# Prove that the sequence converges and find its limit

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 ??

shmoe
Homework Helper
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).

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...

Galileo
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Maybe you could write out the first few terms, that might hint towards an upper bound.

shmoe
Homework Helper
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, $$a_n=3$$. Then what's $$a_{n+1}$$? 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.

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dextercioby
Homework Helper
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...

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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 = $$\sqrt{2-\sqrt{2+\sqrt{2+++}}}$$

So that we get sin^2 + cos^2 = 1.

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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.