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.
 A: 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 $a_{n-1}^2-2<0$
A: Hint
$$a_n^2-2 < 0 \iff 1+\frac{1}{2}a_{n-1}^2 -2 < 0 \iff a_{n-1}^2 -2 <0$$
A: Hint: study the function $$f(x)=\sqrt{1+\frac{1}{2}x^2}$$ and its derivative.
You can easily find a counterexample.
However the assertion is true for $a_{1} \in (-\sqrt{2};\sqrt{2})$
A: Assume $a_0\in{\Bbb R}$. When $n\geq1$ all $a_n$ are $\geq0$. Therefore we may as well study the sequence $b_n:=a_n^2$ $\>(n\geq1)$ with
$$b_{n+1}=1+{1\over2} b_n\qquad(n\geq1)\ .\tag{1}$$
The "master theorem" provides the following general solution of $(1)$:
$$b_n=2+ c\>2^{-n}\qquad(n\geq1),\qquad c\in{\mathbb R}\ .$$
Here $c$ depends on the initial value $b_1=a_1^2$. It turns out that $\lim_{n\to\infty} b_n=2$ whatever $c$, and this proves $\lim_{n\to\infty} a_n=\sqrt{2}$, whatever $a_0$.
A: You should square both of two sides.
Then you get 
$$a_{n+1}^2 = \frac12a_n^2+1$$
then you can think $a_n^2$ as $A_n$. And you can have a sequence
$$A_n^2-2 = \left(\frac12\right)^{n-1}\left(A_1^2-2\right)$$
unless $A_1$ is bigger than $2$, the thing that you want to prove is valid.
A: 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 $<$.
