Series Problem

1. Jul 31, 2007

cybercrypt13

1. The problem statement, all variables and given/known data

a1 = sqrt(3), a2 = sqrt( 3 + sqrt( 3) ), a3 = sqrt( 3 + sqrt( 3 + sqrt( 3) ) )

2. Relevant equations

Notice that each term is inside the sqrt of the previous term. I have no idea how to lay something like this out. Any help would be greatly appreciated.

3. The attempt at a solution

2. Jul 31, 2007

daveb

What are you trying to do? Find an expression for the nth term? Find the sum of the series? What?

3. Jul 31, 2007

cybercrypt13

sorry,

it says find a recursive formula for a n+1 in terms of an. and find its limit

4. Jul 31, 2007

red_dog

$$a_{n+1}=\sqrt{3+a_n}$$

We have $$a_1<a_2$$
Suppose $$a_{n-1}<a_n$$.
Then $$a_n-a_{n+1}=\sqrt{3+a_{n-1}}-\sqrt{3+a_n}<0\Rightarrow a_n<a_{n+1}$$, so $$(a_n)_{n\geq 1}$$ is crescent.
We'll prove that $$\displaystyle a_n<\frac{1+\sqrt{13}}{2}$$
$$a_1<\sqrt{1+\sqrt{13}}{2}$$.
Suppose that $$a_n<\frac{1+\sqrt{13}}{2}$$.
Then $$a_{n+1}=\sqrt{3+a_n}<\sqrt{3+\frac{1+\sqrt{13}}{2}}=\frac{1+\sqrt{13}}{2}$$.
So the sequence is convergent. Let $$l=\lim_{n\to\infty}a_n$$.
Then $$l=\sqrt{3+l}\Rightarrow l=\frac{1+\sqrt{13}}{2}$$

5. Jul 31, 2007

cybercrypt13

Thanks but I'm more confused now that I thought I was before. Where did your sqrt(13) come from?

thanks,

glenn

6. Jul 31, 2007

cybercrypt13

Nevermind, I understand now. Thanks for the help...