# Power series; 2

1. Feb 20, 2012

### arl146

1. The problem statement, all variables and given/known data
Find the radius of convergence and interval of convergence of the series:

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

2. Relevant equations
convergence tests

3. 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 thats how i got it.

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

2. Feb 20, 2012

### vela

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

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→∞.

3. Feb 21, 2012

### 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??

Last edited: Feb 21, 2012
4. Feb 21, 2012

### vela

Staff Emeritus
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.

Last edited: Feb 21, 2012
5. Feb 21, 2012

### Staff: Mentor

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.

6. Feb 21, 2012

### 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

7. Feb 21, 2012

### Staff: Mentor

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

8. Feb 21, 2012

### vela

Staff Emeritus
Why do you think it goes to infinity?

9. Feb 21, 2012

### 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 ....

10. Feb 21, 2012

### Staff: Mentor

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

11. Feb 22, 2012

### 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

12. Feb 22, 2012

### Staff: Mentor

13. Feb 23, 2012

### 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 doesnt ..

Last edited: Feb 23, 2012
14. Feb 23, 2012

### Staff: Mentor

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.

15. Feb 23, 2012

### 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?

16. Feb 23, 2012

### Staff: Mentor

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.
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?

17. Feb 23, 2012

### vela

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

18. Feb 23, 2012

### arl146

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

19. Feb 23, 2012

### Staff: Mentor

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

20. Feb 23, 2012

### arl146

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