Find the limit of the sequence

A sequence is defined by a_{1}=2, a_{n+1}=(2a_{n}+1)^{1/2 }I have proven by induction that a_{n} is bounded above by 1+(2)^{1/2}, and that a_{n} 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?

Re: Find the limit of the sequence

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**Shakarri** A sequence is defined by a_{1}=2, a_{n+1}=(2a_{n}+1)^{1/2 }I have proven by induction that a_{n} is bounded above by 1+(2)^{1/2}, and that a_{n} 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 $\displaystyle L=\sqrt{2L+1}~?$

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?

Re: Find the limit of the sequence

Quote:

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

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 $\displaystyle f(x)=\sqrt{2x+1}$ maps X to X, and (2) prove that f(x) is a contraction on X. Concerning (1), you can choose $\displaystyle X=[2,1+\sqrt{2}]$. From the facts that f(x) increases on X and $\displaystyle f(x)<1+\sqrt{2}$ for $\displaystyle x\in X$ it follows that $\displaystyle f(x):X\to X$. Concerning (2), it is sufficient to show that for some constant q, $\displaystyle |f'(x)|\le q<1$ for all $\displaystyle x\in X$. 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 $\displaystyle a_n$. So, to find the limit of f(x) it is sufficient to solve x = f(x).

Re: Find the limit of the sequence