Finding the Radius and Interval of Convergence of a Series

[arl146]

Homework Statement

Find the radius of convergence and interval of convergence of the series:

as n=1 to infinity: (n(x-4)^n) / (n^3 + 1)

Homework Equations

convergence tests

The Attempt at a Solution

i tried the ratio test but i ended up getting x had to be less than 25/4 ...

when i did it, i ended up with [ (n+1)(x-4)(n^3 + 1) ] / [(n+1)^3 + 1]
take out the x-4 from the limit which leaves you with 4/9 * (x-4) and that's how i got it.

am i right or wrong? if so, how? not good at this stuff

 

[vela]

Homework Statement

Find the radius of convergence and interval of convergence of the series:

as n=1 to infinity: (n(x-4)^n) / (n^3 + 1)

Homework Equations

convergence tests

The Attempt at a Solution

i tried the ratio test but i ended up getting x had to be less than 25/4 ...

when i did it, i ended up with [ (n+1)(x-4)(n^3 + 1) ] / [(n+1)^3 + 1]

Perhaps it's just a typo, but you're missing a factor of n in the denominator.

take out the x-4 from the limit which leaves you with 4/9 * (x-4) and that's how i got it.

am i right or wrong? if so, how? not good at this stuff

You have the right idea, but I don't see where the 4/9 came from. Show us how you took the limit as n→∞.

 

[arl146]

Oh yea the bottom should be n[ ((n+1)^3) +1 ] right?

But I ended up getting the limit goes to 0 =/ idk if that's right ...

I simplified so it looked like this after the first step of the ratio test:

On top: (n+1)*(x-4)*(n^3 + 1)
On bottom: (n^3 + 1)*(3n+1)*(n+1)*n

Then 2 sets of terms cancel. I was left with (x-4)*(limit 1/((3n+1)*n)

And as n goes to infinity, 1/((3n+1)*n) goes to 0, right??

 

[vela]

Hmm, not sure how you got the denominator. You started with [(n+1)³+1]n. You should be able to see that the highest power of n you'll get is n⁴. If you were to multiply (n^3+1)(3n+1)(n+1)n out, the highest power would be n⁶, so those two expressions can't be equal.

 

[Mark44]

Putting the pieces together, here is the limit:
|x - 4|\lim_{n \to \infty}\frac{(n + 1)(n^3 + 1)}{n((n + 1)^3 + 1)}

arl146, you keep omitting the absolute values on the variable. You need them.

Also, the value of the limit expression above is NOT 4/9 or 25/4 or 0.

 

[arl146]

Ok I know what the limit is, I just had a typo in my first post. I just don't know how to find the value of the limit! To me it goes to infinity

 

[Mark44]

Ok I know what the limit is, I just had a typo in my first post. I just don't know how to find the value of the limit! To me it goes to infinity

No, it doesn't. You need to review evaluating limits at infinity of rational functions.

 

[vela]

Why do you think it goes to infinity?

 

[arl146]

Cause I just plugged in infinity for the n's lol.

I know I need to brush up on that stuff its been a year since I've taken this class and I'm trying to finish up work I missed. That's what I'm trying to get help on at this point though ...

 

[Mark44]

You can't just "plug in" infinity. Again, you need to review evaluating limits at infinity of rational functions.

 

[arl146]

Can't you help though? :/ all I know else to do for limits is use l'hopitals rule and that seems too complicated to do with this one

 

[Mark44]

Take a look at this page: http://people.richland.edu/james/lecture/m116/polynomials/rational.html, especially the section on horizontal asymptotes.

 

[arl146]

well when i factored everything out and some stuff cancelled, i got |x-4| lim 1 / (5 + 3n^2) and as n gets closer to infinity, the limit gets closer to 0. which would mean that for any x, the limit will always be 0.

but another approach, looking at what you gave me, the denominator is bigger, so it says that if the numerator is smaller than the denominator (degree wise) then y=0 is a horizontal asymptote.

all that i get from the 2 things above is that the limit goes to 0, but you told me in an above post that it doesn't ..

 

[Mark44]

well when i factored everything out and some stuff cancelled, i got |x-4| lim 1 / (5 + 3n^2) and as n gets closer to infinity, the limit gets closer to 0. which would mean that for any x, the limit will always be 0.

No. I don't know what you did to get the result above, but it's incorrect. From post #6, which I checked again, the rational function in the limit is degree 4 in both the numerator and denominator. If you simplify it, you definitely don't get 1/(5 + 3n²), which is degree 0 in the numerator and degree 2 in the denominator.

but another approach, looking at what you gave me, the denominator is bigger, so it says that if the numerator is smaller than the denominator (degree wise) then y=0 is a horizontal asymptote.

all that i get from the 2 things above is that the limit goes to 0, but you told me in an above post that it doesn't ..

 

[arl146]

Ok yea I get that it's both. Degree 4 .. I don't know what I'm supposed to do after that. So I get for the top: n^4 + n^3 + n +1 and for the bottom: n^4 + 3n^3 + 3n^2 + 1

So idk what to do next? Do I take the derivative of each terms to take the L'hopitals approach and then you end up with 1. So then x should be less than 4... Is that right?

 

[Mark44]

Ok yea I get that it's both. Degree 4 .. I don't know what I'm supposed to do after that. So I get for the top: n^4 + n^3 + n +1 and for the bottom: n^4 + 3n^3 + 3n^2 + 1

So idk what to do next? Do I take the derivative of each terms to take the L'hopitals approach and then you end up with 1.

If you go this way, you'll need to use L'Hopital's Rule 4 times. Another approach that is quicker and simpler is to factor x⁴ out of every term in top and bottom. That way you get
|x - 4|\lim_{n \to \infty}\frac{n^4}{n^4}\frac{1 + <other stuff>}{1 + <other stuff>}

By <other stuff> I mean terms that look like 1/n, 1/n², and so on. When you take the limit, n⁴/n⁴ goes to 1, and the other rational expression goes to 1 because all the terms involving 1/n, 1/n², etc. go to 0.

So then x should be less than 4... Is that right?

No.

What you have been doing is using the Ratio test to determine convergence. The final limit is |x - 4|. What does the Ratio test say about this value for convergence and divergence?

 

[vela]

Ok yea I get that it's both. Degree 4 .. I don't know what I'm supposed to do after that. So I get for the top: n^4 + n^3 + n +1 and for the bottom: n^4 + 3n^3 + 3n^2 + 1

That's close. The last term in the denominator should be 2n, so you should have
$$|x-4|\lim_{n \to \infty} \frac{n^4 + n^3 + n +1}{n^4 + 3n^3 + 3n^2 + 2n}$$

So I don't know what to do next? Do I take the derivative of each terms to take the L'hopitals approach and then you end up with 1. So then x should be less than 4... Is that right?

Using L'Hopital's rule will work, but it's more work than is necessary. The idea here is to divide both the top and bottom by n⁴ because that's the highest-degree term in the numerator:
$$|x-4|\lim_{n \to \infty} \frac{\frac{1}{n^4}(n^4 + n^3 + n +1)}{\frac{1}{n^4}(n^4 + 3n^3 + 3n^2 + 2n)}$$ Multiply out the top and bottom and then take the limit.

What does the ratio test say has to be true about
$$|x-4|\lim_{n \to \infty} \frac{\frac{1}{n^4}(n^4 + n^3 + n +1)}{\frac{1}{n^4}(n^4 + 3n^3 + 3n^2 + 2n)}$$ for the series to converge?

 

[arl146]

Um I don't know, that if it's less than 1 it converges and if greater than 1 it diverges

 

[Mark44]

Um I don't know, that if it's less than 1 it converges and if greater than 1 it diverges

Be specific. If what is less than 1. Don't use "it". Write mathematics equations/inequalities.

 

[arl146]

|x-4|<1 converge
|x-4|>1 diverge

 

[Mark44]

|x-4|<1 converge
|x-4|>1 diverge

OK, what interval are we talking about for convergence.

What about if |x - 4| = 1? The Ratio test doesn't cover this situation, so you have to investigate it as a special case.

 

[arl146]

interval? ... i don't know O_O

 

[Mark44]

interval? ... i don't know O_O

Solve |x - 4| < 1.

 

[arl146]

oh, x < 5 ... so just (-infinity, 5) ?

 

[Mark44]

oh, x < 5 ... so just (-infinity, 5) ?

No. You should get a finite length interval. According to you, -300 would satisfy |x - 4| < 1. Does it?

You should review working with absolute values in equations and inequalities.

 

[arl146]

ohhhhhhhh, i don't know why I KEEP just looking over the absolute values. sooo ... (4, 5) ?

 

[Mark44]

What about 3.5? Doesn't it satisfy |x - 4| < 1?

 

[arl146]

ohh right ... so (3, 5) because since it's not a square bracket it doesn't include the 3 right
but anything after 3 and before 4 would really never actually equal 1 which satisfies the less than sign

 

[Mark44]

ohh right ... so (3, 5) because since it's not a square bracket it doesn't include the 3 right
but anything after 3 and before 4 would really never actually equal 1 which satisfies the less than sign

Between 3 and 5, not 4.
OK, this establishes that the radius of convergence is 1. In the interval (3, 5), the series converges absolutely. If you aren't clear on what this means, look up the definition of absolute convergence in your book.

To finish the problem you need to check the two endpoints: x = 3 and x = 5. Substitute these numbers in your formula for the series and see what you get. Obviously (??) you won't be able to use the Ratio Test, but you should be able to use what other facts you know about series to say whether the series converges (either absolutely or conditionally) or diverges, at each of these values. If you aren't clear on conditional convergence, look up that term as well.

 

[arl146]

I know that, I was explaining how I understood why it is (3,5) instead of (4,5).

I think at x=5 it absolutely converges. It seems to get closer and closer to one and theyre all positive values so Looking for absolute convergence with the absolute values doesn't change that.

For x=3 I think it coverages just conditionally. Not absolute because the values don't seem to get closer to a Specific value when you add the absolute value signs... And I don't think it diverges because The values are -,+,-,+ and so on an they seem to end up going towards a specific number

Although, I've never really understood when something converges or diverges

 

[vela]

You need to be able to prove your conclusions. You can't just look at a few of the partial sums and guess that it's going to converge or not.

What series do you get when you set x=5?

 

[arl146]

i got summation [n(1)^n] / [n^3 + 1] for x=5. and the same for x=3 just with the negative, [n(-1)^n] / [n^3 + 1]

and that's when n=1 to infinity

 

[Mark44]

i got summation [n(1)^n] / [n^3 + 1] for x=5. and the same for x=3 just with the negative, [n(-1)^n] / [n^3 + 1]

and that's when n=1 to infinity

So when x = 5, the series is
\sum_{n = 1}^{\infty}\frac{n\cdot 1^n}{n^3 + 1}
Does that series converge or diverge? (You should simplify it first.)
Why?

What about when x = 3? Same questions.

 

[arl146]

no, not n^3 + 3, its n^3 + 1 ... right

i said that it absolutely converges. but the other person said i need to prove it. i don't know how i do that

 

[Mark44]

no, not n^3 + 3, its n^3 + 1 ... right

That was a typo, which I have now fixed.

i said that it absolutely converges. but the other person said i need to prove it. i don't know how i do that

You must have worked similar problems a little earlier in your course. What techniques do you have to determine whether a series converges?

 

[arl146]

you mean like, root test, alternating series test, integral test .. those kinds? won't the root test work for this?

 

[Mark44]

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]

oh. well we know that 1/x^2 converges so (n+1)*n/n^3+1 must too .. ? does that work

 

[Mark44]

oh. well we know that 1/x^2 converges so (n+1)*n/n^3+1 must too .. ? does that work

You know that \sum \frac{1}{n^2} 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, \sum \frac{1}{n^2} is a reasonable choice.

 

[arl146]

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]

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...

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, \sum \frac{1}{n^2} 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]

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]

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]

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]

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]

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..

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]

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..

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]

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]

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]

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