# Convergence and limit of a recursive sequence $x_{n+1}=\sqrt{p+x_n}$ [duplicate]

Let $p>0$ and suppose that the sequence $\{x_n\}$ is defined recursive as $$x_1 = \sqrt{p}, \quad x_{n+1} = \sqrt{p + x_n},$$ for all $n \in \mathbb{N}$.

How can I show that $x_n$ converges, and find its limit?

I can see that one upper bound is obviously $1 + 2\sqrt{p}$. Can I use this fact somehow?

## marked as duplicate by rtybase, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 6 '18 at 17:50

• So you are telling me that $(x_n) = (\sqrt{p})$ for every $n$? Or do you mean $x_0 = \sqrt{p}$. If it is the latter case, show that the sequence is bounded and monotone. You will have to use induction somewhere. – IAmNoOne Feb 21 '14 at 19:22

Plan. We are going to show that $\{x_n\}_{n\in\mathbb N}$ is an increasing and upper bounded sequence, which shall imply that $\{x_n\}_{n\in\mathbb N}$ converges. Next we shall find the limit using the recursion relation.

A. First, $\{x_n\}_{n\in\mathbb N}$ is an increasing sequence. This can be shown inductively:

For $k=1$: $x_1=\sqrt{p+x_0}>\sqrt{p}=x_0$.

If $x_{k-1}<x_{k}$, then $\sqrt{p+x_{k-1}}<\sqrt{p+x_{k}}$, and thus $x_k<x_{k+1}$.

B. Next, we observe that $\{x_n\}$ is upper bounded by $1+2\sqrt{p}$. Indeed $x_0=\sqrt{p}<1+2\sqrt{p}$. Assume that $x_k<1+2\sqrt{p}$. Then $$x_{k+1}=\sqrt{p+x_k}<\sqrt{p+1+2\sqrt{p}}=1+\sqrt{p}<1+2\sqrt{p}.$$

But an increasing and upper bounded sequence converges.

C. Let $x_n\to x$. Clearly, $x\ge x_0=\sqrt{p}>0$.

Then $x_{n+1}=\sqrt{x_n+p}\to\sqrt{p+x}$.

But $\lim x_{n+1}=x$, and hence $$x=\sqrt{p+x}$$ or $$x^2=x+p,$$ or $$x=\frac{1\pm \sqrt{1+4p}}{2}.$$ We choose the one with the plus sign since the other one is negative.

Hence $$x=\frac{1+ \sqrt{1+4p}}{2}.$$

• Next, we observe that {xn} is upper bounded by 1+2p√.How is this found guessed? – GorillaApe Sep 5 '14 at 17:42
• @Yiorgos S. Symrlis I would also like to know how you arrived at the "obvious" upper bound? – user389056 Dec 1 '16 at 1:31
• @user389056 It is clearly larger than the limit. First, we find the limit and then we observe that the sequence is increasing, and finally we find an upper bound. There are many other choices of course, for example $1+2p$. – Yiorgos S. Smyrlis Dec 2 '16 at 7:03

If I understand, you should get something like $$\sqrt{p + \sqrt{p + \sqrt{p + \ldots}}}$$

Note that if the limit is $x$, then $x^2 = p + x$ which you can solve directly.

• Note that this holds IF the limit exists. You can prove that the limits does exist using either induction or the fixpoint theorem. – EricAm Feb 21 '14 at 19:32
• @EricAm yes, hence the text in the answer "if the limit is $x$" – gt6989b Feb 21 '14 at 19:34
• Yes, I just wanted to clarify that to the OP since it could easily be overlooked. – EricAm Feb 21 '14 at 19:35