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

Question

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?

Answer 1

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

Answer 2

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.