# Proof by induction that a sequence is bounded

Let $$x_1=1$$ and $$x_{n+1}=(2+x_n)^{1/2}$$ for $$n \in \mathbb{N}$$. Prove by induction that the sequence $$(x_n)$$ is monotone and bounded, and determine its limit.

For the base case, we have that n=1, so $$x_{1+1}=x_2=(2+1)^{1/2}=\sqrt(3)\ne x_1$$, so how does this show the base case is satisfied? For the inductive step, I think what we have to prove is $$x_{n+1}=(2+x_n)^{1/2}$$ implies $$x_{n+2}=(2+x_{n+1})^{1/2}$$. I am not quite sure how to do the inductive step since I am bad at recursion. Also, how does this induction prove that the sequence is monotone and bounded? I don't see how induction proves this.

Show that the sequence's bounded above by $$\;2\;$$. Here is the inductive step, assuming $$\;x_n\le 2\;$$ :
$$x_{n+1}=(2+x_n)^{1/2}\le(2+2)^{1/2}=2$$
• How does this use induction? Shouldn't the step be showing that $x_{n+2}=(2+x_{n+1})^{1/2}$? Feb 27 at 16:32
• @Housefire No. The inductive hypothesis is that it is true for $\;n\;$ and then I show for $\;n+1\;$ ...exactly as donde above. Of course, you can assume for any $\;n+p\;$ and prove for $\;n+p+1\;$ , $\;p\in\Bbb N\;$ ...but why to make things messier? Feb 27 at 16:40