Consider the following recursive sequence:
$$ \begin{cases} f_{0}=\sqrt{12}\\ f_{n+1}=\sqrt{12 + f_{n}} \end{cases} $$
for $n \geq 0$. How can I prove that this sequence is bounded above by $4$ and thus its limit is also $4$?
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Try proving that it's increasing and bounded above. That should be enough.. – Tintarn Aug 24 '15 at 18:00
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Proving it's bounded is not enough to prove it converges: the sequence $;(-1)^n$ is bounded. – Bernard Aug 24 '15 at 18:07
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Read carefully: I wrote "increasing and bounded". And that's enough for convergence.. – Tintarn Aug 24 '15 at 18:14
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I don't know what contradiction you're looking for. I edited your title and tags, but I left the contradiction part because you had put it there. – davidlowryduda Aug 24 '15 at 18:20
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3Special case of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ – YuiTo Cheng Jul 11 '19 at 08:03
3 Answers
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First, being bounded above by $4$ neither guarantees convergence nor, if we knew it converges, that it converges to $4$.
It's easier to show that this converges and then to show that it converges to $4$. To show that it converges, you want to
Show that $f_n$ is increasing.
You can do this directly through induction. It's pretty straightforward.
Show that $f_n$ is bounded above.
You will have a much easier time by choosing a clever, lenient upper bound. I recommend showing that $f_n < 1 + 2\sqrt{12}$. This number might seem magical. But notice that in the inductive step, we would argue like this:
$$ f_n = \sqrt{12 + f_{n-1}} \leq \sqrt{12 + 2\sqrt{12} + 1} = 1 + \sqrt 12 \lt 1 + 2\sqrt {12}.$$
This choice simplified that inductive argument quite a bit over what it could have been.
Now that you know it converges, you can use the standard argument. Call $L$ the limit. Then we must have $L = \sqrt{12 + L}$. Simplifying, factoring, and throwing out the negative limit shows the actual limit is $4$. $\diamondsuit$
Steven Stadnicki
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davidlowryduda
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Solid answer ... so a +1. But, just curious here .... Why is the use of $f_n\le 4$, in the inductive step, to conclude that $f_{n+1}\le 4$ viewed as difficult? – Mark Viola Aug 24 '15 at 18:40
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@Dr.MV You know, that's interesting. I thought when I was first writing this down that using $4$ directly was nontrivial. But it's rather easy. Oh well! – davidlowryduda Aug 25 '15 at 01:08
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Hint:
By induction you can prove that the sequence is monoton (strictly) increasing $${ f }_{ 1 }^{ 2 }=12+{ f }_{ 0 }=12+\sqrt { 12 } =12+2\sqrt { 3 } \\ { f }_{ 2 }^{ 2 }=12+12+2\sqrt { 3 } =24+2\sqrt { 3 } \\ { f }_{ 1 }<{ f }_{ 2 }\\ { f }_{ n }<{ f }_{ n+1 }$$
Let say $\lim _{ n\rightarrow \infty }{ { f }_{ n } } =f$ then $${ f }^{ 2 }-f-12=0$$ $$\left( f-4 \right) \left( f+3 \right) =0$$ so monoton increasing and bounded above sequence has limit,and limit is $4$
haqnatural
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Hint:
To prove a recursive sequence is convergent, one strategy is to prove it is non-decreasing and bounded from above, or its non-increasing and bounded from below.
If the sequence is defined with a function $g$,: $f_{n+1}=g(f_n)$, it is non-decreasing if the graph of $g$ is above the line $y=x$ on an interval $I$ and all $f_n$ belong to $I$. If the interval $I$ is bounded from above, the sequence will also be bounded from above, hence it will converge.
Furthermore, if $g$ is continuous, the limit $\ell$ will satisfy the equation $\; \ell=g(\ell)$, in other words, the limit will be a fixed point of $g$. If there is only one fixed point, it will be the limit. If there are several fixed points, you will have to decide (and prove) which fixed point is the limit.
Bernard
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