I came across a question which went like this:
Let $x_1:=1$ and $x_{n+1}=\sqrt{2+x_n}$ for $n\in\mathbb{N}$. Show that $x_n$ converges and find the limit.
I tried to use the Monotone Convergence Theorem. For that, I need to show that $x_n$ is a bounded sequence and that $x_n$ is a monotone.
I showed that $x_n<2$ by mathematical induction and thus, it has an upper bound. But what about the monotone part?
Intuitively, I found out that,
$$x_1=1$$
$$x_2=\sqrt{2+x_1}=\sqrt{3}>1$$
and similarly for $x_3,x_4,…$ I can show it but this is a very vague approach in my opinion to show that this sequence is a monotonically increasing. Can anyone suggest some simple method of showing this?
I was thinking of something like taking the ratio of $\frac{x_{n+1}}{x_n}$ and then showing that $\frac{x_{n+1}}{x_n}\geq1$ but I couldn't go much far with that.
I only found that this ratio is greater than 0 but that was of no help to me.
Later on, I tried to take the difference of the subsequent terms of the sequence and I got stuck at:
$$x_{n+1}^2-x_n=2$$ but it has a square term upon $x_{n+1}$.
Any hints as to how this could be shown will be very much appreciated.
P.S. I found a similar question here where it is said that this can be proved using mathematical induction. However, when I tried to use the mathematical induction but I could only find that $x_{n+1}>x_{n-1}$. What about $x_n$ then?
Best Answer
If $0 \le x_n < 2$
then $0 \le x_n^2 < 2x_n =x_n+x_n < 2+x_n < 4$
so $x_n < \sqrt{2+x_n} < 2$ and you have a monotone increasing sequence