Monotonic sequence help!


by elle
Tags: monotonic, sequence
elle
elle is offline
#1
Nov17-05, 02:38 PM
P: 99
Hi, I was asked to prove a sequence was monotonic but I'm not sure if my answer is right. Can someone please help me check and point out any errors? Thanks very much!
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shmoe
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#2
Nov17-05, 02:55 PM
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You find that

[tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex]

this is correct (though when you find [tex]U_{n+1}[/tex] in the first line, your substitution is wonky). Your next inequality, the

[tex]\frac{1}{n(n+1)}\leq\frac{n-1}{n}[/tex]

while correct when n>1, does not help you show [tex]U_{n+1}\geq U_{n}[/tex]. To show this you just need to prove that [tex]U_{n+1}-U_n\geq 0[/tex] which you've essentially got, right?
elle
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#3
Nov17-05, 03:06 PM
P: 99
Quote Quote by shmoe
You find that

[tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex]

this is correct (though when you find [tex]U_{n+1}[/tex] in the first line, your substitution is wonky). Your next inequality, the

[tex]\frac{1}{n(n+1)}\leq\frac{n-1}{n}[/tex]

while correct when n>1, does not help you show [tex]U_{n+1}\geq U_{n}[/tex]. To show this you just need to prove that [tex]U_{n+1}-U_n\geq 0[/tex] which you've essentially got, right?
Ohh thanks for the pointers!

Erm so basically Ive shown that [tex]U_{n+1}-U_n\geq 0[/tex]? So would it be best to take out the part

[tex]\frac{1}{n(n+1)}\leq\frac{n-1}{n}[/tex]

since like you said, it only proves true for when n > 1. And see [tex]U_{n+1}\geq U_{n}[/tex], is that not monotonic increasing?

shmoe
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#4
Nov17-05, 03:44 PM
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Monotonic sequence help!


Quote Quote by elle
Erm so basically Ive shown that [tex]U_{n+1}-U_n \geq 0[/tex]?
Do you see why this follows from what you've shown:

[tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex]??

Quote Quote by elle
So would it be best to take out the part
[tex]\frac{1}{n(n+1)}\leq\frac{n-1}{n}[/tex]
since like you said, it only proves true for when n > 1.
This only being true when n>1 isn't the reason to leave it out- it doesn't actually help. You want to bound
[tex]U_{n+1}-U_n[/tex] from below by 0, an upper bound for it won't help.


Quote Quote by elle
And see [tex]U_{n+1}\geq U_{n}[/tex], is that not monotonic increasing?
Yes.
elle
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#5
Nov17-05, 03:52 PM
P: 99
Quote Quote by shmoe
Do you see why this follows from what you've shown:

[tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex]??



This only being true when n>1 isn't the reason to leave it out- it doesn't actually help. You want to bound
[tex]U_{n+1}-U_n[/tex] from below by 0, an upper bound for it won't help.




Yes.
Erm I think so....is it because for any natural number n, [tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex] would be equal or greater than 0??

Oh if that's monotonic increasing...does that mean I've made a mistake in my last statement? I wrote it was monotonic decreasing...
shmoe
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#6
Nov17-05, 03:59 PM
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Quote Quote by elle
Erm I think so....is it because for any natural number n, [tex]U_{n+1}-U_n=\frac{1}{n(n+1)}[/tex] would be equal or greater than 0??
Yes! Have some confidence! It wasn't a difficult jump, I just wanted you to explicitly say it

Quote Quote by elle
Oh if that's monotonic increasing...does that mean I've made a mistake in my last statement? I wrote it was monotonic decreasing...
Increasing is correct, you've shown the (n+1)st term is no smaller than the nth term.

By the way, you could also write:

[tex]U_n=1-\frac{1}{n}[/tex]

and increasing should be more immediate, though finding [tex]U_{n+1}-U_n[/tex] explictly like you've done is good too!
elle
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#7
Nov17-05, 04:43 PM
P: 99
Quote Quote by shmoe
Yes! Have some confidence! It wasn't a difficult jump, I just wanted you to explicitly say it
Haha yer, I think I sometimes need a little 'push' but its quite hard coz I tend to be in doubt quite a lot. I've not really done much work on sequences before hence the difficulty I'm having with coursework at the moment Thanks very much for your patience

Is it okay if I ask a few more questions? I've to decide whether the following are bounded above, below or both (bounded). And to also state the supremum and infimum.

I tried the first question out of the three but I'm stuck and dunno how to continue

http://tinypic.com/fubdd3.jpg

For questions 2 and 3, am I suppose to treat the terms in it separately and then combine them at the end?

Question 2 for example, Un = (-1)^n I think is bounded by -1 and 1. And then for (1-1/n), by writing down the nxt few terms in the sequence I get:

{ 0, 1/2, 2/3 ... } so its bounded between 0 and 1?
shmoe
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#8
Nov17-05, 05:07 PM
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I'm happy to help.


In all of those questions it's a good idea to write out the first few terms. You can then try to guess at bounds and sup, inf, then try to prove analytically that your guess is correct.

Try this for #1.

For #2, you are correct that (-1)^n is bounded below by -1 and above 1 (it's just alternating between these two values) and that 1-1/n is bounded below by 0 and above by 1 (though 0 is not the first term in the sequence and you can actually get a sharper lower bound). Can you combine these bounds to get bounds for [tex]U_n[/tex]? Again, you might want to write out the first few terms of [tex]U_n[/tex], not just the 'pieces' of it. It might also be a good idea to plot these points on a number line. Ditto for #3.
elle
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#9
Nov17-05, 05:32 PM
P: 99
hmm okay ive tried writing the terms for the first question but I don't think I'm doing it right...

Oh well here goes:

{ 99, 26, 32.3, 26, 19... }
shmoe
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#10
Nov17-05, 06:43 PM
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Almost...{99, 51, 32.333..., 26, 19, ...}

Now this isn't enough to give a lower bound for this one, but can you guess an upper bound? How about a supremum?

For the lower bound, you're either going to need more terms or do something else to examine the long term behavior of this guy. I'd suggest doing something else. When n is large the 100/n term is very small, so what are the [tex]U_n[/tex] terms close to?
elle
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#11
Nov17-05, 06:59 PM
P: 99
hmm I'm guessing the supremum is 99 and the infimum is 0?
shmoe
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#12
Nov17-05, 07:03 PM
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Good on the supremum, but why 0 for the infimum? What if n=100000001?
elle
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#13
Nov17-05, 07:17 PM
P: 99
hmm but is it not getting closer and closer to 0? Actualli I've noticed if n = 100000001 then Un = 0.000....999 and then if n gets larger than 100000001 it still remains the same answer....

EDITED - Oh wait I think my observation on the decimal digits are wrong....it gets closer and closer to zero does it not?
shmoe
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#14
Nov17-05, 07:22 PM
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Quote Quote by elle
hmm but is it not getting closer and closer to 0? Actualli I've noticed if n = 100000001 then Un = 0.000....999 and then if n gets larger than 100000001 it still remains the same answer....
EDITED - Oh wait I think my observation on the decimal digits are wrong....it gets closer and closer to zero does it not?
You're missing the [tex](-1)^n[/tex] part of [tex]U_n[/tex]...

[tex]U_{10000001}=\frac{100}{10000001}+(-1)^{10000001}[/tex]
elle
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#15
Nov17-05, 07:27 PM
P: 99
ohhh....its -0.9999 oops....

hmm okay...since n can be really really large, can it be possible that infimum doesnt exist? Can it be infinity?
shmoe
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#16
Nov17-05, 07:31 PM
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Quote Quote by elle
ohhh....its -0.9999 oops....
hmm okay...since n can be really really large, can it be possible that infimum doesnt exist? Can it be infinity?
n can be really large, but it's the values of the [tex]U_n[/tex]'s that we're worried about. So you've seen that you can get close to -1, but can you ever be smaller? Can you show that you always have [tex]U_n>-1[/tex]? Is this even true?
elle
elle is offline
#17
Nov17-05, 07:36 PM
P: 99
ahh okies I'll think about it....but now I need to go to sleep. It's 2.30 am still got college tomorrow.

Thanks very much for your help! And of course your patience coz even to myself I feel utterly dumb....

THANKS!!!


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