A sequence is defined by a1 = 1 and an = (2 + an-1) ^ (1/2) for n ≥ 2 Show that the sequence is increasing. Find the limit?

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- June 27th 2014, 04:07 PM #1

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- June 27th 2014, 07:55 PM #2

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## Re: Help

If you are not going to bother learning LaTeX, at least try to use the _ symbol to denote subscripts:

If you wanted to type it into LaTeX, here is the code: [tex]a_1=1[/tex] and [tex]a_n = (2+a_{n-1})^{1/2}[/tex]

Anyway, let . Then for all . So, is increasing showing that . If the limit exists, then it must satisfy . So, implies , so or . Since is increasing and , so you know , so it must be .

- June 28th 2014, 04:58 AM #3

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- June 28th 2014, 05:36 AM #4

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## Re: Help

Personally, I wouldn't use Calculus for a problem on sequences (although that is valid).

Instead, use induction to prove " for any integer n".

We are given that and so the statement is true for n= 1.

Assume that, for some integer, k, the statement is true: . Then and . That is, so the statement is true for k+1 as well. By induction, then, the statement is true for all n.

To find the limit as n goes to infinity, first assume the limit exist- call it "A". Then applying the limit, as n goes to infinity to both sides of we have or . . .

So and . I have taken the "+" because and the sequence is increasing- the limit cannot be 1.

That was, as I said,**assuming**the limit exists. We still need to prove that. Since the sequence is increasing we only need to prove that the sequence has an upper bound. Noticing that the sequence converged to 2, I think 2 would be a convenient upper bound. Prove by induction that " for all n".

so the statement is true for n= 1.

Assume that for some integer k. Then and we are done. The sequence is increasing and has an upper bound, therefore it converges.

- June 28th 2014, 05:47 AM #5

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- June 28th 2014, 06:36 AM #6

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- June 28th 2014, 07:01 AM #7

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- June 28th 2014, 02:45 PM #8

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## Re: Help

- June 29th 2014, 07:15 AM #9

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