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Power series; 2

by arl146
Tags: power, series
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Mark44
#37
Feb28-12, 06:18 PM
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I don't see how you can make the root test work. I'm thinking more along the lines of the comparison test or limit comparison test.
arl146
#38
Feb28-12, 06:30 PM
P: 343
oh. well we know that 1/x^2 converges so (n+1)*n/n^3+1 must too .. ? does that work
Mark44
#39
Feb29-12, 10:20 AM
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Quote Quote by arl146 View Post
oh. well we know that 1/x^2 converges so (n+1)*n/n^3+1 must too .. ? does that work
You know that [itex]\sum \frac{1}{n^2}[/itex] converges (note that the variable is n, not x), but
1) The series you are comparing is not (n+1)*n/n^3+1. What you should be working with is your original series in post #1 evaluated at x = 5.
2) You need to do more than just wave your arms to show convergence. If you are using the comparison test, you need to show that each term of your series is less than the corresponding term of the series you're comparing to. For your problem, [itex]\sum \frac{1}{n^2}[/itex] is a reasonable choice.
arl146
#40
Feb29-12, 07:34 PM
P: 343
yea i meant to change it to n, just slipped my mind after typing.

and oops yea, its n*(x-4)^n / n^3 + 1

but ok, soo like .. how do i show the convergence then with 1/n^2
Mark44
#41
Feb29-12, 09:36 PM
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Quote Quote by arl146 View Post
yea i meant to change it to n, just slipped my mind after typing.

and oops yea, its n*(x-4)^n / n^3 + 1
Try to stay focussed on the problem at hand. For one thing, you're trying to determine the convergence when x = 5 and x = 3. When x = 5, the general term in your series is n/(n3 + 1). Use parentheses!
QUOTE=arl146;3791707]

but ok, soo like .. how do i show the convergence then with 1/n^2[/QUOTE]
I already answered that question...

Quote Quote by Mark44 View Post
If you are using the comparison test, you need to show that each term of your series is less than the corresponding term of the series you're comparing to. For your problem, [itex]\sum \frac{1}{n^2}[/itex] is a reasonable choice.
Your textbook should have some examples where they use comparison. Take a look at them.

Really, you're going to have to step up and show some initiative. This is post #41 on a problem that's not terribly difficult. Instead of continually asking what you should do next, try something and see where it takes you.
arl146
#42
Mar4-12, 09:30 PM
P: 343
Ok umm ..

When x=5 you have summation n/(n^3+1). It is similar to summation 1/n^2. There's a proof in the book that 1/n^p when p>1 converges and when p<1 it diverges. So I don't have to show that right? When I'm writing my homework do I have to include all of that p>1 stuff or can I just put: since we know that 1/n^2 converges and n/(n^3+1) < 1/n^2 that our series n/(n^3+1) also converges. [our series is smaller because of the larger denominator]. So if that's all right, I wasn't exactly asking exactly what to write I guess I just meant I don't know exactly how to present that information, like in what kind of organized manner do I write it all for my homework.

And when x=3 it's [(-1)^n * n]/(n^3+1) ...... Is that right ? I'm going off memory.
So that's just e same thing, same idea so that also converges.

Now, how the heck do you show absolute/conditional convergence or doesn't that matter?
vela
#43
Mar4-12, 09:47 PM
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Quote Quote by arl146 View Post
Ok umm ..

When x=5 you have summation n/(n^3+1). It is similar to summation 1/n^2. There's a proof in the book that 1/n^p when p>1 converges and when p<1 it diverges. So I don't have to show that right? When I'm writing my homework do I have to include all of that p>1 stuff or can I just put: since we know that 1/n^2 converges and n/(n^3+1) < 1/n^2 that our series n/(n^3+1) also converges. [our series is smaller because of the larger denominator]. So if that's all right, I wasn't exactly asking exactly what to write I guess I just meant I don't know exactly how to present that information, like in what kind of organized manner do I write it all for my homework.
You don't need to show that 1/n2 converges because the proof in the book establishes that. It looks like you're using the comparison test, comparing your series to 1/n2. The first thing you need to do is verify that you have a series to which you can the test. You can, in this case, so what you need to show is what you claim to know, that n/(n3+1) < 1/n2. You can't just assert it. Once you establish that the conditions of the test hold, you can conclude that the series converges.

And when x=3 it's [(-1)^n * n]/(n^3+1) ...... Is that right ? I'm going off memory.
So that's just e same thing, same idea so that also converges.
Here, it's not the exact same thing because the conditions required for the test aren't satisfied because of the factor (-1)n. I'll leave it to you to look up what those conditions are.

Now, how the heck do you show absolute/conditional convergence or doesn't that matter?
What are the definitions of absolute and conditional convergence?
arl146
#44
Mar4-12, 10:01 PM
P: 343
Wait wait, why do I have to show that my series is less than the one we are comparing against? Can't you just say that since the degree of the n on the bottom is bigger that the whole fraction is smaller?? I don't get how I would show that .. Do I just plug in different values of n for that?

Ok um I don't see anything in the book that is similar to the x=3 one I don't where else in the book I'd find those conditions you talk about. I don't get it. I mean I get that it won't work since its +,-,+- but how do you show for this one by comparing? And do you still compare with the 1/n^2 ?

Absolute convergence when the value of the limit of the series with absolute value signs is < 1
vela
#45
Mar4-12, 10:27 PM
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Quote Quote by arl146 View Post
Wait wait, why do I have to show that my series is less than the one we are comparing against?
You have to show it because that's one of the conditions required for the comparison test to apply.

Can't you just say that since the degree of the n on the bottom is bigger that the whole fraction is smaller?? I don't get how I would show that .. Do I just plug in different values of n for that?
No, you can't just compare the degree of the denominators. Take the two series ##\sum \frac{1}{n^2}## and ##\sum \frac{n+1}{n^2}##. They both n2 in the denominator, but the first one converges while the second doesn't.

You can't just plug in a few values for n. You have to show that the series you're working with is less than 1/n2 after some point, that is when n>N for some N. I'm sure your book has examples showing how to apply the comparison test.

Ok um I don't see anything in the book that is similar to the x=3 one I don't where else in the book I'd find those conditions you talk about. I don't get it. I mean I get that it won't work since its +,-,+- but how do you show for this one by comparing? And do you still compare with the 1/n^2 ?
The conditions I'm talking about have to do with the test itself, and it's the one you mentioned. The comparison test only works for a non-negative series, and the x=3 series doesn't satisfy that requirement. That means, you can't use the comparison test on that series.

Absolute convergence when the value of the limit of the series with absolute value signs is < 1
No, this is wrong. Look up what it means and what absolute convergence implies. This is the key to figuring out if the x=3 series converges.
arl146
#46
Mar4-12, 10:48 PM
P: 343
Ok well I dont know how to show that it is less than 1/n^2? Literally the book says: "5/(2n^2+4n+2) < 5/(2n^2) because the left side has a bigger denominator. We know that summation 5/(2n^2) = (5/2)* summation (1/n^2) is convergent (p-series with p=2>1). Therefore *the series mentioned for this example* is convergent by the comparison test." it really doesn't show anything else..

Ohhhh I gotcha I can't use te comparison test onthe x=3 one, you should have just said that, that hurts my brain a little less haha (just kidding). So I use the alternating series test right ? (this is all coming together, slowly, but getting there). soooo to be convergent according to the alternating series test, it has to satisfy two things: (i.) b(n+1) <= b(n) [which is really b sub n not b of n]. Which our series does. Because (n+1)/((n+1)^3+1) is less than n/(n^3+1). And has to satisfy (ii.) lim of b sub n must equal 0, which is does!

Also, I did look up absolute convergence in my book. Oh well it just says the series is absolutely convergent if the series of absolute values is convergent. So to me that means nothing, like I get nothing out of that? Can you explain how to apply that. When I start getting values I don't know how to tell if it's convergent.
vela
#47
Mar4-12, 11:21 PM
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Quote Quote by arl146 View Post
Ok well I dont know how to show that it is less than 1/n^2? Literally the book says: "5/(2n^2+4n+2) < 5/(2n^2) because the left side has a bigger denominator. We know that summation 5/(2n^2) = (5/2)* summation (1/n^2) is convergent (p-series with p=2>1). Therefore *the series mentioned for this example* is convergent by the comparison test." it really doesn't show anything else..
Do you understand the logic behind the book's argument here?

Ohhhh I gotcha I can't use te comparison test onthe x=3 one, you should have just said that, that hurts my brain a little less haha (just kidding). So I use the alternating series test right ? (this is all coming together, slowly, but getting there). soooo to be convergent according to the alternating series test, it has to satisfy two things: (i.) b(n+1) <= b(n) [which is really b sub n not b of n]. Which our series does. Because (n+1)/((n+1)^3+1) is less than n/(n^3+1). And has to satisfy (ii.) lim of b sub n must equal 0, which is does!
You have to show (n+1)/((n+1)^3+1) < n/(n^3+1) if you want to use the alternating-series test.

Also, I did look up absolute convergence in my book. Oh well it just says the series is absolutely convergent if the series of absolute values is convergent. So to me that means nothing, like I get nothing out of that? Can you explain how to apply that. When I start getting values I don't know how to tell if it's convergent.
That's the definition of absolute convergence. You need to know that so when the term comes up, you know what's being talked about.

Now look in the book for theorems that apply to absolutely convergent series to see why it might apply to this problem.
arl146
#48
Mar4-12, 11:41 PM
P: 343
Uhhh logic, I mean yes I think I understand it. But I don't understand how that shows anything more than what I was saying.

And that's my problem, how do I show that it is less than?

And I don't know how it applies, I don't even think it does but someone brought it up in a past post. Why does the absolute convergence matter I'm just trying to find the radius of convergence and interval of convergence. The examples in the book don't even mention it. So what's the point in adding that in. Shouldn't I only deal with absolute convergence if the signs of the terms are irregularly switching back and forth?
vela
#49
Mar4-12, 11:49 PM
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You have a tendency to overlook important details. Your claim seems to be that if you have two fractions, the one with the bigger denominator is smaller. But what about 1/10 and 50/100?
arl146
#50
Mar4-12, 11:54 PM
P: 343
Yea yea I get that. But I don't get out of this exactly HOW to show that one is less or more than the other. I just don't see it in the example or how to do it
vela
#51
Mar5-12, 12:01 AM
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Well, take a stab at it proving it and post your attempt.
arl146
#52
Mar5-12, 12:07 AM
P: 343
Everything I already said already was my attempt. That's all I got. I just don't get it. Nothing in the book is any different than what I said about my series being less than the one we're comparing it to. I don't know how to show that it's less than. If it even is less than the series we're comparing it to!
arl146
#53
Mar6-12, 10:47 PM
P: 343
Quote Quote by Mark44 View Post
You know that [itex]\sum \frac{1}{n^2}[/itex] converges (note that the variable is n, not x), but
2) You need to do more than just wave your arms to show convergence. If you are using the comparison test, you need to show that each term of your series is less than the corresponding term of the series you're comparing to. For your problem, [itex]\sum \frac{1}{n^2}[/itex] is a reasonable choice.
like we said, i have the series [itex]\frac{n}{n^3+1}[/itex]

and its from n=1 to infinity .. just doing the first few values, you get 1/2 + 2/9 + 3/28 + ..
which each term is definitely less than the previous term. and just looking at the series you can tell that that will be the case because theres an n with a degree of 1 on top and on bottom theres an n with a degree 3 which shows that. is there another way that im supposed to show this? if thats right, how do i now prove/show that [itex]\frac{n}{n^3+1}[/itex] is less than or equal to the series [itex]\frac{1}{n^2}[/itex] ? can you do it this way :

ok so you have [itex]\frac{n}{n^3+1}[/itex]. pull out an n on top and bottom and youre left with [itex]\frac{1}{n^2+\frac{1}{n}}[/itex]. this is the same as 1/n^2 except it adds the 1/n on the bottom. which makes my series bigger. if this is right, what does that mean? that its not convergent at x=5? so what now?
vela
#54
Mar6-12, 11:11 PM
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Why does that make your series bigger?


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