- #36
arl146
- 343
- 1
you mean like, root test, alternating series test, integral test .. those kinds? won't the root test work for this?
arl146 said:oh. well we know that 1/x^2 converges so (n+1)*n/n^3+1 must too .. ? does that work
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!arl146 said: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
Your textbook should have some examples where they use comparison. Take a look at them.Mark44 said: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.
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.arl146 said: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.
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.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.
What are the definitions of absolute and conditional convergence?Now, how the heck do you show absolute/conditional convergence or doesn't that matter?
You have to show it because that's one of the conditions required for the comparison test to apply.arl146 said:Wait wait, why do I have to show that my series is less than the one we are comparing against?
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.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?
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.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 ?
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.Absolute convergence when the value of the limit of the series with absolute value signs is < 1
Do you understand the logic behind the book's argument here?arl146 said:Ok well I don't 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..
You have to show (n+1)/((n+1)^3+1) < n/(n^3+1) if you want to use the alternating-series test.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!
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.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.
Mark44 said: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.
Now, since n2 + 1/n > n2, for all n >= 1,arl146 said:ok well
we need [itex]\frac{n}{n^3+1}[/itex] < [itex]\frac{1}{n^2}[/itex] for the test
and that can be proven by:
[itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{n(1)}{n(n^2+1/n)}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex]
arl146 said:so now we are looking at [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] for my series to be convergent since [itex]\frac{1}{n^2}[/itex] converges.
looking at the denominators: n2+[itex]\frac{1}{n}[/itex] is > n2
thus making [itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] this true
is that good enough?