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Sum of an infinite alternating series

  1. Sep 8, 2006 #1
    I am having trouble finding [tex]\sum \frac{(-1)^n}{n^2} [/tex]. (from n=1 to n=infinity)

    I know that [tex]\sum \frac{1}{n^2}[/tex] is [tex]\frac{\pi^2}{6} [/tex]. But the sum I need is slightly different than that, it is alternating....

    Any ideas/help would be appreciated! Thx
     
  2. jcsd
  3. Sep 8, 2006 #2

    radou

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    The only hint I can give is that the series do converge, since [tex]lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=lim_{n \rightarrow \infty} \frac{\frac{(-1)^{n+1}}{(n+1)^2}}{\frac{(-1)^n}{n^2}}=lim_{n \rightarrow \infty} \frac{-n^2}{(n+1)^2}=-1 < 1[/tex], considering d'Alembert's criteria...
     
  4. Sep 8, 2006 #3
    EDIT: Forget what I said. Mistake.
     
    Last edited: Sep 8, 2006
  5. Sep 8, 2006 #4

    shmoe

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    Hint-find the sum of the even terms of this series.
     
  6. Sep 8, 2006 #5
    and then subtract the odd terms? I must admit I dont know how to sum that series.... When I took calc 2 we only really tested for convergence and divergence.... And Im pouring through my book (stewert 4th ed) and all I see is convergence and divergence, the only sums in my book are pre given (like the one you quoted me on)
     
    Last edited: Sep 8, 2006
  7. Sep 8, 2006 #6

    JasonRox

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    I feel your pain. That book is terrible for learning anything that is formal.

    EDIT: Follow shmoe's hint and after that a little algebra should do the rest.
     
    Last edited: Sep 8, 2006
  8. Sep 8, 2006 #7
    Ok, I think i got it....

    I find the sum of the even terms, which is the sum [tex]\sum\frac{1}{(2n)^2}[/tex] which equals [tex]\frac{\pi^2}{24}[/tex]. I then subtract the even terms from the total and get [tex]\frac{3\pi^2}{24}[/tex]

    Im thinking this is right, but if someone could verify it for me I would appreciate it, and THX so much for the help

    EDIT- No, I think I did it backwards. My odd terms are negative, so I need to subtract the odd terms sum from the even terms sum, giving me [tex]-\frac{3\pi^2}{24}[/tex]
     
    Last edited: Sep 8, 2006
  9. Sep 8, 2006 #8

    shmoe

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    Close! This is what you get when you subract the whole sum from the sum of just the even terms. But what is left in this sum?

    You took

    [tex]\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots[/tex]

    and subtracted

    [tex]\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots[/tex]

    What terms remain?
     
  10. Sep 8, 2006 #9
    Ok, try again...

    what i did was take the even terms and subtract the whole sum... which is not what i wanted.

    what I wanted was the even terms minus the odd terms. So I find the odd terms by taking the whole sum minus the even terms.

    Then I subtract the odd terms from the even terms and get [tex]\frac{-\pi^2}{12}[/tex]

    God I hope that is right... I have spent alot of time and it is due soon (via internet)

    Thx for your help and feed back!! Is [tex]\frac{-\pi^2}{12}[/tex] right?!?
     
  11. Sep 8, 2006 #10

    shmoe

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    That's good!

    It's not strictly necessary to find the sum of the odd terms, you can also take 2 times the even terms and then subtract the whole series. Nothing wrong with finding the sum of the odd terms of course, but a (very slightly) different view might be usefull to see.
     
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