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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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    How did you compute the cube of $a_n$?2011-03-31
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    I just cubed $a_n$ and got rid of the power's denominator, am I wrong by doing that?2011-03-31
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    Ah. Now that you fixed the TeX it makes sense... Never mind.2011-03-31
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    Oh gosh! $a_n$ does converge to -1. :( I'll be back!2011-03-31
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    While yoyo's counterexample is perfectly fine, it is a bit misleading in that it involves *complex* numbers instead of real ones (the reason why this sequence diverges is similar to the reason of divergence of $s_n = (-1)^n$). Arturo pointed you in the right direction.2011-03-31
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    @enlgmatlc: There is a very important difference between the argument that tries to go from "$(a_n^2)$ converges" to "$(a_n)$ converges"; and the argument that tries to go from "$(a_n^3)$" converges" to "$(a_n)$ converges." Namely, different real numbers may have the same square, but different real numbers *always* have different cubes. The counterexample you have have exploits that feature of the square by taking two numbers that are different ($1$ and $-1$) but with the same square to get a sequence that alternates between two different values, but whose square is constant.2011-03-31
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    @enlgmatlc: Again: does the function $x\mapsto x^3$ have an inverse? Is it continuous (notice the difference with $x\mapsto x^2$, which does *not* have an inverse unless you restrict the domain).2011-03-31
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    I believe it does have an inverse, $\sqrt[3]{x}$ And it is continuous, but I don't2011-03-31
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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.2011-03-31
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    @enlgmatlc: The point is that **if** you have a *continuous* function, then it has to take convergent sequences to convergent sequences. Since $x\mapsto x^2$ is continuous, that tells you that if $(a_n)$ converges then $(a_n^2)$ converges. Because $x\mapsto x^3$ is continuous, if $(a_n)$ converges then $(a_n^3)$ converges. You would like to find a *continuous function* that takes $(s_n^3)$ to $(s_n)$, so that you can go from convergence of former to that of the latter.2011-03-31
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    @enlgmatlc: Intuitively, the reason continuity comes to play is: you know that if $f$ is continuous, then provided that $s_n$ is close enough to $s$, you will have $f(s_n)$ as close as you want to $f(s)$. So if $(s_n^3)$ converges to $s$, if you take $n$ large enough you can make sure that $s_n^3$ is close enough to $s$ so that $f(s_n^3)$ will be as close as you want to $f(s)$. So if you can find a continuous function to transform $a_n^3$ into $a_n$...2011-03-31
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    @englmatlc: use `\sqrt[n]{x}` to get $\sqrt[n]{x}$.2011-04-01
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    @englmatlc: Yes, you are correct. The argument will be to try to use the fact that $s_n$ is very close to $a$ in order to show that $\sqrt[3]{s_n}$ is very close to $\sqrt[3]{a}$. It is going to be a bit trickier, though, and as user6312 suggests, you may have to consider the case of $a=0$ and of $a\neq 0$ separately. This because instead of the nice identity $(b^3-c^3)=(b-c)(b^2-bc+c^2)$, which is easy to exploit, here you will probably use $(b^{1/3}-c^{1/3}) = (b-c)/(b^{2/3}-b^{1/3}c^{1/3}+c^{2/3})$, so you'll have to bound the denominator away from $0$ somehow.2011-04-01
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    @englmatlc: Your argument is incorrect: you seem to be trying to use $$(a^{1/3}-b^{1/3})=(a-b)(a^{2/3}+a^{1/3}b^{1/3} + b^{2/3})$$but that's false. What is true is that $(a^{1/3}-b^{1/3})(a^{2/3}+a^{1/3}b^{1/3}+c^{2/3}) = a-b$, hence$$a^{1/3}-b^{1/3} = \frac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}.$$That is, you have *quotient*, not a product. That's why you'll want to bound the denominator *away* from zero: if $|a|\gt L$, then $\frac{1}{|a|}\lt \frac{1}{L}$.2011-04-01
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    @englmatlc: Your first inequality is incorrect. Since $|s_n^2 + s_ns + s^2|\leq |s_n|^2 + |s_n||s|+|s|^2$, then taking reciprocals you have$$\frac{1}{|s_n^2+s_ns+s^2|}\geq \frac{1}{|s_n|^2+|s_n||s|+|s|^2}.$$Multiplying through by $|s_n^3-s^3|$ gives the opposite inequality from what you have. (Your denominator on the second fraction is *larger* than the denominator on the first fraction, so the quotient is *smaller*, not larger).2011-04-10

4 Answers 4

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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...2011-03-31
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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!2011-03-31
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    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)2011-03-31
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    The class is only working with real numbers.2011-03-31
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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!2011-03-31
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    @enlgmatlc: What section? You don't know what "continuous" means yet?2011-03-31
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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?2011-03-31
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    @enlgmatlc: If you remember it from Calculus, then the "intuitive" explanation I gave in the other comment should make sense to you, which should in turn tell you what the answer *ought* to be (whether the converse does or does not hold). The fact that you haven't covered continuity yet just means you would not be able to simply invoke the result I mentioned above to justify the answer.2011-03-31
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    I basically know that a function is continuous if it doesn't have any "exceptions" like a hole or a jump in the graph of the function.2011-03-31
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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?2011-03-31
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    Would it not be approaching s as well?2011-03-31
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    @enlgmatlc: $s$ in the *domain*. The value $f(a_n)$ are up there in the "graph". How could the $f(a_n)$ possibly be approaching $s$?2011-03-31
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    I believe I know what the answer is based on your previous comments, but I honestly do not know why its the answer. Is it supposed to be f(s)? I guess it would be since it is continuous and $a_n$ -> s2011-03-31
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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 $y=\sqrt[3]{x}$.2011-03-31
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    I'm on it!!!!!!2011-03-31
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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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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!!!2011-04-15