# Find a value for c such that a sequence is monotone

I am working on practice problems for an upcoming midterm in my introduction to real analysis course, and there is one practice problem involving sequences that is troubling me.

Let $s_1 = c$ and define $s_{n+1} = \sqrt(4s_n - 1)$. Determine what values of c will make the sequence $s_n$ monotone increasing or monotone decreasing.

Here's what I thought to try:

Let the sequence $s_n$ be a constant sequence. That is, $s_n = s_{n+1}$. Then $s_n = \sqrt(4s_n - 1)$, or $s_n^2 = 4s_n - 1$, or $s_n^2 - 4s_n + 1 = 0$. If $s_n = s_1$, then $c^2 - 4c + 1 = 0$. Finding the roots of $c^2 - 4c + 1 = 0$, we get $c = 2 + \sqrt3$ and $c = 2 - \sqrt3$. Thus, if $c = 2 + \sqrt(3)$ or $c = 2 - \sqrt(3)$, then $s_n = s_{n+1} = s_{n+2} = ...$

I also found that

$s_1 = 2$, $s_{n+1} = \sqrt(4s_n - 1)$

is monotone increasing. Can I then say that:

If $k \in [2-\sqrt3, 2+\sqrt3]$, then sequence is monotone increasing. If $k \in [1/4, 2-\sqrt3] \cup k \in [2+\sqrt3, \infty)$, then sequence is monotone decreasing.

Or do I have to show something else? Thanks in advance.

• It is definitely not immediately clear why the answer is true. Mar 7, 2017 at 3:05

What you did is correct but it should be better explained.

Say you want to know when $(s_n)$ is monotone increasing. Then you have to solve $s_{n+1}>s_n$, that is

$$\sqrt{4s_n-1}>s_n\iff4{s_n}-1>{s_n}^2.$$

Here we have assumed the $s_n$ are all non-negative. Indeed, if some $s_n$ were less than $1/4$, then $s_{n+1}$ would be ill-defined. Well, what we have hence is

$${s_n}^2-4s_n+1<0.$$

This looks like your quadratic equation! Indeed, since the coefficient of the quadratic term is positive, the inequality will be satisfied when $s_n$ lies strictly between the roots of the corresponding equality. More explicity:

$$2-\sqrt{3}<s_n<2+\sqrt{3}$$

In particular, $s_1=c$ must also satisfy this.

Are we finished? If we were to be super rigorous, I'd say no. We'd still have to show that if the inequality above is satisfied, then $s_{n+1}<2+\sqrt{3}$. This way, we guarantee that it never increases past the mark. Thankfully, showing this is easy. We have that:

$$s_{n+1}<2+\sqrt{3}\iff{s_{n+1}}^2<7+4\sqrt{3}\iff4s_n-1<7+4\sqrt{3}\\ \iff4s_n<8+4\sqrt{3}\iff s_n<2+\sqrt{3}.$$

Ah, it's all good then! Do you think you can use similar reasoning to show the $s_{n+1}<s_n$ case?