Prove that the sequence converges and find its limit

In summary, the sequence given by the recurrence relation a_1 = 2^(1/2) and a_n = (2 + a_n-1)^1/2 is an increasing and bounded sequence. By using the "cosine trick" and induction, it can be shown that the sequence converges to a limit of 2.
  • #1
mansi
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given a recurrence relation, a_1 =2^(1/2) and a_n = (2 +a_n-1)^1/2 ...prove that the sequence converges and find its limit..
are we supposed to begin by guessing the limit and the bounds ??
 
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  • #2
Yes, try to guess the limit. You can make an educated guess though, if you assume that your sequence converges you should be able to find what it's limit must be (hint:use the recurrance relation).

Once you've got this 'guess' limit in hand, you should be able to prove convergence (hint:show it's an increasing & bounded sequence).
 
  • #3
By mere observation, it's quite clear that a_1 < a_2 <...<a_n...
so, it's an increasing sequence...
but i can't think of how we can show it's bounded...i mean,how do we use the recurrence relation?..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...
 
  • #4
Maybe you could write out the first few terms, that might hint towards an upper bound.
 
  • #5
mansi said:
By mere observation, it's quite clear that a_1 < a_2 <...<a_n...
so, it's an increasing sequence...

You don't think you need any conditions on the terms to deduce this? What if, at some point, [tex]a_n=3[/tex]. Then what's [tex]a_{n+1}[/tex]? Can you guarantee this won't hapen?

mansi said:
..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...

If you happened to find the least upper bound, then yes.
 
  • #6
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  • #7
Robert,it can't be.The #-s on the right (the square roots) are increasing,while the sinus-es are decreasing (their arguments tends to 0)...

Daniel.

EDIT:You edited,and put "cos".Now it makes sense.Both are increasing.This "cosine" trick is really elegant... :approve:
 
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  • #8
dextercioby: This "cosine" trick is really elegant.:

Wow! Thanks for the complement. I at first mistakenly put in the sin, which has this form:

2sin x = [tex]\sqrt{2-\sqrt{2+\sqrt{2+++}}}[/tex]

So that we get sin^2 + cos^2 = 1.
 
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  • #9
Induction is a pretty simple way to show that the sequence converges:
A1 = 2^(1/2) < 2
Assume An < 2
A(n+1) = (An + 2)^(1/2) < (2 + 2)^(1/2) = 2
By the principle of mathematical induction, An < 2 for all n.
 

Related to Prove that the sequence converges and find its limit

1. What is a convergent sequence?

A convergent sequence is a sequence of numbers that approaches a specific value, known as its limit, as the sequence progresses towards infinity. This means that as more and more terms are added to the sequence, the values of the terms get closer and closer to the limit value.

2. How do you prove that a sequence converges?

In order to prove that a sequence converges, one must show that for any given positive number, there exists a point in the sequence after which all the terms are within that given distance from the limit. This is known as the epsilon definition of convergence.

3. What is the limit of a convergent sequence?

The limit of a convergent sequence is the value that the terms of the sequence approach as the sequence progresses towards infinity. It is the final value that the sequence "converges" to.

4. Can a sequence have more than one limit?

No, a sequence can only have one limit. If a sequence has more than one limit, then it is not a convergent sequence.

5. What are some common methods for finding the limit of a sequence?

Some common methods for finding the limit of a sequence include using the epsilon definition of convergence, using the monotone convergence theorem, or using algebraic manipulations to simplify the sequence and identify a pattern that leads to the limit.

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