1. Not finding help here? Sign up for a free 30min 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!

Series, it's a riddle to me !

  1. Nov 26, 2008 #1
    Hey, guys !
    I have to calculate:

    [tex]\sum 1/(2n + 1)^2 [/tex]
    n from 0 to infinite

    Knowing that 1/n² = (pi)²/6

    The answer is (pi)²/8

    I don't know why I'm stuck here but I just don't get it...

    Greetings
     
    Last edited: Nov 26, 2008
  2. jcsd
  3. Nov 26, 2008 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Once you have what

    [tex]\sum_{n=1}^{ \infty } \frac{1}{n^2}[/tex] is, you should be able to find what

    [tex]\sum_{n=1}^{ \infty } \frac{1}{(2n)^2}[/tex] is pretty easily. And then use both of these to find the series sum that you're looking for
     
  4. Nov 26, 2008 #3
    It is given ...
    Calculate:
    [tex]\sum^{\infty}_{0} 1/(2n +1)^2 [/tex]

    if you know that:

    [tex]\sum^{\infty}_{1} 1/n^2 = (pi)^2 /6 [/tex]


    But I can't make it work :(
     
  5. Nov 26, 2008 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you calculate this?


    [tex]
    \sum_{n=1}^{ \infty } \frac{1}{(2n)^2}
    [/tex]

    Then what is


    [tex]
    \sum_{n=1}^{ \infty } \frac{1}{(2n)^2} + \sum_{n=0}^{ \infty } \frac{1}{(2n+1)^2}
    [/tex]
     
  6. Nov 26, 2008 #5
    haha I'm sorry for wasting your time !
    It was infact not hard, but had a few drinks too much last night.

    [​IMG]


    thank you anyway !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?