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: Calculating a Summation

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    It can be shown that ∑(n=1) to (n=∞) of 1/n² = π²/6

    use this fact to show that ∑(n=1) to (n=∞) of 1/(2n-1)² = π²/8

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 24, 2010 #2

    Mark44

    Staff: Mentor

    What have you tried? You need to make an attempt at a solution before anyone can give you any help.
     
  4. Feb 24, 2010 #3
    Hint: It is so easy, you can do it in your head in less than ten seconds.
     
  5. Mar 1, 2010 #4
    My attempt which needs to be shown on paper...

    First I wrote out the first few terms of ∑(n=1) to (n=∞) of 1/n². So 1+1/4+1/9+1/16+...+1/n² = π²/6.

    I then just tried to simply substituted (2n-1) into the previous summation in place of just n to somehow show that ∑(n=1) to (n=∞) of 1/(2n-1)² = π²/8... but I have no idea how to accurately demonstrate the rest of the proof.
    Any additional help would be appreciated.
     
  6. Mar 1, 2010 #5

    Mark44

    Staff: Mentor

    What are the first few terms of
    [tex]\sum_{n = 1}^{\infty}\frac{1}{(2n - 1)^2}[/tex]? How does this series relate to the other series?
     
  7. Mar 2, 2010 #6
    When writing out the first few terms of Σ1/(2n-1)², I noticed that this represents the terms when n is some odd number from the first summation, ∑1/n², but I'm stuck as to the proof with setting that equal to 8.
     
    Last edited: Mar 2, 2010
  8. Mar 2, 2010 #7

    Mark44

    Staff: Mentor

    Show me the first five terms of Σ1/(2n-1)².
    Show me the first five terms of Σ1/n².

    Pretend for the time being that you don't what the answer is supposed to be. It seems to me that you are too focused on the pi^2/8 result, and not focused enough on how to get there.
     
  9. Mar 2, 2010 #8
    Σ1/(2n-1)²: 1+1/9+1/25+1/49+1/81+...

    Σ1/n²: 1+1/4+1/9+1/16+1/25+...
     
  10. Mar 2, 2010 #9

    Mark44

    Staff: Mentor

    And which terms are missing from the first series that are in the second series? What is Σ1/n² - Σ1/(2n-1)² ?
     
  11. Mar 2, 2010 #10
    1/4+1/16+1/36+...

    Σ1/n² - Σ1/(2n-1)² then equals Σ1/(2n)² from n=1 to infinity
     
  12. Mar 2, 2010 #11

    Mark44

    Staff: Mentor

    Now, can you manipulate this series--Σ1/(2n)²--to get to something you know?
     
  13. Mar 2, 2010 #12
    Thank you, I understand all of that perfectly... The issue I am having with the problem, however, is that I don't recall ever being taught in my calculus class exactly how to determine the Sn of a series like Σ1/(2n)²... If there is some technique that I could use please let me know.
     
  14. Mar 2, 2010 #13
    Just simplify 1/(2n)²
     
  15. Mar 2, 2010 #14
    How exactly?
     
  16. Mar 2, 2010 #15

    Mark44

    Staff: Mentor

    Write 1/(2n)² in a different way. C'mon, this ain't rocket science...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook