How to prove {$a_n$ } is increasing where $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{ 2+ a_n}$

Question

I already found out that this sequence is bounded above and $a_n <2 \forall n \in \mathbb Z_+ $

I think I'm missing a point as I can't think of a way to prove that the sequence is increasing.

Answer 1

$a_{n+1} = \sqrt{2 + a_n} \geq \sqrt {a_n +a_n} \geq \sqrt{2a_n} \geq \sqrt{a_n^2}$.

Answer 2

Hint:You need to prove $$a_{n+1}-a_n>0$$

Answer 3

Assume $\exists n$ such that $a_{n+1}-a_n\leq0$.

Then:

$$\sqrt{2+a_n}-a_n\leq0\\ a_n^2-a_n-2=(a_n-2)(a_n+1)\geq 0$$

So $a_n$ is increasing for all $n$ such that $-1<a_n<2$, where the inequality does not hold, and thus provides a contradiction to our assumption of $n$ being non-decreasing at a point.

Answer 4

Another way to see that $a_n < 2$ is the function $f(x) = \sqrt{2+x}$ is increasing, and further: $a_{n+1} - a_n = \sqrt{2+a_n} - a_n > \sqrt{2a_n} - a_n = \sqrt{a_n}\cdot \left(\sqrt{2} - \sqrt{a_n}\right) > 0$