Prove Limit of Sequence: a_n ≥ 1 for n ≥ N

In summary, the conversation discusses the sequence an+2 = sqrt(an+1) + sqrt(an) for positive real numbers a0 and a1. The participants discuss how to prove that for all n greater than or equal to some value N, an is greater than or equal to 1. They also discuss the sequence converging and provide hints for proving convergence.
  • #1
HLUM
3
0
1. Let a0 and a1 be positive real numbers, and set an+2 = sqrt(an+1) + sqrt(an) for n [tex]\geq[/tex] 0.
(a) Show that there is N such that for all n [tex]\geq[/tex] N, an [tex]\geq[/tex] 1.
(b) Let en = |an −4|. Show that en+2 [tex]\leq[/tex](en+1 +en)/3 for n[tex]\geq[/tex] N.
(c) Prove that this sequence converges.




Can someone please give me some hints to start with a)? Thank you in advanced.
 
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  • #2
If [itex]0<x<1[/itex], is [itex]\sqrt{x}[/itex] bigger or smaller than [itex]x[/itex]?
 
  • #3
if 0 < an < 1 then an x an < an ==>
an < sqrt(an)

So you mean i should prove part a by contradiction...
 
  • #4
Perhaps. I don't know. But you can definitely show now that [itex]a_{n+2} > a_{n+1}+a_n[/itex] if [itex]a_n, a_{n+1} < 1[/itex], which may be useful.
 
  • #5
So
a) Assume that for all n [tex]\geq[/tex] 0, 0 < an < 1
then sqrt(an) > an

ie, an+2 > an+1 + an
an is increasing sequence

I don't know how to show the contradiction here, but there is no assumption of increasing sequence if you choose a0 to start with

---> there is N st aN [tex]\geq[/tex] 1
Assume for all n [tex]\geq[/tex] N, an+1 = an + an - 1 > 1

therefore, by induction it is true for all n [tex]\geq[/tex] N, an [tex]\geq[/tex] 1
 
  • #6
Edit: My previous post was so incomprehensible that I don't think that it would have been much help. I'll post again later if I can get my thoughts together, but anyway, good luck!
 
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FAQ: Prove Limit of Sequence: a_n ≥ 1 for n ≥ N

What does the notation "a_n ≥ 1" mean in the context of a limit of a sequence?

The notation "a_n ≥ 1" means that the terms of the sequence starting at index n and beyond are greater than or equal to 1. In other words, the sequence is bounded below by 1.

How is the limit of a sequence defined?

The limit of a sequence is defined as the value that the terms of the sequence approach as the index approaches infinity. In other words, it is the value that the terms of the sequence get closer and closer to, but may not necessarily reach.

Why is it important to prove the limit of a sequence?

Proving the limit of a sequence is important because it allows us to determine the behavior of the sequence as the index approaches infinity. This can help us understand the long-term trend of a sequence and make predictions about its values.

How is the limit of a sequence with a lower bound of 1 proven?

The limit of a sequence with a lower bound of 1 can be proven by using the definition of a limit. This involves showing that for any arbitrarily small positive number, there exists an index N such that all terms of the sequence beyond N are within that small distance from the limit of the sequence, which in this case is 1.

Can a sequence have a lower bound of 1 and still approach a limit other than 1?

Yes, a sequence can have a lower bound of 1 and still approach a limit other than 1. For example, the sequence 1, 1/2, 1/3, 1/4, ... has a lower bound of 1 but approaches a limit of 0 as the index approaches infinity.

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