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: QM Value of an infinite sum

  1. Nov 17, 2005 #1
    In the course of doing my quantum homework I ran into a bit of a snag.

    In one of my calculations I need to replace the sum from n = 1 to infinity of 1/n^2 (for odd n only) with its number value.

    My book instructs me to get the information from a table and actualy gives the value (for odd n) of the infinite series for 1/n^4 = (pi^4)/96.

    In calculus the only series I evaluated to get numbers were geometric and telescoping.

    I tried to rewrite my own series as a telescoping and run from there, but I feel like I'm wasting time and trying to reinvent the wheel.

    A google search gave me little, and I think I'm using the wrong key words in my wikapedia search.

    Does anyone know where I can find my missing piece of info?

  2. jcsd
  3. Nov 17, 2005 #2
    The sum of 1/(2n-1)^2 for 1,2,3,4... is Pi^2/8 :)
  4. Nov 17, 2005 #3
    :blushing: hey, thanks a bunch!

    Do you happen to know how it was derived, and if so, was my telescoping series idea correct?
  5. Nov 17, 2005 #4

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    You can break the sum [tex] \sum^\infty_{n=1} \frac{1}{n^2} [/tex] into its even and odd terms. The odd series is the one you wanted. Now focus on the even series for second. Since each term is even, it can be written as [tex] n = 2 m [/tex] where [tex] m [/tex] runs over all the integers. This means that the even sum is actually just
    \sum^\infty_{n=2m} \frac{1}{n^2} = \sum^\infty_{m=1} \frac{1}{2^2}\frac{1}{m^2} = \frac{1}{4} \sum^\infty_{m=1} \frac{1}{m^2},
    but the last term is just the full sum over evens and odds. Thus we can say the even sum is one fourth of the full sum. This means that the odd sum is three fourths of the full sum. All you need to know now is the value of the full sum.

    You can obtain the value of the full sum by considering the Fourier series for the function [tex] x^2 [/tex] on the interval [tex] [-\pi,\pi] [/tex]. It turns out that the fourier coeffecients are related to [tex] \frac{1}{n^2} [/tex] which enables you to sum the series. The result, which you can also find in any table, is [tex] \pi^2/6 [/tex]. There are other ways to do it, but this way is one of the easiest in my opinion. That gives the value of the odd sum, the one you wanted, as
    [tex] \frac{3}{4} \frac{\pi^2}{6} = \frac{\pi^2}{8}. [/tex]
    Last edited: Nov 17, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook