Exploring the Pattern of a1, a2, and a3

[cybercrypt13]

Homework Statement

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

Homework 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.

The Attempt at a Solution

 

[daveb]

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

 

[cybercrypt13]

sorry,

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

 

[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 $$l=\sqrt{3+l}\Rightarrow l=\frac{1+\sqrt{13}}{2}$$

 

[cybercrypt13]

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

thanks,

glenn

 

[cybercrypt13]

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