Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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 $(s_n^3)$ = $s^3$

share|cite|improve this question
Try using "s_n" to get $s_n$ instead of $sn$. – Alex Becker Mar 31 '11 at 0:53
Do you mean $s_n$ and $s_n^2$? I assume so. To show why $s_n^3$ converges, consider the identity $$s_n^3-s^3=(s_n-s)(s_n^2+s\cdot s_n+s^2).$$ (or prove the multiplicative property of limits) What exactly is confusing you in the proof you posted? Everything looks correct to me. – Eric Naslund Mar 31 '11 at 0:59
Sorry, I'm very NEW to posting here, and I post and edit, post and edit... I'm slow! Thanks for helping with the s_n notation! – enlgmatlc Mar 31 '11 at 1:08
Here are some TeX-hints: \leq gives $\leq$, \varepsilon gives $\varepsilon$ and \cdot gives $\cdot$, while \mathbb{R} and \mathbb{N} give $\mathbb{R}$ and $\mathbb{N}$. – t.b. Mar 31 '11 at 2:42
This looks very good, congratulations! Just some very small things: Say immediately that $M \gt 0$ instead of $M \in \mathbb{R}$ (to avoid division by $0$ when writing $E/(3M^2)$). Then take $N \in \mathbb{N}$ instead of $\mathbb{R}$. Finally, I'd include one more minor step: $|s^2| = |s|^2 \leq M^2$ which you carry out implicitly. – t.b. Mar 31 '11 at 3:32
up vote 3 down vote accepted

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 $\displaystyle \frac{\epsilon}{(k+1)M^k}$.

share|cite|improve this answer
Could you explain how you got $$|s_n^3-s^3|\le|s_n-s|(|s_n|^2+|s_n||s|+|s|^2)\le |s_n-s|3M^2.$$ I'm having difficulty understanding how you converted the $a^2+ab+b^2$ to $3M^2$ I understand how you got E/$3M^2$, I guess I'm confused on how we defined M to make the polynomial convert to the $3M^2$ – enlgmatlc Mar 31 '11 at 1:35
I edited the answer to clarify the step you asked about. Let me know if you need additional details. – Andrés E. Caicedo Mar 31 '11 at 3:00
Ah I see! Based on $|s_n|\leq M$ as well as $|s|\leq M$ as Mr. Andres pointed out. We are able to simply say $M^2$ + M*M + $M^2$ by replacing |s| and |$s_n$| with M due to the definition of convergence. Overall, gives us the $3M^2$ – enlgmatlc Mar 31 '11 at 3:04
Awesome! Thanks!!! I'm going to add my final proof here in a few minutes, please check it out and let me know if I did it correctly. :) – enlgmatlc Mar 31 '11 at 3:06

I really appreciate how you requested that no one tell you the proof because you want to solve your own HW problems, so +1 for that. I'm not sure if this will help you, but here goes anyway!

Definition: We say that a sequence $\{s_k\}$ converges to a limiting value $s$ if for every real number $\epsilon >0$, no matter how small, it is always possible to find a sufficiently large $N$ so that for every $M>N$: $|s_M-s|<\epsilon$.

The thing we really care about is $|s_n^2-s^2|$, and the idea is basically this:

since we can factor $s_n^2-s^2=(s_n-s)(s_n+s)$, then it follows from the convergence of $\{s_n\}$ that as n goes to infinity, the first factor will go to zero, and the second factor will be converge to 2s. Therefore the whole right hand side converges to zero, which shows that $\{s_n^2\}$ converges.

To make everything rigorous, you have to rescale your $\epsilon$ because the ordinary sequence and the squared sequence converge at a different rate.

We want to say $|s_n^2-s^2|<\epsilon '$, and this is not the same $\epsilon$ as in $|s_n-s|<\epsilon$, but the two are related. The important thing is once someone gives us the $\epsilon '>0$, how to find the N that satisfies the definition of convergence. We can do this by saying if some given N satisfies the definition for an ordinary sequence for some $\epsilon >0$, then the same N will work for the squared sequence, so long as we rescale our $\epsilon '$ to $\epsilon '=\epsilon(M+|s|)$.

In other words, whatever N works for $\epsilon=\frac{\epsilon '}{M+|s|}$, will work for $|s_n^2-s^2|<\epsilon '$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.