1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coverging or diverging series

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data
    the sum of (n sin n)/(n^3+1)
    n from 1 to infinity

    2. Relevant equations



    3. The attempt at a solution
    what I have done so far is using the comparison method by comparing this with n sin n/ n^3, but I don't know what's the next step and even if my comparison is correct or not
     
  2. jcsd
  3. Mar 5, 2007 #2

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Have you heard of L'Hopital's Rule?

    The comparing method your using is also another way.

    So, what happens to [n*sin(n)]/n^3 when n goes to infinity?
     
  4. Mar 5, 2007 #3
    that's where I was stuck. I think it doesn't go anywhere therefore it doesn't exist. Am I right??
     
  5. Mar 5, 2007 #4

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    What's your argument for it not going anywhere?
     
  6. Mar 5, 2007 #5
    well because sin n varies between -1 and 1. This what makes me think that the limit does not converges to some value
     
  7. Mar 5, 2007 #6

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Yes, but where does the denominator go?
     
  8. Mar 5, 2007 #7
    it goes to infinity, so therefore it goes to 0?? Am I right?
     
  9. Mar 5, 2007 #8

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Not so quick.

    So, the denominator goes to infinity, and the numerator bounces around -1 and 1.

    Can you replace sin(n) by something else? Like say, -1 or 1?
     
  10. Mar 5, 2007 #9
    yeah If I replace it with either -1 or 1 both will most likely to reach 0 from both sides, from the negative sides and the positive sides.
     
  11. Mar 5, 2007 #10

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    So, why not just pick one?

    So, we have 1/n^2, and [n*sin(n)]/n^3, and the original one.

    Write down the comparisons, and you should be good to go.
     
  12. Mar 5, 2007 #11
    well we can simplify [n * sin n]/n^3 into sin n/n^2 right?? therefore the answer is 0?
     
  13. Mar 5, 2007 #12

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Yeah, the answer is 0, but you have to deduce it now. Or I think you do. Sometimes they just ask for the solution.
     
  14. Mar 5, 2007 #13
    what do you mean by deduce?
     
  15. Mar 5, 2007 #14

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Like, actually show the limit is 0.
     
  16. Mar 6, 2007 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Excuse me, but the original question asked for the SUM of [itex]\frac{nsin(n)}{n^3+1}[/itex], not the limit of the sequence.
     
  17. Mar 6, 2007 #16

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Whoa! My mistake.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Coverging or diverging series
  1. Series divergence (Replies: 3)

  2. Divergence of a series (Replies: 12)

Loading...