For example, if you had a series like:

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

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

Printable View

- Apr 5th 2010, 11:09 PMMattpdHow do you know if a sequence is monotone and bounded?
For example, if you had a series like:

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

what steps would you take to test if it is monotone and bounded? - Apr 5th 2010, 11:51 PMharish21
Its good to keep in mind the definition of Monotone sequence and bounded sequence.

A sequence $\displaystyle {a_n}$ is:

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

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

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 $\displaystyle {a_n} \leq M$ for any $\displaystyle n \geq 1$. M is called the upper bound.

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

Can you tackle your question now? - Apr 6th 2010, 04:21 AMHallsofIvy
Is it true that $\displaystyle a_{n+1}\le a_n$?

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

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

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

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

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

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 $\displaystyle 2n^2+ 6n+ 7> 0$ for all n. Crucially, everystep is "reversible" so we could go from $\displaystyle 2n^2+ 6n+ 7> 0$ back to $\displaystyle a_{n+1}> a_n$. This is a strictly increasing sequence.

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

Again, you could start from the obvious fact that $\displaystyle 8< 9$ and reverse the steps to get $\displaystyle \frac{3n^2+ 4}{2n^2+ 3}< 3/2$. 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". - Apr 6th 2010, 08:49 AMMattpd
I think I have it now, thanks.