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Convergence Test Help

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Ʃ cos(k*pi)/k from 1 to infinity.
    This is a test for convergence.

    and when is the proper time to use the alternating series test
    like using it on (-1)k(4k/8k) would result to divergence
    since lim of (4k/8k) is infinity and not 0 but the function is really
    convergent by the geometric series?

    2. Relevant equations



    3. The attempt at a solution
    Is it right to do this:

    for k is odd cos is negative and for k is even cos is positive
    then
    cos(k*pi)\k < (-1)k/k
    and by alternating series test ;
    (-1)k*(1/k) since 1/k is decreasing and lim as 1/k approaches infinity is 0 then
    (-1)k1/k converges thus cos(k*pi)/k converges by direct comparison test.
    [
     
  2. jcsd
  3. Sep 19, 2013 #2

    arildno

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    That is perfectly OK for the alternating 1/k-series! :smile:

    "since lim of (4^k/8^k) is infinity"
    Is it?
     
  4. Sep 19, 2013 #3
    Thanks for the reply
    So rechecking limit of (4/8)^k as k approaches infinity is 0
    Then if it is sine or cosine over a variable, the comparison test for it is (-1)^k?
    and if it is sine or cosine squared, the comparison is 1 only?
     
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