Register to reply

Subsequences and boundedness

by rudders93
Tags: boundedness, subsequences
Share this thread:
rudders93
#1
Sep4-11, 04:15 AM
P: 46
1. The problem statement, all variables and given/known data
Without using logarithms, prove that [itex]a^{n}\rightarrow0[/itex] as [itex]n\rightarrow\infty[/itex] for [itex]|a|<1[/itex] by using the properties of the sequence [itex]u_{n}=|a|^{n}[/itex] and its subsequence [itex]u_{2n}[/itex].

3. The attempt at a solution


I'm really stuck on this. One thing I thought of doing was to prove that the sequence is strictly decreasing as [itex]u_{n+1}-u_{n}=|a|^{n+1}-|a|^{n}=|a|^{n}(|a|^{n}-1)<0[/itex] as [itex]|a|<1[/itex]. Also we know it's bounded within interval (0,1) and so by the Monotonic Sequence theorem it must converge to some limit. Then using standard limits we can prove that it converges to 0, but I think that's kind of cheating the proof

However, not sure how to do this using the properties of subsequences. How can I do it using that?

Thanks!
Phys.Org News Partner Science news on Phys.org
Apple to unveil 'iWatch' on September 9
NASA deep-space rocket, SLS, to launch in 2018
Study examines 13,000-year-old nanodiamonds from multiple locations across three continents
HallsofIvy
#2
Sep4-11, 06:28 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,544
Yes, that is a decreasing sequence bounded below by 0 and so, as you say, converges. The point of "subsequences" is this- a sequemce converges, to limit L, if and only if every subseqence converges to L. You know this sequence converges so if any subsequence converges to 0, the entire sequence does. Can you show that [itex]a_{2n}[/itex] converges to 0? What is special about that sequence? In particular, why "2n"?
Dickfore
#3
Sep4-11, 06:33 AM
P: 3,014
You should use this:

http://en.wikipedia.org/wiki/Bernoulli%27s_inequality

and:

http://en.wikipedia.org/wiki/Squeeze_theorem

in your proof.

Hint: If [itex]|a| < 1[/itex], we can represent it as:
[tex]
|a| = \frac{1}{1 + h}
[/tex]
What can you say about h?

rudders93
#4
Sep4-11, 06:40 AM
P: 46
Subsequences and boundedness

Hi, thanks for reply.

Yeah I tried to see find the reason behind the 2n but the only thing I could think of was that it converges at a faster rate than just n, as [itex]u_{2n}=(|a|^{n})^{2}[/itex] but I still couldn't see what I could do to show it converges to 0, other than the standard limits :(

An [itex]\epsilon-N[/itex] could I guess show it converges to 0, but then again, it could do that for just [itex]u_{n}[/itex]?

Thanks again for reply
Dickfore
#5
Sep4-11, 06:43 AM
P: 3,014
I think the point is that [itex] u_{2 n} = (u_{n})^{2}[/itex]. Since the sequence is convergent, so is any of its subsequences and it has the same limit. Take the limit in the above equality.
rudders93
#6
Sep4-11, 06:50 AM
P: 46
Ah i see that, thanks Dickfore! So can do:

[itex]|a|=\frac{1}{1+h}[/itex] so therefore if we take [itex]u_{2n}=|a|^{2}=(\frac{1}{1+h})^{2}=\frac{1}{(1+h^{2})}[/itex] which tends to 0 as n tends to infinity by standard limits. Is that it? But nevertheless, won't just [itex]u_{n}[/itex] also tend to infinity using the same argument, although it may take longer to do so?

Thanks again!
Dickfore
#7
Sep4-11, 06:51 AM
P: 3,014
No, that is not what I meant. My last post was not connected with my previous hint, since you are supposed to follow a different method.

Also, you have a mistake:
[tex]
\left(\frac{1}{1 + h}\right)^{2} \neq \frac{1}{1 + h^{2}}
[/tex]
rudders93
#8
Sep4-11, 06:55 AM
P: 46
Sorry, I'm not sure how [itex]u_{2n}=(un)^{2}[/itex] can help me :( Nor how the squeeze theorem/bernoulli's can help with in proving that [itex]u_{2n}[/itex] converges without using standard limits :(
Dickfore
#9
Sep4-11, 06:58 AM
P: 3,014
Disregard post #3.

1) Did you prove that [itex]\lbrace u_{n} \rbrace[/itex] is convergent?

2) What does that tell you about the convergence of the subsequence [itex]\lbrace u_{2 n}\rbrace[/itex]

3) Can you prove this [itex]u_{2 n} = (u_{n})^{2}[/itex] for the particular sequence we are discussing?

4) What will happen if we take the limit in the equality:

[tex]
\lim_{n \rightarrow \infty}{u_{2 n}} = \lim_{n \rightarrow \infty}{(u_{n})^{2}}
[/tex]
rudders93
#10
Sep4-11, 06:59 AM
P: 46
ooops sorry that was me being incompetent in latex. [itex]u_{2n}=|a|^{2}=(\frac{1}{1+h})^{2}=\frac{1}{(1+h)^{2}}[/itex]. But that nevertheless converges to infinity also by standard limits?
Dickfore
#11
Sep4-11, 07:00 AM
P: 3,014
Quote Quote by rudders93 View Post
ooops sorry that was me being incompetent in latex. [itex]u_{2n}=|a|^{2}=(\frac{1}{1+h})^{2}=\frac{1}{(1+h)^{2}}[/itex]. But that nevertheless converges to infinity also by standard limits?
What you wrote is a constant sequence, so to speak of its convergence properties is trivial. However, this has nothing to do with the problem we are discussing.
rudders93
#12
Sep4-11, 07:06 AM
P: 46
Hmm. Well:

1) Yes [itex]{u_{n}}[/itex] is convergent as it is decreasing and it is bounded as it must converge.

2) Subsequences of convergent sequences must converge, so yes this also converges.

3) Yes. [itex]u_{2n}=|a|^{2n}=(|a|^{n})^{2}=(u_{n})^{2}[/itex]

4) Not sure, they're equivalent? If so, how does that help with proving it converges using the subsequence

Thanks again for your patience!
Dickfore
#13
Sep4-11, 07:08 AM
P: 3,014
If the limit of a sequence is L, what is the limit of any of its subsequences?
rudders93
#14
Sep4-11, 07:12 AM
P: 46
It will also be L. But I can see how to prove that it converges to L=0 by the limit theorems with [itex]u_{2n}[/itex] but I think you could still prove that same thing with just [itex]u_{n}[/itex]?
Dickfore
#15
Sep4-11, 07:14 AM
P: 3,014
Quote Quote by rudders93 View Post
But I can see how to prove that it converges to L=0 by the limit theorems with [itex]u_{2n}[/itex] ...
How?
rudders93
#16
Sep4-11, 07:21 AM
P: 46
Well:

[itex]u_{2n}=|a|^{2}=(\frac{1}{1+h})^{2}=\frac{1}{(1+h)^{2}}[/itex]

So: [tex]\lim_{n \rightarrow \infty}{(u_{n})^{2}}=\lim_{n \rightarrow \infty}\frac{1}{(1+h)^{2}}=0[/tex] by standard limits? Or is that incorrect, as that would work just fine using the same arguement just for [itex]u_{n}[/itex]?
Dickfore
#17
Sep4-11, 07:24 AM
P: 3,014
No, this is wrong.
rudders93
#18
Sep4-11, 07:32 AM
P: 46
Hmm, I see that yep my post was incorrect with the limits. I do not know then :(


Register to reply

Related Discussions
Boundedness: Prove that M'-m' =< M-m Calculus & Beyond Homework 3
Subsequences of Xn Calculus & Beyond Homework 2
Would it be true that if a set is bounded Calculus 1
Total boundedness in R^n Differential Geometry 7