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Homework Help: Partial Sum Approximation for Alternating Harmonic Series

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a value for n for which the nth partial sum is ensured to approximate the sum of the alternating harmonic infinite series to three decimal places.

    2. Relevant equations

    Sn = Ʃ(-1)^k+1*1/k = 1 - 1/2 + 1/3 - 1/4 + 1/5 - . . .

    S1 = 1
    S2 = 1 - 1/2
    S3 = 1 - 1/2 + 1/3
    S4 = 1 - 1/2 + 1/3 -1/4


    Sn = ?

    3. The attempt at a solution

    An attempt was made to derive a formula that would permit finding the value of the nth partial sum (e.g., S1000 = ?), but without success. It appears as though such a formula might not exist. Any help in this regard would be most appreciated.
  2. jcsd
  3. Feb 24, 2013 #2

    Ray Vickson

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    There is a formula, but it involves the non-elementary function "digamma" function Psi(x), defined as
    [tex] \Psi(x) = \frac{d}{dx} \ln \left( \Gamma(x) \right)
    = \frac{\Gamma^{\prime}(x)}{\Gamma(x)}.[/tex] Here, ##\Gamma## is the standard Gamma function. According to Maple 11 the sum is:
    Sum((-1)^(k-1)/k,k=1..N):lprint(%);Sum((-1)^(k-1)/k,k = 1 .. N) <--echo the input

    S:=value(%): lprint(S);
    ln(2)+1/2*(-1)^(N+1)*Psi(1+1/2*N)+1/2*(-1)^N*Psi(1/2*N+1/2). <--- output

    That is,
    [tex] \sum_{k=1}^N (-1)^{k-1} \frac{1}{k}
    = \ln(2) + \frac{1}{2}(-1)^N \left[ \Psi\left(\frac{N}{2}+\frac{1}{2}\right) -
    \Psi\left(\frac{N}{2}+1\right) \right].[/tex]
  4. Feb 24, 2013 #3
    I doubt you know digamma functions, so that's not going to be the way to solve it.

    Let [itex]s=\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n}[/itex] and let [itex]s_n[/itex] be the n-th partial sum.

    Can you show that [itex]s_{2n+1}[/itex] is decreasing? And that [itex]s_{2n}[/itex] is increasing? Can you conclude from that that [itex]s_{2n}\leq s \leq s_{2n+1}[/itex]?

    What can you say about [itex]|s_{2n+1}-s_{2n}|[/itex]?? What does this say about [itex]|s-s_n|[/itex]?? Do you obtain a formula for how well the approximation is?
  5. Feb 24, 2013 #4

    Ray Vickson

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    I agree about it not being the way to solve it, but he/she did ask. If one is going to use a computer, one might as well just do a recursive computation (although avoiding subtractive roundoff errors is an issue for getting accurate numbers).
  6. Feb 24, 2013 #5
    Sure. I was not criticizing your post as it was very informative! I was merely informing the OP of a more elementary method.
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