Find whether the sequence is convergent . Find whether the sequence $(a_n)$ given by $a_{n+1}= \sqrt{a_n}+\sqrt{a_{n-1}}$, where $a_1=1$ and $a_2=2$, is convergent.
So , $a_{n+1}-a_{n}= \sqrt{a_n} + \sqrt{a_{n-1}} -a_n \implies         \sqrt{a_n}(1-\sqrt{a_n})+ \sqrt{a_{n-1}}.$
Now I assumed the sequence is $>1$ and I showed it by induction  then ,
$a_{n+1}-a_{n} < \sqrt{a_{n-1}}$.Any help from here ?
 A: We show that the sequence is convergent by showing that it's bounded and monotone.

Claim 1. $2 \le a_n \le 4$ for all $n \ge 2.$
Proof. We prove this via induction. For $n = 2, 3$, it is manually verified.
Let $P(n)$ denote the statement "$2 \le a_n \le 4$".
Assume that $n \ge 4$ and that $P(k)$ is true for all $2 \le k \le n-1$. We prove that $P(n)$ is true.
By hypothesis, we have
\begin{align}
a_n &= \sqrt{a_{n-1}} + \sqrt{a_{n-2}}\\
&\ge \sqrt{2} + \sqrt{2} = 2\sqrt{2}\\
&\ge2.
\end{align}
Similarly, we have
\begin{align}
a_n &= \sqrt{a_{n-1}} + \sqrt{a_{n-2}}\\
&\le \sqrt{4} + \sqrt{4}\\
&=4.
\end{align}
This proves the statement.

Claim 2. $a_n \le a_{n+1}$ for all $n \ge 1$.
Proof. Let $P(n)$ denote the statement "$a_n \le a_{n+1}$".
$P(n)$ can be manually verified for $n = 1, 2, 3.$
Assume that $n \ge 4$ and that $P(k)$ is true for all $1 \le k \le n-1$. We prove that $P(n)$ is true.
Using the hypothesis, we see that $$a_n \ge a_{n-1} \ge a_{n-2}.$$
(Note that $n \ge 3$, so all these terms are defined.)
By the previous claim, we also see that all the terms are positive and thus, we can conclude
$$\sqrt{a_n} \ge \sqrt{a_{n-2}}. \quad (*)$$
Now, we have
\begin{align}
a_{n+1} - a_n &= \sqrt{a_n} + \sqrt{a_{n-1}} - \sqrt{a_{n-1}} - \sqrt{a_{n-2}}\\
&= \sqrt{a_n} - \sqrt{a_{n-2}}\\
&\ge 0.
\end{align}
The last inequality followed using $(*)$.
This proves this claim as well.

By Claim 1, the sequence is bounded and by Claim 2, the sequence is monotone. Thus, the sequence converges. (The value of the limit can be found to be $4$.)
