How do you know if a sequence is monotone and bounded?

**Mattpd** · April 5th 2010, 11:09 PM

For example, if you had a series like:

an= $(3n^2+4)/(2n^2+3)$ for n>=1

what steps would you take to test if it is monotone and bounded?

**harish21** · April 5th 2010, 11:51 PM

Originally Posted by Mattpd

For example, if you had a series like:

an= $(3n^2+4)/(2n^2+3)$ for n>=1

what steps would you take to test if it is monotone and bounded?

Its good to keep in mind the definition of Monotone sequence and bounded sequence.

A sequence ${a_n}$ is:

(a) increasing if and only if ${a_n}<{a_{n+1}}$ for any $n \geq 1$

(b) decreasing if and only if ${a_n}>{a_{n+1}}$ for any $n \geq 1$

The sequence is MONOTONE if either one of these properties holds.

So check for the terms in you sequence, and you'll be able to find out if its monotone or not!

************************************************** ************************************************** ***************************

The sequence is BOUNDED ABOVE if there exists a number M such that ${a_n} \leq M$ for any $n \geq 1$ . M is called the upper bound.

The sequence is BOUNDED below if there exists a number m such that ${a_n} \geq m$ for any $n \geq 1$ . m is called the lower bound.

Can you tackle your question now?

**HallsofIvy** · April 6th 2010, 04:21 AM

Originally Posted by Mattpd

For example, if you had a series like:

an= $(3n^2+4)/(2n^2+3)$ for n>=1

what steps would you take to test if it is monotone and bounded?

Is it true that $a_{n+1}\le a_n$ ?

That is, is $\frac{3(n+1)^2+ 4}{2(n+1)^2+ 3}\le \frac{3n^2+ 4}{2n^2+ 3}$ ?

Since both denominators are positive, that would be the same as $(3(n+1)^2+ 4)(2n^2+ 3)\le (2(n+1)^2+ 3)(3n^2+ 4)$

$(3n^2+ 6n+ 7)(2n^2+ 3)\le (2n^2+ 4n+ 5)(3n^2+ 3)$

$6n^4+ 12n^3+ 23n^2+ 18n+ 21\le 6n^3+ 21n^2+ 12n+ 15$

$2n^2+ 6n+ 7\le 0$

The discriminant of that quadratic is 36- 4(2)(7)< 0 so the quadratic is never 0. In fact, it is easy to see that it is always positive so what is really true is that $2n^2+ 6n+ 7> 0$ for all n. Crucially, everystep is "reversible" so we could go from $2n^2+ 6n+ 7> 0$ back to $a_{n+1}> a_n$ . This is a strictly increasing sequence.

To see that it is bounded, note that $\frac{3n^2+ 4}{2n^2+ 3}< 3/2$ leads to $2(3n^2+ 4)< 3(2n^2+ 3)$ or 6n^2+ 8< 6n^2+ 9[/tex] and then $8< 9$ .

Again, you could start from the obvious fact that $8< 9$ and reverse the steps to get $\frac{3n^2+ 4}{2n^2+ 3}< 3/2$ . The sequence has 2 as an upper bound.

It should be obvious that this sequence converges to 3/2 and so has 3/2 as "least upper bound".

**Mattpd** · April 6th 2010, 08:49 AM

I think I have it now, thanks.

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