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Convergence of alternating series

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Let s be the sum of the alternating series [tex]\sum[/tex](from n=1 to [tex]\infty[/tex])(-1)n+1an with n-th partial sum sn. Show that |s - sn| [tex]\leq[/tex]an+1



    2. Relevant equations
    I know about Cauchy sequences, the Ratio test, the Root test


    3. The attempt at a solution
    I really have no idea where i'm meant to start with this question. I didn't know whether I thne have to take more partial sums i.e. partial sums of the n-th partial sum and work with it from there, or something else. I would be really grateful if someone could help me out.
     
  2. jcsd
  3. Mar 24, 2009 #2

    CompuChip

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    I don't think your statement is true as you gave it.
    For example, let
    [tex]\begin{cases}
    a_0 & {} = 1 \\
    a_1 & {} = 0 \\
    a_n & {} = 2^{-n} \qquad (n > 1)
    \end{cases}[/tex]

    Then s = 3/2, |s - s0| = |5/6 - 1| = |-1/6| = 1/6 which is not smaller than a1 = 0.

    [edit]I had s = 5/6, I think it should be 3/2. Doesn't affect my argument though.[/edit]
     
    Last edited: Mar 24, 2009
  4. Mar 24, 2009 #3
    sorry i don't really understand that - how did you work out that s was 5/6? And did you just choose random values for a0, a1 and an? I have rechecked my homework question and that is exactly what it said!
     
  5. Mar 24, 2009 #4

    Dick

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    CompuChip just gave you a counterexample. To prove your conclusion, you also need to assume that a_n is a nonnegative sequence decreasing to zero monotonically. That might be what you are missing.
     
  6. Mar 24, 2009 #5
    okay, but i still don't understand how i'm meant to show the result, sorry. This question has got me completely flummoxed.
     
  7. Mar 24, 2009 #6

    Dick

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    Start by looking at the difference of two partial sums |s_m-s_n| for m<n. Try to show that's less than or equal to a_m+1. Write down the terms making up the difference and regroup them. This would show the sequence is Cauchy, right?
     
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