Proving convergence of a series and then finding limit I need to show that $\sqrt{2}$, $\sqrt{2+\sqrt{2}}$, $\sqrt{2+\sqrt{2+\sqrt{2}}},\ldots$
converges and find the limit. 
I started by defining the sequence by $x_1=\sqrt{2}$ and then $x_{n+1}=\sqrt{2+x_n}$. Then I proved by induction that the sequence is increasing and that it is bounded. Then I use the Monotone Convergence Theorem to prove that it converges.
Now, I claim that the limit of the sequence is $2$. So I need to show that for all $\epsilon>0$, there exists $N$ such that for all $n\geq N$ we have $|x_n-2| < \epsilon$. This is kind of where I'm stuck. I don't know how to proceed to prove the convergence to $2$. 

Let $f(x)=\sqrt{2+x}$, $x\geq 0$. Note that $f(x)-f(y)=\sqrt{2+x}-\sqrt{2+y}=\frac{x-y}{\sqrt{2+x}+\sqrt{2+y}}$. So $|f(x)-f(y)\leq \frac{1}{2\sqrt{2}}|x-y|$.Then $f$ in an contraction and there is $\lim_n f^n(x)$, independently of $x$. If $x=x_n$ you have $f(x)=x_{n+1}$. Given $x=x_0$ you have $x_n=f^n(x_0)$. This implies that $x_n$ is convergent.Your limit satisfy $L^2=2+L$. 
