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.

If you want a proof by contradiction: If $a_{n+1}^2 - 2 \geq 0$, then $1 + {1 \over 2} a_n^2 \geq 2$, which after a little algebra is the same as $a_n^2 - 2 \geq 0$. And a direct proof is obtained by doing these steps in the opposite direction, with the $\geq$ sign replaced by $<$.
