# Monotone increasing/decreasing sequence

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 $$k$$, $$s_2=k$$: $$k^2=4k-1$$, $$k=2+\sqrt{3}$$

• Should one of those "increasing"s in the second sentence be "decreasing"? Nov 13, 2014 at 2:45
• @CameronBuie yes Nov 13, 2014 at 2:53

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

• Sorry I don't follow.. Nov 13, 2014 at 3:06
• @MathMajor: What happens if $k>\frac12$? That's not the whole story, but it should get you started. Nov 13, 2014 at 3:13

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 $k\le 2-\sqrt3$.

• 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. Nov 13, 2014 at 3:22
• Also, I think you should mention that the magic number $2+\sqrt{3}$ is the fixed point of the iteration. Nov 13, 2014 at 3:24
• @martycohen Thanks for the comments, you are right, but please see the first word of my answer ;-) Nov 13, 2014 at 3:26
• @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? Nov 13, 2014 at 3:53
• @MathMajor rationalise $\sqrt a-\sqrt b$ by multiplying by $\sqrt a+\sqrt b$ in numerator and denominator, then do a bit of easy algebra. Nov 13, 2014 at 3:54