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Alternating series (Leibniz criterion)

  1. Nov 30, 2011 #1
    I read that an alternating series [tex] \Sigma (-1)^n a_n [/tex] converges if "and only if" the sequence [tex]a_n [/tex] is both monotonous and converges to zero.

    I tried with this series:

    [tex] \Sigma_{n=1}^{\infty} (-1)^n | \frac{1}{n^2} \sin(n)| [/tex]

    in the wolfram alpha and seems to converge to -0.61..., even if [tex] a_n = |\frac{1}{n^2} \sin(n)| [/tex] is not monotonous decreasing.

    What am I doing wrong? Is the monotone condition necesary for this test, but the fact that a_n is not monotonous does not guarantee if the series converges or not?

    Thanks.
     
  2. jcsd
  3. Nov 30, 2011 #2

    Office_Shredder

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    The conditions given are sufficient, not necessary. It is possible for the terms to not be decreasing in magnitude, and for the whole series to still converge.

    For your particular series, you can prove it converges absolutely by comparison with 1/n2
     
  4. Nov 30, 2011 #3
    You are right, thanks! I confused because in class we saw this statement:

    Let [tex] a_k > 0 [/tex] be a sequence monotonous decreasing. Then [tex] \Sigma_{n=1}^{\infty} (-1)^k a_k [/tex] converges [tex] \Leftrightarrow a_k \to 0 [/tex]

    The [tex] \Rightarrow [/tex] part of the proof is simply the necesary condition for any series to converge that [tex] a_k \to 0 [/tex].

    But I thoght the [tex] \Leftarrow [/tex] part would imply everything besides it: both that the series converges and that [tex] a_k [/tex] is monotonous decreasing.

    So it's a logical question now: when we have a double implication like above, the "leftarrow" implication only implies what the thesis says, but the hipothesis is the same. Am I right now?

    Thanks!
     
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