1. ## Help

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?

2. ## 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: $$a_1=1$$ and $$a_n = (2+a_{n-1})^{1/2}$$

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

3. ## Re: Help

Tanks alot for the solution and fot the tip how to write the indeks, but you did mean f(2) in the solution not f(1)...

Tanks again

Best regards
Haytham

4. ## Re: Help

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

Instead, use induction to prove " $a_{n+1}> a_n$ for any integer n".

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

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

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

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

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

5. ## Re: Help

Originally Posted by Haytham1111
Tanks alot for the solution and fot the tip how to write the indeks, but you did mean f(2) in the solution not f(1)...
No, he didn't. Slip Eternal's $f(x)= \sqrt{2+ x}$ is your $a_{x+1}$. $a_2= f(1)$

Tanks again

Best regards
Haytham

6. ## Re: Help

Whats about A^2 - A + (1/4) = (A-(1/2))^2

That is completely wrong

7. ## Re: Help

No, it's not. Multiply it yourself.

8. ## Re: Help

Originally Posted by Haytham1111
Whats about A^2 - A + (1/4) = (A-(1/2))^2

That is completely wrong
HallsofIvy completed the square rather than factoring or using the Quadratic Formula. Completing the square is actually the method used to prove the Quadratic Formula, so it is certainly a valid approach to solving any quadratic.

9. ## Re: Help

Ja, its right I understood it now. Thanks to you two