How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ Question:
Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}&gt;0$, 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 $\displaystyle\lim_{n\to\infty}a_{n}=x&gt;0$;
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!
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 $x$.
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.