Proving ∑ n = 1 , 3 , 5.. ∞ 4 k sin 2 ⁡ ( n k ) n 2 = πWhere k any number greater than 0

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Proving      ∑          n      =      1      ,      3      ,      5..              ∞                  4      k              sin        2            ⁡              (                  n          k                )                    n      2        =  πWhere k any number greater than 0

a1p2ung1ls6t7 · Accepted Answer

Using (in the interval   (  0  ,  π  ) )            2      π      x      −              x                  2                      8    =      ∑          n      ≥      1                  sin      ⁡                        (          n                      x            2                    )                          2                            n              2                  (  1  )we have            2      π      x      −              x                  2                      8    =      ∑          n      ≥      1                  sin      ⁡                        (                      (            2            n            −            1            )                                x            2                    )                          2                                    (        2        n        −        1        )                    2              +      1    4        ∑          n      ≥      1                  sin      ⁡                        (          n          x          )                          2                            n              2            and using (1) again we get      1    4        ∑          n      ≥      1                  sin      ⁡                        (          n          x          )                          2                            n              2              =            π      x      −              x                  2                      8  so            π      x        8    =      ∑          n      ≥      1                  sin      ⁡                        (                      (            2            n            −            1            )                                x            2                    )                          2                                    (        2        n        −        1        )                    2            now put  x  =      2    k  and we have      π          4      k        =      ∑          n      ≥      1                  sin      ⁡                        (                      (            2            n            −            1            )                                1            k                    )                          2                                    (        2        n        −        1        )                    2            as wanted.

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