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.
Best Answer
Show that the sequence's bounded above by $\;2\;$. Here is the inductive step, assuming $\;x_n\le 2\;$ :
$$x_{n+1}=(2+x_n)^{1/2}\le(2+2)^{1/2}=2$$
and we're done.