How to prove ∑ k = 1 n cos ⁡ ( 2 π k n ) = 0 for any n&gt;1?

Question

How to prove       ∑          k      =      1        n    cos  ⁡  (            2      π      k        n    )  =  0 for any n&amp;gt;1?

cefflid6y · Accepted Answer

We have      ∑          k      =      1              n        cos  ⁡      (                  2        π        k            n        )    =      Re        (          ∑              k        =        1                    n                    e              2        π        i        k                  /                n              )  and so      ∑          k      =      1              n            e          2      π      i      k              /            n        =                    e                  2          π          i                      /                    n                            (        1        −                  e                      2            π            i                          )                    1      −              e                  2          π          i                      /                    n                    and notice that      e          2      π      i        =  cos  ⁡      (    2    π    )    +  i  sin  ⁡      (    2    π    )    =  1so the claim follows.

fythynwyrk0 · Accepted Answer

One approach is to write  cos  ⁡      (                  2        π        k            n        )    =      1          2      sin      ⁡              (                              2            π                    n                )                  (    sin    ⁡          (                        2          π          (          k          +          1          )                n            )        −    sin    ⁡          (                        2          π          (          k          −          1          )                n            )        )  Now, we have converted the sum into a telescoping sum, which we can evaluate directly as                              ∑                      k            =            1                    n                cos        ⁡                  (                                    2              π              k                        n                    )                                    =                  1                      2            sin            ⁡                          (                                                2                  π                                n                            )                                                (          sin          ⁡                      (                                          2                π                (                n                +                1                )                            n                        )                    +          sin          ⁡                      (                                          2                π                (                n                )                            n                        )                    )                                                                −                  1                      2            sin            ⁡                          (                                                2                  π                                n                            )                                                (          sin          ⁡                      (                                          2                π                (                2                −                1                )                            n                        )                    +          sin          ⁡                      (                                          2                π                (                1                −                1                )                            n                        )                    )                                                                =                  1                      2            sin            ⁡                          (                                                2                  π                                n                            )                                                (          sin          ⁡                      (                                          2                π                (                n                +                1                )                            n                        )                    −          sin          ⁡                      (                                          2                π                (                2                −                1                )                            n                        )                    )                                                                =                  1                      2            sin            ⁡                          (                                                2                  π                                n                            )                                                (          sin          ⁡                      (                                          2                π                            n                        )                    −          sin          ⁡                      (                                          2                π                            n                        )                    )                                                                =        0

How to prove <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> k

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