Given: $a_0=4$, $a_{n+1}=\sqrt{2+a_n}$.
Show that $(a_n)$ converges and determine the limit.
I don't know where to start.
I have tried determining the limit: I know that $a_n\to A$, so $a_{n+1}\to A$. This gives $A=\sqrt{2+A}$, so A=2 or A=-1, but a limit must be unique... And I don't even know where to start to show that $(a_n)$ converges. I hope somebody can help me. Thank you in advance!
Answer
First. Show tha $\{a_n\}$ is decreasing, using induction. (Straight-forward.)
Second. As $a_n>0$ and decreasing, then $\{a_n\}$ converges.
Third. Let $a_n\to a$, then $a_{n+1}=\sqrt{a_n+2}\to \sqrt{a+2}$. But $a_{n+1}\to a$, as well. Thus $a=\sqrt{a+2}$, and hence $a^2-a-2=0$, which means that $a=2$ or $a=-1$. The $a=-1$ is eliminated as $a_n>0$. Thus $a_n\to 2$.
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