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

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- Jun 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 $\displaystyle f(x) = (2+x)^{1/2}$. Then $\displaystyle f'(x) = \dfrac{1}{2(2+x)^{1/2}}>0$ for all $\displaystyle -2 < x$. So, $\displaystyle f(x)$ is increasing showing that $\displaystyle a_n>a_{n-1}$. If the limit exists, then it must satisfy $\displaystyle x = f(x)$. So, $\displaystyle x = (2+x)^{1/2}$ implies $\displaystyle x^2-x-2 = (x-2)(x+1) = 0$, so $\displaystyle x = 2$ or $\displaystyle x = -1$. Since $\displaystyle f(x)$ is increasing and $\displaystyle f(1) = \sqrt{3}>1$, so you know $\displaystyle \lim_{n \to \infty} a_n \ge 1$, so it must be $\displaystyle \lim_{n \to \infty} a_n = 2$.

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

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- Jun 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 "$\displaystyle a_{n+1}> a_n$ for any integer n".

We are given that $\displaystyle a_1= 1$ and $\displaystyle a_2= \sqrt{2+ a_1}= \sqrt{2+ 1}= \sqrt{3}> 1$ so the statement is true for n= 1.

Assume that, for some integer, k, the statement is true: $\displaystyle a_{k+1}> a_k$. Then $\displaystyle 2+ a_{k+1}> 2+ a_k$ and $\displaystyle \sqrt{2+ a_{k+1}}> \sqrt{2+ a_k}$. That is, $\displaystyle a_{k+2}= \sqrt{2+ a_{k+1}}> \sqrt{2+ a_k}= a_{k+1}$ 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 $\displaystyle a_{n+1}= \sqrt{2+ a_n}$ we have $\displaystyle A= \sqrt{2+ A}$ or $\displaystyle A^2= 2+ A$. $\displaystyle A^2- A= 2$. $\displaystyle A^2- A+ \frac{1}{4}= (A- \frac{1}{2})^2= 2+ \frac{1}{4}= \frac{9}{4}$.

So $\displaystyle A- \frac{1}{2}= \pm\frac{3}{2}$ and $\displaystyle A= \frac{4}{2}= 2$. I have taken the "+" because $\displaystyle a_1= 1$ 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 "$\displaystyle a_n< 2$ for all n".

$\displaystyle a_1= 1< 2$ so the statement is true for n= 1.

Assume that $\displaystyle a_k< 2$ for some integer k. Then $\displaystyle a_{k+1}= \sqrt{2+ a_k}< \sqrt{2+ 2}= \sqrt{4}= 2$ and we are done. The sequence is increasing and has an upper bound, therefore it converges.

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

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

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

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

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## Re: Help

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

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