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Limit of a sequence

  1. May 20, 2012 #1
    The problem statement, all variables and given/known data

    infƩn=0 cos(m*n*pi)/(n+1)
    where m is a fixed integer. Determine the values of m, such that the series converges. Explain your reasoning in detail.

    The attempt at a solution

    I have figured out that cos(n*pi)/(n+1) can be represented as ((-1)^(n+1))/(n+1) (as it bounces back and forth from -1 to 1) and by the alternating series test, this converges.

    However, I am unsure how to explain what m would do this.

    I believe as m is an integer the cos(n*m*pi) term can't equal zero (need m = 1/2 and n=1 ect) so no matter what value m is, the sequence would converge, as the cos term can't be greater than -1 or 1? so m is any real integer (bit like n, though n is positive). Is this a bit trivial?



    ***And sorry, I understand this is a Series not a sequence***
     
    Last edited: May 20, 2012
  2. jcsd
  3. May 20, 2012 #2

    sharks

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    Gold Member

    Hi Tomp
    [tex]\sum^{\infty}_{n=0}\frac{\cos (mn\pi)}{n+1}[/tex] is actually a series, not a sequence.
     
    Last edited: May 20, 2012
  4. May 20, 2012 #3
    yeah sorry, can't change the heading. hoping people would pick up on my correction in my edit
     
  5. May 20, 2012 #4

    sharks

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    Gold Member

    You could prove that the limit has to be zero if the series converges, according to the nth-term test.

    Try the squeeze theorem to show that the sequence is equal to zero, then solve for m.

    I get [itex]m=\frac{1}{n\pi}[/itex]
     
  6. May 20, 2012 #5
    I have never learnt about the squeeze theorem sorry. We have learnt about the alternating, ration, comparison tests.

    And m has to be an integer :/
     
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