Proving cinvergence by showing that sequence is monotone and bounded

Question

$a(1)=1, a(n+1)=\sqrt{2+a(n)}$ for all n greater than or equal to $1$

I can prove that it is a strictly increasing function and can find the limit as $2$ if i take $a(n+1)=a(n$) for large $n$ but this needs to be proved. But how to prove?

Answer 1

$$a_{n+1}=\sqrt{2+a_n}$$ 1) $a_1=1, a_2=\sqrt3$

$a_2>a_1$

If $a_n>a_{n-1}$ then $a_{n+1}>a_{n}$. Really,

$a_{n+1}^2-a_n^2=2+a_n-2-a_{n-1}=a_n-a_{n-1}>0$.

Hence, it is a strictly increasing sequence.

2) $\forall n$ $a_n<2$. Really $$a_1<2$$ Let $a_n<2$. Then $$a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=2$$ Hence, it is a bounded sequence.

3) Let $$\lim_{n \rightarrow \infty}=a$$ Then $$a=\sqrt{2+a} (a>0)$$ $$a^2=2+a$$ $$a=2$$