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Is the sum of this series correct?

  1. Feb 28, 2006 #1
    I am given the following series and its sum
    [tex]
    \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8}
    [/tex]



    I need to find the sum of this series
    [tex]
    \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}}
    [/tex]

    Is the method I used below this the correct approach? Is there a better or different way?

    From the first series where n=1 we have,

    [tex]
    a_n = \frac{1}{{\left( {2n - 1} \right)^2 }}
    [/tex]

    [tex]
    = {\rm{\{ 1, 1/9, 1/25, }}...{\rm{\} }}
    [/tex]

    Therefore I concluded that we can subtract off the first two terms of
    the above sequence to get the sum when n starts at 3. Hence,

    [tex]
    \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8} - 1 - \frac{1}{9}
    [/tex]
     
  2. jcsd
  3. Feb 28, 2006 #2
    Your reasoning is correct.
     
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