$a_1\in(0,3)$, $a_{n+1}=\sqrt{a_n+6}$. Calculate $$\lim_{n\to\infty}6^n(a_n-3). $$
What I've done:
$a_n<3$;
$a_n<a_{n+1}$;
$$\lim_{n\to\infty}a_n=3;$$
$$\lim_{n\to\infty}\frac{a_n-3}{a_{n+1}-3}=6;$$
The answer is not $0$.
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rushusuixing
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Set $L$ to be the limit. Then it will satisfy $L=\sqrt{L+6}$. Use this to solve for $L$. Note that you have already have that it's bounded above and monotone increasing so it converges. – CyclotomicField Oct 24 '23 at 11:11
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1Emm... Not the limit of $a_n$@CyclotomicField – rushusuixing Oct 24 '23 at 11:13
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The rate of growth for $6^n$ is much higher than the rate at which $a_n-3$ approaches zero. Calculate the first few terms then see if the sequences is unbounded. Then try to prove what you observe. – CyclotomicField Oct 24 '23 at 11:21
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If you let $b_n= 6^n(a_n-3)$ then $a_n = \frac{b_n}{6^n}+3$ and you can get a recurrence for $b_n$. Empirically $a_1 \in (0,3)$ gives $b_1 \in (-18,0)$ and $b_n \in (-20.2,0)$ with $\lim b_n$ existing but appearing to depend on $b_1$ and so $a_1$. – Henry Oct 24 '23 at 11:32