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We have a sequence $$a_1=\sqrt{6}\\a_2=\sqrt{6+\sqrt{6}}\\a_3=\sqrt{6+\sqrt{6+\sqrt{6}}}\\...$$a)Find a recursion formula for $a_{n+1}$
b)Find a limit

Attempt: a) Tried finding the recursion formula: $$a_{n+1}=\sqrt{6+a_n}$$ I am not not sure about it because the problem does not say where n starts. So if n starts at zero $a_0$ is not defined. Or is it implied that n starts at 1 since the sequence starts with $a_1$.

b)Right away I assume that $a_{n}$ converges to some L which I will find. $n+1$ goes to infinity when n goes to infinity. So I asumme that $a_{n+1}$ converges to the same L. As a result I obtain the following: $$L=\sqrt{6+L}\\-L^2+L+6=0$$ Solving for L i get $L=3,-2$. What would be an argument that L converges to 3, but not to 2. Thanks.

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It's generally considered implicit that your sequence starts with some specified boundary condition; in your case they would be $a_1=\sqrt{6}$, since you were given that. – Steven Stadnicki Oct 2 '12 at 1:24
up vote 6 down vote accepted

Detailed Hint:

Note that $6<3^2$; therefore, $a_1<3$.

Suppose that $a_k<3$, then $a_{k+1}=\sqrt{6+a_k}<\sqrt{6+3}=3$. Thus, we have shown that $a_k<3$ for all $k\ge1$ by induction.

Define $f_0(x)=\sqrt{x}$ and $f_{n+1}(x)=\sqrt{6+f_n(x)}$. Note that each $f_n$ is monotonic increasing and that $a_n=f_n(0)$ and $a_{n+1}=f_n(6)$. Thus, $a_n<a_{n+1}$. Therefore, $a_n$ is an increasing sequence, bounded above, which means that $a=\lim\limits_{n\to\infty}a_n$ exists.

Now that you know that the limit exists, you should be able to evaluate it.

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First observe that the $a_n$ increase. Now you need to confect the right induction argument to show $a_n < 3$ for all $n$. To do this, you need $a_n < 3 - q_n$ for the right sequence $q_n$.

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