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 $s_n \to s$.
Best Answer
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 $f(a_n)\to f(a)$...