Show that the sequence $\sqrt{2}, \ \sqrt{2+\sqrt{2}}, \ \sqrt{2+\sqrt{2+\sqrt{2+}}}...$ converges and find its limit. Setting $a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\ $ I get that $$a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.$$ Clearly all numbers in the sequence are positive and we see that $a_n<a_{n+1} \ \forall \ n$ which implies that the sequence is strictly increasing. To show that it's convergent, I need to show that it's bounded above.
We can use the help-function $f(x)=\sqrt{2+x}$ such that $a_{n+1}=f(a_n).$ But since $$f'(x)=\frac{1}{2\sqrt{2+x}}=0\Leftrightarrow\text{no real solutions,}$$
$f(x)$ never flattens out or decreases, so it can't be convergent?
 A: I'm not sure why solving $f'(x)=0$ should be significant. 
You can instead think like this. Let's do a little exploring.
If $a_n$ converges to a limit $a$, then $f(a_n)$ should converge to $f(a)$ (by continuity of $f$). 
But then $f(a_n)=a_{n+1}$ should also converge to $a$. 
So $f(a)=a$, which you can solve, for a solution that is $> a_1$.
Having the actual solution for $a$ in your hands, you can then attempt to prove by induction that $a_n \le a$ for all $n$. 
From that you conclude that $a_n$ converges. And from that, together with continuity of $f$ and uniqueness of the solution $> a_1$, you can conclude that $a_n$ converges to $a$.
A: Your sequence is an increasing sequence of elements of $\left[\sqrt2,2\right)$. Since it is increasing and bounded, it converges. And it must converge to an $x$ such that $f(x)=x$. It is easy to see that such $x$ must be $2$.
A: Show, by induction that $a_n < 2$. i.e.


*

*$a_1=\sqrt{2} < 2$

*let's assume $0<a_n < 2 \Rightarrow 2<2+a_n < 4 \Rightarrow 0<a_{n+1}=\sqrt{2+a_n} < 2$


So, you have an increasing and bounded above sequence.

The sequence is increasing because 


*

*$f'(x)>0, \forall x\geq0$, i.e. the help-function is increasing

*then $a_1 < a_2 \Rightarrow f(a_1)\leq f(a_2)$ which is $a_2\leq a_3$ and by induction $a_n \leq a_{n+1}$.

