TITLE: How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$
QUESTION [11 upvotes]: Question:
Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}>0$, and for all $n \in \mathbb{N}$ we have:

$$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$

Prove that :

$\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.

My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x>0$;
then we have
$$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$
But I can't see how to prove the limit of $\{a_{n}\}$ exists.
Thank you for you help!

REPLY [2 votes]: and I have consider other solution
if $a_{0}>0,a_{1}>0,$,and $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$,then $a_{n}\to 4$
pf: if $0<a_{0}\le a_{1}\le 1$,then $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{1}}\ge a_{1}$$
we note
$$a_{n+2}-a_{n+1}=(\sqrt{a_{n+1}}+\sqrt{a_{n}})-(\sqrt{a_{n}}+\sqrt{a_{n-1}})=(\sqrt{a_{n+1}}-\sqrt{a_{n}})+(\sqrt{a_{n}}-\sqrt{a_{n-1}})\ge 0$$
so $\{a_{n}\}$is  Monotone increasing,
$$a_{n}\le 4,\mbox{since } a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}\le 2+2=4$$
(2):if $0<a_{1}\le a_{0}\le 1$,then 
$$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}\ge a_{0}$$
and
$$a_{n+2}-a_{n+1}=\sqrt{a_{n+1}}-\sqrt{a_{n-1}}\ge 0$$.the simalar we have $\{a_{n}\}$ is increasing and $a_{n}\le 4$.
(3):if$a_{0}>1$(or $a_{1}>1$),then we have
$$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}>1,\Longrightarrow a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}>1$$
let $x_{n}=|a_{n}-4|$,then we have
$$x_{n+1}=|a_{n+1}-4|<|\sqrt{a_{n}}-2|+|\sqrt{a_{n-1}}-2|=\dfrac{|a_{n}-4|}{\sqrt{a_{n}}+2}+\dfrac{|a_{n-1}-4|}{\sqrt{a_{n-1}}+2}<\dfrac{1}{3}x_{n}+\dfrac{1}{3}x_{n-1}$$
so
$$x_{n+1}-\dfrac{1-\sqrt{13}}{6}x_{n}<\dfrac{1+\sqrt{13}}{6}(x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1})<\cdots<\left(\dfrac{1+\sqrt{13}}{6}\right)^n\left(x_{1}-\dfrac{1-\sqrt{13}}{6}x_{0}\right)\to 0,n\to\infty$$
so
$$0<x_{n}<x_{n}+\dfrac{\sqrt{13}-1}{6}x_{n-1}=x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1}\to 0$$

REPLY [2 votes]: First note that since $a_n\gt0$,
$$
\begin{align}
a_n-4
&=(\sqrt{a_{n-1}}-2)+(\sqrt{a_{n-2}}-2)\\
&=\frac{a_{n-1}-4}{\sqrt{a_{n-1}}+2}+\frac{a_{n-2}-4}{\sqrt{a_{n-2}}+2}\tag{1}
\end{align}
$$
Furthermore,
$$
a_n\gt\sqrt{a_{n-1}}\tag{2}
$$
Therefore, since $a_0\gt0$, there is an $n_0$ so that for $n\ge n_0$, $a_n\gt\frac14$. Thus, for $n\ge n_0+2$, $(1)$ says
$$
|a_n-4|\lt\tfrac25\big(|a_{n-1}-4|+|a_{n-2}-4|\big)\tag{3}
$$
Since both roots of $x^2-\tfrac25x-\tfrac25=0$ have absolute value less than $1$, by comparison with any positive sequence so that $b_n=\tfrac25(b_{n-1}+b_{n-2})$, $(3)$ implies that
$$
\lim_{n\to\infty}a_n=4\tag{4}
$$