Find and Prove the supremum of this set $X_1=2^{1/2}$ and $X_{n+1}=(2+X_n)^{1/2}$ for each $n\ge 1$.
Find the supremum of the set $\{ X_n \}$ and prove it.
I figured that the supremum should be $2$ by just examining the elements of the set and from the fact that each element is less than $2$.
Is there a proper way to find and prove the supremum of this set? 
 A: *

*Prove that $X_n \leq 2$ for all $n$. This is immediate from induction.

*Prove that $X_n$ is monotone.

*Hence, by monotone sequence theorem, $\lim_{n} X_n = L$ exists.

*This gives us $L = \sqrt{2+L} \implies L = 2$.

*Conclude that $2$ is the supremum.

A: We can reduce finding the supremum of $x_n$ to finding the limit. This follows because this is an increasing sequence, given the initial condition.
Firstly, as you mentioned, $x_n < 2$ for all $n$. This can be demonstrated by induction. Moreover, all the terms are positive.
Now we would like $$x_{n+1} > x_n$$ this gives us the equation: $$0 \ge (x_n +1)(x_n -2).$$ This is satisfied provided that $$-1 < x_n < 2,$$ which we have already established.
We need to find the limit of this sequence. It is a bounded monotonic sequence, so there is a limit. Let $L$ be this limit. We can find an algebraic equation that this limit must satisfy:
$$L=\lim_{n\to\infty} x_n = \lim_{n\to\infty} (2+x_n)^{1/2} = (2+L)^{1/2}$$ the last step follows by continuity of $(2+x)^{1/2}.$ Finally this means that $L$ satisfies: $$L^2 - L - 2 = 0$$ and hence $L=-1$ or $L=2$. $L\neq -1$ since it is the limit of a sequence of positive real numbers, and $[0,\infty)$ is closed. Therefore $L=2$. Since this sequence is increasing to $L$, that is our supremum.
