I believe I have the gist of how to prove this. My professor worked out a problem similar to this one only, instead of ($s_n^3$), he used ($s_n^2$), and I am slightly confused as to how he came up with certain portions of his proof. The following is the proof he gave us for ($s_n^2$). I believe after understanding his proof better, I can prove the original problem more easily. So please do not post the solution to the original question.
Proof {the convergence of the sequence ($s_n$) implies the convergence of ($s_n^2$)}
Since the lim ($s_n$)=s, we know ($s_n$) is bounded.
That is there exists $M\in R$ such that $|s_n|$ $\le$ M for all $n\in N$
Now, for every $\varepsilon >0$ we have lim ($s_n$)$=s$. Working on $\varepsilon/(M+|s|)>0$, there exists $N\in R$ such that $|s_n-s| \le \varepsilon /(M+|s|)$ whenever $n>N$, therefore for all $n>N$, $|s_n^2 – s^2| = |s_n – s|*|s_n + s| \le |s_n – s|(|s_n|+|s|) \le |s_n – s|*(M + |s|)< \varepsilon $
Which proves lim $(s_n^2)$ = $s^2$.
The following is my proof for the current problem (that is in the title).
Let me know if I did anything incorrect.
Proof
Since the lim ($s_n$)=s, we know ($s_n$) is bounded.
That is there exists $M > 0$ such that $|s_n|\le M$ for all $n\in \mathbb{N}$
Now, for every $\varepsilon >0$ since lim ($s_n$)=s, working on $\varepsilon /(3M^2)>0$,
there exists $N\in \mathbb{N}$ such that
$|s_n-s| < \varepsilon /3M^2$ whenever $n>\mathbb{N}$
Therefore, for all $n>\mathbb{N}$
$|s_n^3 – s^3|$ = $|s_n – s|$ $|s_n^2 + s_n*s + s^2| \le $ $|s_n – s|$ $(|s_n^2|+|s_n||s|+ |s^2|) \le $ $(|s_n|^2+|s_n||s|+ |s|^2) \le $ $|s_n – s|*(M^2 + M*M + M^2) \le $ $|s_n – s|*(3M^2)< \varepsilon $
Which proves lim $(s_n^3)$ = $s^3$
Best Answer
As you point out, if $s_n\to s$, then $|s_n|\le M$ for some $M>0$. Note that this implies that $|s|\le M$ as well. In this case, you have $$ s_n^3-s^3=(s_n-s)(s_n^2+s_ns+s^2). $$ Note that $$|s_n^2+s_ns+s^2|\le|s_n^2|+|s_ns|+|s^2|=|s_n|^2+|s_n||s|+|s|^2\le M^2+M\cdot M+M^2=3M^2,$$ where in the first step we use the triangle inequality, and in the third step we use that both $|s_n|\le M$ and $|s|\le M$. But then $$|s_n^3-s^3|=|s_n-s|(|s_n^2+s_ns+s^2|)\le |s_n-s|3M^2.$$ Using the definition of convergence, given $\epsilon>0$, pick $N$ so that if $n\ge N$ then $|s_n-s|<\epsilon/3M^2$. Then $$|s_n^3-s^3|<\epsilon.$$
In general, $$s_n^{k+1}-s^{k+1}=(s_n-s)\sum_{j=0}^ks_n^{k-j}s^k,$$ so $s_n^{k+1}\to s^{k+1}$, and the argument is the same, only that now, instead of $\epsilon/3M^2$, you want something like $\displaystyle \frac{\epsilon}{(k+1)M^k}$.