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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.

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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.

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  • $\begingroup$ How does this use induction? Shouldn't the step be showing that $x_{n+2}=(2+x_{n+1})^{1/2}$? $\endgroup$
    – Housefire
    Feb 27 at 16:32
  • $\begingroup$ @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? $\endgroup$
    – DonAntonio
    Feb 27 at 16:40

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