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Prove the sum of this series =pi/3

  1. Jan 25, 2008 #1
    prove the sum of this series =pi/3

    [tex]1 +\frac{ 1}{5} -\frac{ 1}{7} - \frac{1}{11 }+\frac{1}{13} + \frac{1}{17} - ...[/tex]
     
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  3. Jan 25, 2008 #2

    CompuChip

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    Surely you must have some idea.
    Why don't you start by writing it out in sum notation ([tex]\sum_{n = 0}^\infty \cdots[/tex])?
     
  4. Jan 25, 2008 #3

    Gib Z

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    Cheeky CompuChip :P You know he knows the solution, he just wants to tease us :P Look at the rest of his posts to see what I mean.

    I've got a nice solution for this though =] I'll post it when someone else asks for it, because I want to let some other guys give this a try. It's not actually that hard, this one.
     
  5. Jan 26, 2008 #4
    a twisted Gregorian pi series , missing fractions of odd multiples of 3, is all what it is.





    i Am not sure whether he know all the solutions. all his posts where only questions no replies or answers.
    even if he knows all the answer , he should be interested in knowing different method of doing it.which usually people do less they are descendent's of newton or Einstein .
     
    Last edited: Jan 26, 2008
  6. Jan 26, 2008 #5
    Pi happens to be a transcendal no. meaning-you can't get its value by any algebraic method. If this a puzzle, you could get a nobel prize for the answer.
     
  7. Jan 26, 2008 #6
    That's not what transcendental means. Transcendental just means that it is not the root (zero) of any polynomial with integer coefficients. This has nothing to do with infinite sums such as this one, and there are many such sums (and products as well) that are quite well-known. The Wikipedia article on Pi and this MathWorld article list quite a lot of them.
     
  8. Jan 26, 2008 #7

    CompuChip

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    A Nobel Prize in what? Mathematics?
     
  9. Jan 26, 2008 #8

    Gib Z

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    Not to mention, The Nobel Prize is not award to Mathematicians unless the work has had a serious application in one of the awarded fields, such as the case of John Nash. I doubt this one would.

    Anyone made any progress? It sounds like some of you already have a solution, just can't be bothered to post it >.<" Ill post mine tomorrow.

    EDIT: CompuChip beat me to it !
     
  10. Jan 26, 2008 #9

    CompuChip

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    Hmm, can't be that hard if even I solved it in a little under 5 minutes. (Though I must admit, having the answer already helped me think of arctan a lot sooner :smile:)
     
  11. Jan 26, 2008 #10
    believe me it wont take more than a five min or even less than a min
     
  12. Jan 26, 2008 #11

    Gib Z

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    Aww dont say that! I wanted to sound really smart for having a solution!! Ruin my fun :(
     
  13. Jan 27, 2008 #12
    thanks for all

    from these http://mathworld.wolfram.com/PiFormulas.html


    let[tex] S=1- \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ... = \frac{\pi}{4}.[/tex]

    thus [tex] \frac{1}{3} S= \frac{1}{3} - \frac{1}{9}+ \frac{1}{15} - \frac{1}{21} + ... = \frac{\pi}{12}.[/tex]

    series = [tex] S + \frac{1}{3} S =\frac{\pi}{3} [/tex]
     
  14. Jan 27, 2008 #13
    Sorry for my answer! I am just a grade10 student and I just had read of the word in the library. I will remember not to reply about thing I dont know. Sorry Gib Z and thanx Moo of Doom!!
     
    Last edited: Jan 27, 2008
  15. Jan 27, 2008 #14
    Heh, interesting.
     
  16. Jan 28, 2008 #15
    that mine method too
     
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