$\displaystyle a_{2}>a_{1}$

because $\displaystyle \sqrt{2}<2$

$\displaystyle a_3>a_2$

because

$\displaystyle 2>$$\displaystyle \sqrt{2 +\sqrt {2}}$

for $\displaystyle a_4>a_{3}$

$\displaystyle 2>\sqrt{2+\sqrt{2 +\sqrt{2}}}$

Therefor it is true for all $\displaystyle n \in N$

<..... you still need to check until GOD KNOWS WHERE.

Now I want to find where it converges too.

Since the $\displaystyle \lim [a_{n+1}] = \lim [a_n]$

=some real number <.... we can't say that until we are sure that the limit exists.

we have

$\displaystyle a = \sqrt{2+a}$

then

$\displaystyle a^2 = 2+a$

then

$\displaystyle a^2 -a-2=0$

then

$\displaystyle a = 0.5 \pm \sqrt{\frac{1}{2}+2}$

$\displaystyle a = 0.5 \pm 1.5$

$\displaystyle a_1 = -1$

and

$\displaystyle a_2 = 2$

Since a can not be negative cause$\displaystyle \sqrt{2} <a_n<a_{n+1}\leq {2}$

Then it must converge towards $\displaystyle \sqrt {2+2} = 2$

Does this make sence or is my prof going to \Rightarrow