Sequence $a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$ I am trying but cant figure out anything.
$a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$
I am trying to proove that $a_n^2-2<0$.
Getting $$a_{n+1} -a_n=\dots=\frac{2-a_n^2}{2\left(\sqrt{1+\frac{1}{2}a_n^2} +a_n\right)}$$
Then I have no clue how to proove it since I am not given $a_1$.Induction doesnt seem to work nor any contradiction.

Hints:
1) $\dfrac{1}{2}a_{n}^2-a_{n+1}^2=\dfrac{1}{2}a_{n+1}^2-a_{n+2}^2=\ldots=\dfrac{1}{2}a_{n+i}^2-a_{n+i+1}^2=-1$
Therefore...
2) Proving that $a_n^2-2<0$ is the same as proving that $a_{n+1}^2-2<0$, and
3) Proving that $a_n^2-2<0$ is the same as proving that $a_{n-1}^2-2<0$
