How do I evaluate the sum ∑ n = 1 , 3 , 5 , . . . ∞ 1 n 4

Question

How do I evaluate the sum       ∑          n      =      1      ,      3      ,      5      ,      .      .      .              ∞            1          n              4

elladanzaez · Accepted Answer

Define Riemann zeta function as  ζ  (  s  )  =      ∑          n      ≥      1                  1              n        s            Then we have                              ∑                                                                                          n                    ≥                    1                                                                                        n                     odd                                                                                                            1                          n              s                                                          =                  ∑                      n            ≥            1                                                1                          n              s                                      −                  ∑                                                                                          n                    ≥                    1                                                                                        n                     even                                                                                                            1                          n              s                                                                        =        ζ        (        s        )        −                  ∑                      n            ≥            1                                                1                          (              2              n                              )                s                                                                                      =        (        1        −                  2                      −            s                          )        ζ        (        s        )            This indicates that                                          480                          ℏ                              2                                                                        π                              4                                      m                          a                              2                                                    ⋅                  ∑                      n            =            1            ,            3            ,            5            ,            .            .            .                                ∞                                    1                      n                          4                                                          =                              480                          ℏ                              2                                                                        π                              4                                      m                          a                              2                                                                          15            16                          ζ        (        4        )                                          =                                            450                              ℏ                2                                                                    π                4                            m                              a                2                                                    ζ        (        4        )            Finally, using the fact that   ζ  (  4  )  =                    π        4            90      , we conclude            480              ℏ                  2                                    π                  4                    m              a                  2                      ⋅      ∑          n      =      1      ,      3      ,      5      ,      .      .      .              ∞            1          n              4              =                    5                  ℏ          2                            m                  a          2

Cooper Krause · Accepted Answer

Another method is by using Fourier coefficients of the function f on   [  −  π  ,  π  ] defined by  f  (  x  )  =      |    x      |  Then using integration by parts and using Fourier coefficient definition, one can obtain            f      ^        (  0  )  =      π    2  and for   n  ≠  0            f      ^        (  n  )  =      {                            0                                       if                           n                         is even                                                                          −              2                                                      n                2                            π                                                             if                           n                         is odd                                  Now, the famous Parseval&#039;s identity (which relates function with it&#039;s Fourier coefficients) states that if   f  ∈      L    2    (  [  −  π  ,  π  ]  ), then      ∑          n      =      −      ∞              ∞            |              f      ^        (  n  )            |        2    =      1          2      π            ∫          −      π              π            |    f  (  x  )            |        2    d  xFor our case, we can write      |              f      ^        (  0  )            |        2    +      ∑          n      ∈              Z            ∖      {      0      }            |              f      ^        (  n  )            |        2    =      1          2      π            ∫          −      π              π            x    2    d  xThus,            π      2        4    +      ∑          n      =      ±      1      ,      ±      2      ,      .      .      .            4                  n        4                    π        2              =      1          2      π                  2              π        3              3    =            π      2        3  which implies      4          π      2        ⋅  2      ∑          n      =      1      ,      3      ,      .      .      .            1          n      4        =            π      2        12  and hence      ∑          n      =      1      ,      3      ,      .      .      .            1          n      4        =            π      2        12    ⋅            π      2        8    =            π      4        96  And hopefully you can use it in your computation.

How do I evaluate the sum <munderover> ∑ <mrow class="MJX-TeXAtom-ORD">

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