0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It is definitely not immediately clear why the answer is true. $\endgroup$ – Fimpellizieri Mar 7 '17 at 3:05
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.