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!

Determine Convergent or Divergent

  1. May 6, 2012 #1
    an = [cos (3n) )] /n.cos(1/n)


    My solution is , I wrote an as

    liman→∞ [cos (3n) )] /[ cos(1/n) / (1/n) ) .

    We get liman→∞ cos(3n) / ( 1/0 ) .

    I think solution of this limit is zero but ı'm not sure cos(3n) .I think cos(3n) as a number and number/infinity is zero .As a result of this ı say it is a convergent.

    Is the correct approach to the question ?
     
  2. jcsd
  3. May 6, 2012 #2
    You know that cos n ≤ 1 for any n, so cos(3n) ≤ 1 and cos(3n)/n ≤ 1/n

    For the denominator I wouldn't write it as cos(1/n) / (1/n). Try thinking of 1/(n·cos(1/n)) as (1/n)·(1/cos(1/n))

    What's the limit of 1/cos(1/n) as n→∞?
     
  4. May 7, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I don't think it is as simple as the OP indicates. Of course, for large n we can consider cos(1/n) to be essentially 1, so we almost have cos(3^n)/n --> 0. However, the sign changes due to cos(3^n) is very important; as n grows there is increasingly rapid oscillation in sign.

    RGV
     
    Last edited: May 7, 2012
  5. May 7, 2012 #4

    sharks

    User Avatar
    Gold Member

    Use the squeeze theorem on [itex]\frac{\cos(3^n)}{n}[/itex], then evaluate [itex]\frac{1}{\cos(1/n)}[/itex]. Using the properties of limits: [itex]\lim (a.b)=\lim a . \lim b[/itex], you should get the answer.
     
  6. May 7, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, it won't. The series sum 1/n diverges, but the series sum (-1)^n/n converges, so the sign changes are vital. In the current problem there are rapid sign changes (due to oscillation in cos(3^n)), but it is not clear whether the series alternates in any sense (that is, with any kind of nice pattern). I really do believe the problem is non-trivial.

    RGV
     
  7. May 7, 2012 #6

    sharks

    User Avatar
    Gold Member

    But the problem involves sequence [itex]a_n[/itex] not series.
     
  8. May 7, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    OK, then it's easy. I guess I thought the OP was doing what so many others have done, viz., saying sequence they mean series. I should just have taken him at his word.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook