Is the converse of {Prove that the convergence of the sequence ($s_n$) implies the convergence of (${s_n}^3$)} T or F? In a previous post, I proved (with help) that the convergence of the sequence ($s_n$) implies the convergence of (${s_n}^3$). Here is a link to it: Prove that the convergence of the sequence ($s_n$) implies the convergence of ($s_n^3$)
Now, I want to either prove the converse or determine a counter-example.  
OK, I believe I understand now. Since the function $\sqrt[3]{x}$ is continuous there is nothing to "exploit." In other words, the converse of the original statement is actually true. 
I believe I can prove it similar to how I proved the original conjecture by showing that the convergence of ${s_n}^3$ implies the convergence of $s_n$. Please let me know if I did something wrong.   
The following is my proof:  
Proof
Assume ${s_n}^3 \to s^3$.
Then we know ${s_n}^3$ is bounded.
Hence, there exists $M > 0$ such that ${|s_n|}^3 \le M$ for all $n\in \mathbb{N}$
Now, for every $\varepsilon >0$ since ${s_n}^3 \to s^3$, working on $\varepsilon * 3\sqrt[3]{M^2}>0$,
there exists $N\in \mathbb{R}$ such that 
$|{s_n}^3 - s^3| < \varepsilon * 3\sqrt[3]{M^2}$ whenever $n>\mathbb{N}$
Therefore, for all $n>\mathbb{N}$
$|s_n - s|$ = $$\frac{|{s_n}^3 - s^3|}{|{s_n}^2 + s_n*s + s^2|} \le $$ $$\frac{|{s_n}^3 - s^3|}{|{s_n}^2| + |s_n||s| + |s^2|} \le $$ $$\frac{|{s_n}^3 - s^3|}{|{s_n}|^2 +|s_n|*|s|+ {|s|}^2} \le $$ $$\frac{|s_n - s|}{\sqrt[3]{M^2} + \sqrt[3]{M}*\sqrt[3]{M} + \sqrt[3]{M^2}} \le $$ $$\frac{|s_n - s|}{3\sqrt[3]{M^2}} < \varepsilon $$
which proves $s_n \to s$.  
 A: Your counterexample is incorrect: your assertion that $a_n = (-1)^{(2n+1)/3}$ diverges is incorrect. The value of $a_n$ is $-1$ for all $n$: by definition, you have
$$a_n = \sqrt[3]{(-1)^{2n+1}} = \sqrt[3]{-1} = -1.$$
So $a_n$ converges as well.
Here's a hint: is the function $f(x) = \sqrt[3]{x}$ continuous? Because if it is continuous, it better take convergent sequences to convergent sequences, since a function $f$ is continuous at $a$ if and only if for every sequence $a_n$ such that $a_n\to a$, you have $f(a_n)\to f(a)$...
A: Hint: Before looking for a counterexample, or a proof, suppose the sequence $(a_n^3)$ converges.  Converges to what? Let us call the limit $a^3$.  It is a nice name, perfectly permissible since every real number is the cube of something.
We know that for $n$ large, $a_n^3$ is close to $a^3$.  Does that force $a_n$ to be close to something nice?  Once we have a concrete understanding of what we are looking for, the details should not be hard. (When it comes time to work with the formal definition, you may wish to treat $a=0$ and $a \ne 0$ separately.)
A: You are having trouble handling the denominator. If you've never seen the technique used to deal with that sort of thing it may be very mysterious and confusing to figure out how to proceed. So let me do this: I'll give an $\epsilon$-$N$ proof for:

Let $(s_n)$ be a sequence of real numbers. If $s_n\to s\neq 0$, then the sequence $\left(\frac{1}{s_n}\right)$ converges to $\frac{1}{s}$.

Let $\epsilon\gt 0$. The idea is that if $s_n$ is very close to $s$, then $\frac{1}{s_n}$ will be very close to $\frac{1}{s}$. The problem is that
$$\frac{1}{s_n}-\frac{1}{s} = \frac{s-s_n}{s_ns}.$$ So while we can make the numerator small, the denominator is also changing, so that presents an obstacle to a simple chain of inequalities.
So: the first thing to do is to show that we can make sure that $s_ns$ is not too small; since we want to say that the expression is "less than or equal" to something, we need to be able to say $$\left|\frac{1}{s_ns}\right|\leq k$$
for some $k$. But in order to ensure that this quotient is small, we need to make sure that $s_ns$ is large. That is, we need to find some lower bound to $|s_ns|$. 
Let $\ell = |s|\neq 0$.  We know that there exists $N_1\gt 0$ such that if $n\geq N_1$, then $|s_n-s|\leq \frac{\ell}{2}$. Since
$$|s|-|s_n| \leq |s_n-s| \leq \frac{\ell}{2} = \frac{|s|}{2},$$
then we have that if $n\geq N_1$, then
$$|s| - \frac{|s|}{2} \leq |s_n|,$$
that is,
$$|s_n| \geq \frac{|s|}{2};$$
and therefore
$$|s_n|\,|s| \geq \frac{|s|^2}{2},$$
so if $n\geq N_1$, then
$$\frac{1}{|s_n|\,|s|} \leq \frac{2}{|s|^2}.$$
So we have succeded in bounding above $\frac{1}{|s_ns|}$, by bounding $|s_ns|$ below ("away from $0$").
Now: let $\epsilon\gt 0$. We need to show that there exists an $N\gt 0$ such that for all $n\geq N$, $|\frac{1}{s_n} - \frac{1}{s}|\lt \epsilon$.
What we will want to do is manipulate the latter inner product as:
$$\left|\frac{1}{s_n} - \frac{1}{s}\right| =\left|\frac{s-s_n}{s_ns}\right| = \frac{1}{|s_ns|}\left|s-s_n\right|.$$
We know that if $n\geq N_1$, then $\frac{1}{|s_ns|}\leq \frac{2}{|s|^2}$. We also know that there exists $N_2\gt 0$ such that for all $n\geq N_2$, we have
$$|s-s_n| \lt \frac{|s|^2\epsilon}{2}.$$
We know this because $s_n\to s$. So let $N=\max(N_1,N_2)$. Then if $n\geq N$, we will have
$$\text{both}\quad \frac{1}{|s_ns|}\leq \frac{2}{|s|^2}\qquad\text{and}\qquad |s-s_n|\lt \frac{|s|^2\epsilon}{2}.$$
Therefore, if $n\geq N$, then
$$\begin{align*}
\left|\frac{1}{s_n} - \frac{1}{s}\right| &= \left|\frac{s-s_n}{s_ns}\right| \\
&= \frac{|s-s_n|}{|s_ns|}\\
&= \frac{1}{|s_ns|}|s-s_n|\\
&\leq \left(\frac{2}{|s|^2}\right)|s-s_n|\\
&\lt \left(\frac{2}{|s|^2}\right)\left(\frac{|s|^2\epsilon}{2}\right)\\
&= \epsilon,
\end{align*}$$
and so we have shown that $\frac{1}{s_n}\to \frac{1}{s}$.

Now, think about what you have here. Your problem is that you have a denominator that depends on $s_n$ and $s$; so you want to bound that denominator "away from $0$"; this will give you control over the denominator. Then you can also bound $|s_n^3 - s^3|$, which will give you control over the numerator. Having control over both will give the desired result.
A: a counter example: $(\zeta_3^n)_n$ diverges but the constant sequence $1$ converges (here $\zeta_3=e^{2\pi i/3}$).  if you work over $\mathbb{R}$, then there is a unique cube root.
