\lim_{n\to\infty}z_n^{\,3}=\infty\iff\lim_{n\to\infty}z_n=\infty$ Let $z_n$ be a complex sequence.
How would I go about proving $\lim_{n\to\infty}z_n^{\,3}=\infty\iff\lim_{n\to\infty}z_n=\infty$?
Is there a formal way to prove divergence other than the fact it contradicts the definition of convergence?
 A: Hint: Use the fact that $\lim_{n\to\infty}z_n=\infty\iff\lim_{n\to\infty}|z_n|=\infty$.
A: For a complex sequence $\{z_n \}$,
$$\lim\limits_{n \to \infty} z_n = \infty$$ means that $\lim\limits_{n \to \infty} \vert z_n \vert = \infty$.
As the real maps $x \mapsto x^3$ is continuous on $[0,\infty)$ as well as $x \mapsto \sqrt[3]{x}$, the required equivalence is immediate.
A: Hint: if $|z_n^3|=|z_n|^3\ge M$, then $|z_n|>\lfloor\sqrt[3]M\rfloor$, and conversely if $|z_n|\ge M$, then $|z_n^3|=|z_n|^3>M^3$.
A: What does it mean for $\lim_{n\to\infty}z_n=\infty$?
Well, for any open neighborhood of $\infty$ all terms of the sequence are eventually contained in it. So we have, given any $R>0$, we define the open set $U_R=\{z\in\mathbb{C}:|z|>R\}$ and we know that there exists $N_0\in\mathbb{N}$ such that $z_k\in U_R$ for every $k\geq N_0$.  What does the cubing operation in $\mathbb{C}$ do to modulus? This gives one direction.  For the other you can argue in a similar way but one needs to be mindful of multivaluedness of the cube root.
