How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ 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!
 A: Obviously, it is enough to show that
$$
\limsup a_n \leq 4 \leq \liminf a_n.
$$
To prove this, first remark that the sequence $M_n = \max\{4,a_n, a_{n-1}\}$ is non-increasing.
Indeed, the trivial lower bound $M_n \geq 4$ yields $a_{n+1} \leq 2\sqrt{M_n} \leq M_n$ ; we conclude with
$$
M_{n+1} = \max\{4, a_{n+1}, a_n\} \leq \max\{4, M_n, M_n\} = M_n.
$$
As a consequence, an upper bound for $a_n$ is $\max\{4, a_1, a_2\}$. In the same way, we can prove the lower bound $a_n \geq \min\{4, a_1, a_2\}$.
These bounds show that both $\liminf a_n$ and $\limsup a_n$ are finite and positive. Furthermore we deduce from $a_{n+1} = \sqrt{a_n} + \sqrt{a_{n-1}}$ that
$$
\liminf a_n \geq   2\sqrt{\liminf a_n},\qquad \limsup a_n \leq 2\sqrt{\limsup a_n}
$$
The conclusion follows because $x \geq 2\sqrt{x} \implies x \geq 4$ as well as $x \leq 2\sqrt{x}\implies x \leq 4$ for every positive real number $x$.
A: 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
$$0<x_{n}<x_{n}+\dfrac{\sqrt{13}-1}{6}x_{n-1}=x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1}\to 0$$
A: 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
$$
\lim_{n\to\infty}a_n=4\tag{4}
$$
A: other idea:
Now I have edit:
if $c_{0}>0,c_{1}>0.c_{n+1}=\sqrt{c_{n}}+\sqrt{c_{n-1}},n\ge 1$, then $\lim_{n\to\infty}c_{n}$ is exsit.
pf:let $$a_{0}=\min(c_{0},c_{1},4),b_{0}=\max(c_{0},c_{1},4)$$
then
$$a_{n}=2\sqrt{a_{n-1}},b_{n}=2\sqrt{b_{n-1}}$$
then $$a_{0}\le a_{1}\le\cdots\le 4,b_{0}\ge b_{1}\ge \cdots\ge 4$$
and since
$a_{0}\le c_{0},a_{0}\le c_{1}$,so
$$a_{0}\le min(c_{0},c_{1})$$
Aussmu that for $n-1$,we have
$$a_{n-1}\le\min(c_{2n-2},c_{2n-1})$$
then
$$c_{2n}=\sqrt{c_{2n-1}}+\sqrt{c_{2n-2}}\ge 2\sqrt{a_{n-1}}=a_{n}$$
$$c_{2n+1}=\sqrt{c_{2n}}+\sqrt{c_{2n-1}}\ge\sqrt{a_{n}}+\sqrt{a_{n-1}}\ge 2\sqrt{a_{n-1}}=a_{n}$$
so According to the mathematical induction
we have
$a_{n}\le \min(c_{2n},c_{2n+1})$.and simaler we have
$$b_{n}\ge \max(c_{2n},c_{2n+1})$$
so
$$a_{n}\le c_{2n}\le b_{n},a_{n}\le c_{2n+1}\le b_{n}$$
By done.
A: we have
$$\sqrt{\min\left(\dfrac{a_{n+1}}{4},\dfrac{a_{n}}{4}\right)}\le\dfrac{a_{n+2}}{4}=\dfrac{\sqrt{\dfrac{a_{n+1}}{4}}+\sqrt{\dfrac{a_{n}}{4}}}{2}\le\sqrt{\max\left(\dfrac{a_{n+1}}{4},\dfrac{a_{n}}{4}\right)}$$
so
$$\left(\min\left(\dfrac{a_{0}}{4},\dfrac{a_{1}}{4},\dfrac{a_{2}}{4}\right)\right)^{\frac{1}{2^{n+1}}}\le\dfrac{a_{n+2}}{4}\le\left(\max\left(\dfrac{a_{0}}{4},\dfrac{a_{1}}{4},\dfrac{a_{2}}{4}\right)\right)^{\frac{1}{2^{[\dfrac{n+1}{2}]}}}$$
so
$$\lim_{n\to\infty}\dfrac{a_{n+2}}{4}=1$$
so
$$\lim_{n\to\infty}a_{n+2}=4$$
I hope this methods is no problem.
