Monotone Sequences

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Nearly finished studying for my exam tomorrow;

Prove that the sequences ${a_n}$ , with a_n

, are monotone or eventually monotone.

$a_{n}=\frac{n}{2^n}$
eventually monotone since $\frac{a_n}{a_{n+1}}>1$ for n > 1. Is decreasing. How do I determine the lower bound on this?

$a_n=\frac{n!}{1*3*5*...*(2n-1)}$
When I do $a_{n+1}$ do I get $a_{n+1}=\frac{(n+1)n!}{1*3*5*...*(2n+1)}$ ? Also need to determine the bound on this for a future problem.

Edit (Shadow): Fixed dangerous LaTeX formula

Shadow · **Posted:** Wed, 21 Mar 2012 02:21:31 UTC

doodlenoodle wrote:

Nearly finished studying for my exam tomorrow;

Prove that the sequences ${a_n}$ , with a_n

, are monotone or eventually monotone.

$a_{n}=\frac{a}{2^n}$
eventually monotone since $\frac{a_n}{a_{n+1}}>1$ > 1 for n > 1. Is decreasing. How do I determine the lower bound on this?

$a_n=\frac{n!}{1*3*5*...*(2n-1)}$
When I do $a_{n+1}$ do I get $a_{n+1}=\frac{(n+1)n!}{1*3*5*...*(2n+1)}$ ? Also need to determine the bound on this for a future problem.

Edit (Shadow): Fixed dangerous LaTeX formula

For the first one, how do you know it's increasing? What if a=-1?

For the second one: write in the evens on the top and on the bottom.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

the first one is suppose to read:
$a_n=\frac{n}{2^n}$
Does that make it so?

For the second one, is that a suggestion to help me get the answer, or is it not written correctly? Can I say the denominator is (2n+1)!?

Shadow · **Posted:** Wed, 21 Mar 2012 02:31:03 UTC

doodlenoodle wrote:

the first one is suppose to read:
$a_n=\frac{n}{2^n}$
Does that make it so?

For the second one, is that a suggestion to help me get the answer, or is it not written correctly? Can I say the denominator is (2n+1)!?

For the first one, look at the ratio of the n+1st term to the nth term, and for the second of course the denominator is not 2n+1, it's 2n+1 times a bunch of stuff which is not 1, it is a suggestion as to how to get the answer.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

For the first, where does summing mean anything? I am just to prove if its monotone. Isn't it so that if a_n / a_n+1 that it is decreasing since a_n > a_n+1?

For the second, I meant, can the denominator be: (2n+1)!, was it written correctly on my first post?

Shadow · **Posted:** Wed, 21 Mar 2012 02:34:56 UTC

doodlenoodle wrote:

For the first, where does summing mean anything? I am just to prove if its monotone.

For the second, I meant, can the denominator be: (2n+1)!, was it written correctly on my first post?

Yes, after you put the evens in on the top and bottom, that's exactly what you get.

Reread my post, just take the ratio.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

For another problem:
Give an example of a sequence a_n

where $$\lim_{x\to\infty} |a_{n+1}-a_n|=0$ . but the sequence diverges.

Been trying for some time now, tried a lot of (-1)^n... etc. yet no luck

Shadow · **Posted:** Wed, 21 Mar 2012 03:09:59 UTC

doodlenoodle wrote:

For another problem:
Give an example of a sequence a_n

where $$\lim_{x\to\infty} |a_{n+1}-a_n|=0$ . but the sequence diverges.

Been trying for some time now, tried a lot of (-1)^n... etc. yet no luck

What have you tried?

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

I think everything dealing with a_n = (-1)^n, all sorts of variations of it, along with sin and cos. Just trying to think of other ways.

Shadow · **Posted:** Wed, 21 Mar 2012 03:20:13 UTC

doodlenoodle wrote:

I think everything dealing with a_n = (-1)^n, all sorts of variations of it, along with sin and cos. Just trying to think of other ways.

think back to simpler functions from calculus. There is an easy choice with strictly positive functions

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Deals with ln?

Shadow · **Posted:** Wed, 21 Mar 2012 03:26:34 UTC

doodlenoodle wrote:

Deals with ln?

That's one.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Can I do ln(1/n)? For this, when functions go out to infinity, is it always considered diverge?

Shadow · **Posted:** Wed, 21 Mar 2012 03:34:39 UTC

doodlenoodle wrote:

Can I do ln(1/n)? For this, when functions go out to infinity, is it always considered diverge?

you realize log(1/n)=-log(n) right?

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Now I do,

Really not sure where to go on this problem.

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