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Let $s_1=k$ and define $s_{n+1}=\sqrt{4s_n-1}$ for $n\ge\ 1$. Determine for what values of k the sequence $(s_n)$ will be monotone increasing and for what values of k it will be monotone decreasing.

Ok, so I know that for a sequence to be monotone it must be increasing or decreasing, and $(s_n)$ is increasing if $s_n\le\ s_{n+1}$ and decreasing if $s_n\ge\ s_{n+1}$.

So for this I have to find which values of k result in $s_n\le\ s_{n+1}$ and which result in $s_n\ge\ s_{n+1}$.

Starting with $s_1=k$ then $s_{1+1}=s_{2}=\sqrt{4s_1-1}=\sqrt{4k-1}$.

So for it to be increasing, $\sqrt{4k-1}\ge\ k$ and for it to be decreasing, $\sqrt{4k-1}\le\ k$.

And I am kind of stuck from here:

So I found for what values of $k$, $s_2=k$: $k^2=4k-1$, $k=2+\sqrt{3}$

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    $\begingroup$ Should one of those "increasing"s in the second sentence be "decreasing"? $\endgroup$ Nov 13, 2014 at 2:45
  • $\begingroup$ @CameronBuie yes $\endgroup$
    – Math Major
    Nov 13, 2014 at 2:53

2 Answers 2

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HINT: If $0<x<1$, then $\sqrt x>x$, and if $x>1$, then $\sqrt x<x$. Also, for what value of $k$ is $s_2=k$?

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  • $\begingroup$ Sorry I don't follow.. $\endgroup$
    – Math Major
    Nov 13, 2014 at 3:06
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    $\begingroup$ @MathMajor: What happens if $k>\frac12$? That's not the whole story, but it should get you started. $\endgroup$ Nov 13, 2014 at 3:13
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Hint. We have $$\eqalign{s_{n+1}-(2+\sqrt3) &=\sqrt{4s_n-1}-(2+\sqrt3)\cr &=4\frac{s_n-(2+\sqrt3)}{\sqrt{4s_n-1}+(2+\sqrt3)}\ .\cr}$$ This shows that if $s_n>2+\sqrt3$ then $$s_{n+1}-(2+\sqrt3)>0$$ and $$s_{n+1}-(2+\sqrt3)<s_n-(2+\sqrt3)\ ,$$ that is, $$2+\sqrt3<s_{n+1}<s_n\ .$$ So if we start with $k>2+\sqrt3$ then $s_n$ is always greater than $2+\sqrt3$ and always decreasing.

See if you can work out the other cases for yourself. You will need to think carefully about what happens if $k\le 2-\sqrt3$.

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    $\begingroup$ To show that $s_n$ is decreasing, you need to show that $4 < \sqrt{4s_n-1}+2+\sqrt{3}$. Not difficult, but not obvious. $\endgroup$ Nov 13, 2014 at 3:22
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    $\begingroup$ Also, I think you should mention that the magic number $2+\sqrt{3}$ is the fixed point of the iteration. $\endgroup$ Nov 13, 2014 at 3:24
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    $\begingroup$ @martycohen Thanks for the comments, you are right, but please see the first word of my answer ;-) $\endgroup$
    – David
    Nov 13, 2014 at 3:26
  • $\begingroup$ @David thanks for your explanation, I guess I am a little bit confused sitll about the second line in your first equation, could you explain to me how you got that? $\endgroup$
    – Math Major
    Nov 13, 2014 at 3:53
  • $\begingroup$ @MathMajor rationalise $\sqrt a-\sqrt b$ by multiplying by $\sqrt a+\sqrt b$ in numerator and denominator, then do a bit of easy algebra. $\endgroup$
    – David
    Nov 13, 2014 at 3:54

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