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Math Help - Series involving summation of arctan()

  1. #1
    Member Pranas's Avatar
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    Series involving summation of arctan()

    I am very new to this kind of thing, so I would really appreciate at least a little hint:

    Main goal is to evaluate

    \displaystyle \[\sum\limits_{n = 1}^{ + \infty } {\arctan \left( {\frac{1}{{2 \cdot {n^2}}}} \right)} \]

    and in the process to come up with the partial summation formula.


    I actually figured out the nature of adding two arctan() results

    \displaystyle \[\arctan \left( x \right) + \arctan \left( y \right) = \arctan \left( {\frac{{x + y}}{{1 - x \cdot y}}} \right)\]

    but still not able to move forward...
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  2. #2
    Super Member girdav's Avatar
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    Put x:=\frac 1{n-1} and y:=-\frac 1{n+1}.
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by girdav View Post
    Put x:=\frac 1{n-1} and y:=-\frac 1{n+1}.
    I believe that would give me

    \displaystyle \[\arctan \left( {\frac{2}{{{n^2}}}} \right)\]

    so maybe a little correction would produce

    \displaystyle \[{\arctan \left( {\frac{1}{{2 \cdot {n^2}}}} \right)}\]

    precisely?


    By the way, how do you choose this kind of substitution? I am not mentally able to comprehend it
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  4. #4
    Super Member girdav's Avatar
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    Second attempt x= \frac 1{2n-1} and y = \frac 1{2n+1}.
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  5. #5
    Member Pranas's Avatar
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    Quote Originally Posted by girdav View Post
    Second attempt x= \frac 1{2n-1} and y = \frac 1{2n+1}.
    I believe x= \frac 1{2n-1} and \[y = \frac{{ - 1}}{{2n + 1}}\]

    But how to comprehend it other than come and ask in MHF forum?

    I understand that that I need x and y such that

    \[\frac{{x + y}}{{1 - x \cdot y}} = \frac{1}{{2 \cdot {n^2}}}\]

    is true for every n. But how to know the form of x and y?
    Last edited by Pranas; April 13th 2011 at 10:48 AM.
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  6. #6
    Super Member girdav's Avatar
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    You can hope that you will find a telescopic series so you try to find x_n such that \frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}. We can try x_n =\frac 1{an+b} and find a and b.
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  7. #7
    Member Pranas's Avatar
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    Quote Originally Posted by girdav View Post
    You can hope that you will find a telescopic series so you try to find x_n such that \frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}. We can try x_n =\frac 1{an+b} and find a and b.
    Thanks for help. Apparently that telescopes pretty flawlessly and produces

    \displaystyle \[\mathop {\lim }\limits_{n \to \infty } \left( {\frac{\pi }{4} - \arctan \left( {\frac{1}{{2n + 1}}} \right)} \right) = \frac{\pi }{4}\]

    P.S. Is anything related to

    \displaystyle \[Arc\tan \left( \chi  \right) = \frac{i}{2} \cdot Ln\left( {\frac{{i + \chi }}{{i - \chi }}} \right)\]

    is a no-go while evaluating similar sums?
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