Sequence of real numbers | Proof of convergence

Click For Summary

Homework Help Overview

The discussion revolves around proving the convergence of a sequence of real numbers, specifically focusing on parts a, b, and c of the problem. The original poster expresses uncertainty in formalizing their intuition regarding the behavior of square roots of numbers between 0 and 1 and their implications for convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the sequence's behavior when values are between 0 and 1, questioning how to formally prove convergence and establish necessary conditions for the sequence. There are discussions about selecting appropriate values for m and N to demonstrate convergence.

Discussion Status

Some participants have offered hints and partial insights into the proof structure, particularly regarding the relationships between terms in the sequence. However, there remains a lack of consensus on the formal setup for parts a and c, with ongoing inquiries about the iteration process and the conditions needed for convergence.

Contextual Notes

Participants note the challenge of proving convergence under specific conditions, particularly when initial terms are less than 1. There is also mention of imposed homework rules that may limit the extent of assistance provided.

kingwinner
Messages
1,266
Reaction score
0

Homework Statement


ra12.JPG


Homework Equations


N/A

The Attempt at a Solution


Assuming the truth of part a, I proved part b.
But now I have no idea how to prove parts a & c.
Part a seems true intuitively. The sqaure root of a number between 0 and 1 is will be larger than that number, and if we take more and more square roots, it will get close to 1, and then if we add two numbers that are close to 1, it must be ≥1.
But how can we write a FORMAL proof of it? How can we find/construct N and demonstrate exactly that there exists an N such that n≥N => an≥1?

Can someone help me, please?
Any help is much appreciated!



[note: also under discussion in Math Links forum]
 
Last edited:
Physics news on Phys.org
For part a, I have some more idea...if a(n) and a(n+1) are positive then we can surely pick an m such that 1/(2m) is less than them both: just select a sufficiently large m, e.g. select m such that 2m ≥ 1/min{a(n),a(n+1)}.

But how can I find N such that n≥N implies a(n)≥1 ?

Any help is much appreciated!
 
Since you seem to have spent quite some time on the problem i will try to give you some hints, i hope i don't get another warning from pf moderators for offering too much help(solving >90% of the problem for the op) :(

This might not be the nicest proof in the world, but i think it works.

As you have figured out the main problem is when 0<a_1<1 and 0<a_o<1. So we will deal with this case only, since others are trivial.
Let:
[tex]0<a_0<1,0<a_1<1[/tex]
then:
[tex]a_o<\sqrt{a_o}...and...a_1<\sqrt{a_1}[/tex]

adding these together we get:
[tex]a_3=\sqrt{a_0}+\sqrt{a_1}>a_0+a_1[/tex]

If we continue in this fashion, after n-2 steps we would get something like:

[tex]a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_n}>a_0+a_1+...+a_n>n*min\{a_0,a_1,...,a_n\}=n*a[/tex]

So, now you see that if we let n>N=1/a we get our result. where a=min{a_o,...,a_n}

cheers!
 
Last edited:
Thanks.
Using part a, I proved part b.

Any hints about part c?
 
For part c, I'm stuck with using the hint.

From part b, en+2 ≤ (en+1 + en)/3 for n≥N.

In part c, I think I need to end up proving something like
en ≤ (2/3)some exponent involving n max(eN,eN+1)
If the RHS tends to 0, then by squeeze theorem en->0.

But I'm not sure how to SET UP the iteration. From part b, en+2 ≤ (en+1 + en)/3 for n≥N. Is it also true that en+1 ≤ (en + en-1)/3? Why or why not?
And how can I find that "some exponent involving n"?

May someone help me, please?
Thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K