# Thread: Find the limit of the sequence

1. ## Find the limit of the sequence

A sequence is defined by a1=2, an+1=(2an+1)​1/2

I have proven by induction that an is bounded above by 1+(2)1/2, and that an is monotone increasing. By the monotone convergence theorem I can deduce that the sequence converges but I cannot figure out how to prove what it converges to. I know that it converges to 1+(2)1/2 but I don't know how to show that. Could someone help me finish this question?

2. ## Re: Find the limit of the sequence

Originally Posted by Shakarri
A sequence is defined by a1=2, an+1=(2an+1)​1/2
I have proven by induction that an is bounded above by 1+(2)1/2, and that an is monotone increasing. By the monotone convergence theorem I can deduce that the sequence converges but I cannot figure out how to prove what it converges to. I know that it converges to 1+(2)2 but I don't know how to show that. Could someone help me finish this question?
Can you solve $L=\sqrt{2L+1}~?$

3. ## Re: Find the limit of the sequence

Actually that's what I did to find the limit but I believe this doesn't work in the general case and so cannot be considered a proof. Are you sure it is a valid proof?

4. ## Re: Find the limit of the sequence

Originally Posted by Shakarri
Actually that's what I did to find the limit but I believe this doesn't work in the general case and so cannot be considered a proof. Are you sure it is a valid proof?
Validity depends upon what you are trying to do.

I understood you to be asking how to find the limit, not to prove anything.

Monotone, increasing, bounded sequences converge to its least upper bound.

I think it tedious to prove that $1+\sqrt2$ is the LUB of that sequence. But it can be done (I will not do it).

5. ## Re: Find the limit of the sequence

One way to prove that the limit is the fixpoint is to use the Banach fixed-point theorem. To apply it, you need to do two things: (1) find a closed segment X such that $f(x)=\sqrt{2x+1}$ maps X to X, and (2) prove that f(x) is a contraction on X. Concerning (1), you can choose $X=[2,1+\sqrt{2}]$. From the facts that f(x) increases on X and $f(x)<1+\sqrt{2}$ for $x\in X$ it follows that $f(x):X\to X$. Concerning (2), it is sufficient to show that for some constant q, $|f'(x)|\le q<1$ for all $x\in X$. Then the fact that f(x) is a contraction follows by the the mean value theorem.

The Banach fixed-point theorem says that the fixpoint of f(x) is unique and is the limit of $a_n$. So, to find the limit of f(x) it is sufficient to solve x = f(x).

Весело -)