1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Sure. I was not criticizing your post as it was very informative! I was merely informing the OP of a more elementary method.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Partial Sum Approximation for Alternating Harmonic Series
Loading...