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!

Sequences and Convergence or Divergence?

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following sequence, whose nth term is given, converges or diverges. Find the limit of each convergent one.

    n[1 - cos(2/n)]

    2. Relevant equations
    I have made a solid attempt and obtained an answer but I am convinced I made a mistake and have missed something.

    3. The attempt at a solution
    lim n[1 - cos(2/n)]
    lim [ n - n[cos(2/n)] ]
    lim(n) - lim (n[cos(2/n)])
    lim(n) - lim ( [cos(2/n)] / (1/n) ) <--- Now using L'Hopital's Theorem:
    lim(n) - lim ( [(-2/n^2)sin(2/n)] / (-1/n^2) )
    lim(n) - lim 2sin(2/n)
    lim(n) - 2sin(lim [2/n])
    Infinity - 0 = Infinity, hence divergent!

    Note: Each limit (lim) shown about is the limit as n approaches infinity, just didn't write it every time to make things a bit neater.

    Now for some reason that seems to simple and incorrect to me. Confirmation or any sort of help would be much appreciated.
    Sorry about the confusing working above, thanks guys.
  2. jcsd
  3. Oct 2, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    You can't apply l'Hospital's theorem to:
    [tex]\frac{\cos (\frac{2}{n})}{\frac{1}{n}}[/tex]
    since it's not of a suitable form. (The numerator goes to 1.)
  4. Oct 2, 2007 #3
    Of course but I'm a bit stumped as to what to do with this one. Any suggestions? I presume that means what I did before applying L'Hopital's was right. Thanks for the reply.
  5. Oct 2, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    [tex]\frac{1-\cos (\frac{2}{n})}{\frac{1}{n}}[/tex] is in a good form to apply L'Hopital to, isn't it?
  6. Oct 2, 2007 #5


    User Avatar
    Homework Helper

    How about this:

    if [itex]cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...[/itex] replace all the x's with 2/n and you'll get the series expansion for cos(2/n)

    then simplify [itex]n(1-cos(\frac{2}{n}))[/itex] and then check what happens as n tends to infinity....

    I did this just to make the 1 in the problem disappear
  7. Oct 3, 2007 #6
    Very true. Well I gave that a try and this is how I went:

    lim [(-2/n^2)sin(2/n)] / (-1/n^2)

    lim 2sin(2/n)

    2sin (lim [2/n])


    0, hence convergent.

    Please tell me I'm right! Thank you so much for the replies guys.
  8. Oct 3, 2007 #7


    User Avatar
    Science Advisor
    Homework Helper

    That's what I would do. And that's the answer I would get.
  9. Oct 3, 2007 #8


    User Avatar
    Homework Helper

    There's nothing wrong with your method, but if Illusionist is in a typical calculus course introducing infinite series, you are anticipating things a little. Generally, convergence of infinite series and problems of the sort posted appear a while before Taylor series are discussed. (Otherwise, yeah, Taylor series certainly are a "nice big hammer" for dealing with problems of this sort...)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Sequences and Convergence or Divergence?