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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 $s_n \to s$.

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I just cubed $a_n$ and got rid of the power's denominator, am I wrong by doing that? –  enlgmatlc Mar 31 '11 at 4:22
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Oh gosh! $a_n$ does converge to -1. :( I'll be back! –  enlgmatlc Mar 31 '11 at 4:37
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I believe it does have an inverse, $\sqrt[3]{x}$ And it is continuous, but I don't –  enlgmatlc Mar 31 '11 at 20:32
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I'm not sure I understand what your trying to convey. I understand that $x^2$ does not have an inverse whereas $x^3$ does. But what am I supposed to deduct from this? Currently, all that is coming to mind is that a cubed root will be involved in my counter-example, which I had already assumed with my original FAILED attempt of a counter-example. –  enlgmatlc Mar 31 '11 at 21:18
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@englmatlc: use \sqrt[n]{x} to get $\sqrt[n]{x}$. –  Arturo Magidin Apr 1 '11 at 15:15
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4 Answers

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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.

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This was extremely helpful! My teacher has actually provided me with a proof for the problem, and I honestly didn't understand a portion of the proof, but after reading through your example here, I have a better understanding of the proof my professor gave out. Thank you very much!!! –  enlgmatlc Apr 15 '11 at 20:34
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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 $f(a_n)\to f(a)$...

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Perhaps I do not fully understand the meaning of a continuous function. However, we have not covered that particular section in class yet. I will add the definitions we are using from the sections this problem came from at the top. Do be patient with me, I am trying to understand! This particular problem is driving me crazy! –  enlgmatlc Mar 31 '11 at 21:24
    
@enlgmatlc: What section? You don't know what "continuous" means yet? –  Arturo Magidin Mar 31 '11 at 21:28
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I only know from what I learned in my Calculus classes, I just checked my syllabus for this class (real-analysis) and we aren't supposed to cover continuous functions until next week. Is that bad? –  enlgmatlc Mar 31 '11 at 21:30
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So, imagine you have a sequence $(a_n)$ on the "domain", and it's approaching a point $s$. And you look at the value of $f$ on the sequence, that is, $(f(a_n))$. If the graph of the function doesn't have any holes or jumps or breaks, what do the values $(f(a_n))$ approach, if anything? –  Arturo Magidin Mar 31 '11 at 21:33
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@enlgmatlc: Yes, it is. Draw a picture! Draw the graph of a some continuous function. Then draw the points of a sequence $(a_n)$ that is approaching $s$ on the domain (the $x$-axis). Then look at the values of $f$ on the points $a_n$, $f(a_n)$; draw the points on the graph. See what they are doing and what they are approaching. Think about why the lack of jumps and holes is important for this to happen. And then, think about what this means if the graph of the function you have is the graph of $y=\sqrt[3]{x}$. –  Arturo Magidin Mar 31 '11 at 21:46
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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.)

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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.

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Given that the question is tagged [real-analysis], it seems reasonable to assume that the question is about real numbers and real sequences, not complex ones... –  Arturo Magidin Mar 31 '11 at 4:50
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Please do not post any counter-examples yet. I'm trying to figure it out. However, I'm happy to accept hints! –  enlgmatlc Mar 31 '11 at 4:57
    
I immediately though of this answer too! I am not really sure if the question indicates which field we are working in. (Rudin 3E would always use $\mathbb{C}$ if I remember) –  Eric Naslund Mar 31 '11 at 20:15
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The class is only working with real numbers. –  enlgmatlc Mar 31 '11 at 21:09
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