real analysis - How do I show a sequence is bounded, use for one part of the monotone convergence theorem?

So the monotone convergence theorem states that if a sequence is bounded and monotone then it converges.

Now I am trying to prove that the sequence defined recursively as $x_1 = \sqrt 2$, $x_{n+1}=\sqrt {2x_n}$ converges and to find it's limit.

I am able to show that the sequence is monotone by determining if the ratio of $\frac{x_{n+1}}{x_n}$ is greater than and equal to or less than or equal to 1. I am even know what the limit is.

My dilemma is I am not sure how to determine if this sequence is bounded and I am not quite sure what I need to do to show that its bounded. Can anyone offer a few hints or strategies? I know that it is bounded below by $\sqrt2$; however the sequence is increasing so I need to show that there is an upperbound as well.

Answer

Hint: Use Induction. Suppose $x_n < 2$, what can you say about $x_{n+1}$?