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!

Homework Help: Sum of infinite series

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find the sum of a given infinite series when [itex]|x|<1[/itex] (which is the radius of this series)

    2. Relevant equations

    [itex]\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n+1}}{4n^2-1} [/itex]

    3. The attempt at a solution
    I've tried to do the following:

    [itex]S'(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n}}{2n-1} \\
    S''(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{2nx^{2n-1}}{2n-1}\\
    S'''(x) = 2\sum_{n=1}^{∞}(-1)^{n+1}nx^{2(n-1)}\\[/itex]

    And I was thinking about substitution [itex]t = x^2[/itex], but I had no success.
  2. jcsd
  3. Jun 4, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    hi stefaneli! :smile:
    [itex]= x\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}[/itex] :wink:
  4. Jun 4, 2012 #3
    Thanks tiny-tim. I don't know how I haven't noticed. Stupid.:) But can it be done the way I started? Just curious.
  5. Jun 4, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    i suppose so, but you'd need to be very careful when you change the variable :frown:
  6. Jun 4, 2012 #5
    Try doing partial fractions. IE 1/(4n^2-1)= 1/2*(1/(2n-1)-1/(2n+1))
  7. Jun 4, 2012 #6
    Your answer will be something like 1/2*((x^2-1)arctan(x)-x), just use partial fractions, and get it into a more manageable form
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook