Evaluating a summation of inverse squares over odd indices ∑ n = 1 ∞ 1 n 2 = π 2 6

Question

Evaluating a summation of inverse squares over odd indices      ∑          n      =      1        ∞        1          n      2        =            π      2        6

Savion Chaney · Accepted Answer

Note that      ∑          n       is even                  1              n        2              =      ∑          k      =      1              ∞                  1              (        2        k                  )          2                      =            1      4            ∑          k      =      1              ∞                  1              k        2              =                    ζ        (        2        )            4      Also,      ∑          k      =      1              ∞                  1              k        2              =      ∑          k       is odd                  1              k        2              +      ∑          k       is even                  1              k        2            Hence,      ∑          k       is odd                  1              k        2              =            3      4        ζ  (  2  )

Adrian Brown · Accepted Answer

This can be shown in a similar way to Euler&#039;s proof of   ζ  (  2  )  =            π      2        6  , which starts with the function             sin      ⁡      (      x      )        x   (i.e. the sinc function). Here we start with the cosine function which can be expressed as the infinite product                    cos        ⁡        (        x        )                            =                  ∏                      n            =            1                    ∞                          (          1          −                                    4                              x                2                                                                    π                2                            (              2              n              −              1                              )                2                                              )                                                  =                  (          1          −                                    4                              x                2                                                    π              2                                )                          (          1          −                                    4                              x                2                                                    9                              π                2                                              )                          (          1          −                                    4                              x                2                                                    25                              π                2                                              )                .        .        .                                          =        1        −                  x          2                          (                      4                          π              2                                +                      4                          9                              π                2                                              +                      4                          25                              π                2                                              +          .          .          .          )                +        .        .        .            cos(x) can also be expressed by the following Maclaurin series expansion:  cos  ⁡  (  x  )  =      ∑          n      =      1        ∞              (      −      1              )        n                    (      2      n      )      !            x          2      n        =  1  −            x      2              2      !        +            x      4              4      !        −            x      6              6      !        +  .  .  .Comparing the       x    2   coefficients gives:  −      1          2      !        =  −      4          π      2            (    1    +          1      9        +          1      25        +    .    .    .    )  Thus,      ∑          n      =      1        ∞        1          (      2      n      −      1              )        2              =            π      2        8

Evaluating a summation of inverse squares over odd indices sum_(n=1)^infty 1/n^2 =pi^2/6

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