Limit with a summation and sine: how to calculate lim n → ∞ n 2 ∑ k = 0 n − 1 sin ⁡ ( 2 π k n ) ?

Question

Limit with a summation and sine: how to calculate       lim          n      →      ∞            n    2        ∑          k      =      0              n      −      1        sin  ⁡      (                  2        π        k            n        )   ?

laure6237ma · Accepted Answer

∑          k      =      0              n      −      1        sin  ⁡  (  2  π  k      /    n  ) is the imaginary part of       ∑          k      =      0              n      −      1            e          i      2      π      k              /            n       . We can evaluate the latter sum directly since it&#039;s a truncated power series:      ∑          k      =      0              n      −      1            z    k    =                    z        n            −      1              z      −      1      provided that   z  ≠  1, so      ∑          k      =      0              n      −      1            e          i      2      π      k              /            n        =                    e                  i          2          π                    −      1                      e                  i          2          π                      /                    n                    −      1        =            1      −      1                      e                  i          2          π                      /                    n                    −      1        =  0Hence also      ∑          k      =      0              n      −      1        sin  ⁡  (  2  π  k      /    n  )  =  0for every positive integer n. So your limit is      lim          n      →      ∞            n    2        ∑          k      =      0              n      −      1        sin  ⁡  (  2  π  k      /    n  )  =      lim          n      →      ∞        (      n    2    ⋅  0  )  =      lim          n      →      ∞        0  =  0

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