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Homework Help: Checking divergence/convergence for alternating series

  1. Apr 23, 2009 #1
    i am given te series:


    1st thing i check is

    lim |an| = 0

    so i know that the series might converge and might diverge,

    since this is a series with an alternating sign, i checked with lebniz rule,
    an>an+1 ==> not true for this series since -1<cos(n)<1 i cannot know if the denominator of an is bigger or smaller than that of an+1

    so using leibniz i cannot prove conditional convergence, (and i know from the answer that this series converges),

    can i use the second comparative test, ie

    if an/bn=K ( not 0, not \infty) then an, bn both diverge/converge

    if i take bn=(-1)n-1/n which i know converges(conditionally)

    lim an/bn= 1

    so an converges

    which is the right answer BUT i think that this comaprison is only for positive series,

    how else can i solve alternating series
  2. jcsd
  3. Apr 24, 2009 #2


    Staff: Mentor

    This is a tough one, that's for sure. A thought that might or might not help is this:
    n - 50 <= n + cos(n) <= n + 50

    so 1/(n + 50) <= 1/(n + cos(n)) <= 1/(n - 50) (assuming that n > 50)

    Looking at the expressions at the two ends above, can you convince yourself that the two alternating series whose general terms are given by these two expressions both converge? If that's the case, it seems to me that the alternating series whose general term is given by the middle expression has to converge as well.
  4. Apr 24, 2009 #3
    i thought of that too, but i was told that these comparative tests are only good for positive series,

    what i wanted to do was use the 2nd comparative test that says

    lim An/Bn =K (not 0 not infinity) then the 2 series converge/diverge together,

    i wanted to take my Bn as(-1)^(n-1)/n
    so that the division would give me

    lim (n+50*cos(n))/n =1

    now i know that Bn converges therefore An must too converge, But these tests are apparently no good for alternating signs.

    any other thoughts?
  5. Apr 24, 2009 #4


    Staff: Mentor

    What I've suggested is more along the lines of the "squeeze play" theorem that is used to evaluate limits. I'm suggesting that if, term by term, Sigma an <= Sigma bn <= Sigma cn, and you know that the outer two series converge, it seems to be that the one in the middle is forced to converge as well.
  6. Apr 25, 2009 #5
    yes i get that, but from what i understand this is incorrect since you are taking the abs values of this series and not taking into acount the -1 , which changes the sign, so i get (Sigma an <= Sigma bn <= Sigma cn) and then (Sigma an+1 >= Sigma bn+1 >= Sigma cn+1),

    {also this is only correct from n=50, but i can overcome that}
  7. Apr 25, 2009 #6
    is it alright to only compare the positive series of these?
  8. Apr 25, 2009 #7


    User Avatar
    Gold Member

    I would say yes; by letting the (-1)n-1=1 you would be attempting to prove absolute convergence. If the series does converge absolutely, then it converges automatically with the (-1)n-1 term.

    I'm not sure how comparing the absolute series an to a series that converges conditionally would affect the convergence vis-a-vis absoluteness of conditionality; I would say that would result in an converging conditionally.

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