I tried to evaluate the limit of the following sequence: 1 n ( sin ⁡ ϕ n + sin ⁡ 2 ϕ n + … + sin ⁡ ( n − 1 ) ϕ n )

Question

I tried to evaluate the limit of the following sequence:      1    n        (    sin    ⁡          ϕ      n        +    sin    ⁡                  2        ϕ            n        +        …        +    sin    ⁡                  (        n        −        1        )        ϕ            n        )

Cristopher Barrera · Accepted Answer

You could use the following  sin  ⁡  (  x  )  +  sin  ⁡  (  2  x  )  +  ⋯  +  sin  ⁡  (  k  x  )  =                    sin        ⁡                  (                                    k              +              1                        2                    x          )                ⋅        sin        ⁡                  (                                    k              x                        2                    )                            sin        ⁡                  (                      x            2                    )                    Then we taking   x  =      ϕ    n   and   k  =  n  −  1 we have                              1          n                          (          sin          ⁡                      ϕ            n                    +          sin          ⁡                                    2              ϕ                        n                    +          ⋯          +          sin          ⁡                                    (              n              −              1              )              ϕ                        n                    )                                    =                  1          n                                                    sin              ⁡                              (                                  n                  2                                                  ϕ                  n                                )                            ⋅              sin              ⁡                              (                                                      (                    n                    −                    1                    )                                    2                                                  ϕ                  n                                )                                                    sin              ⁡                              (                                  ϕ                                      2                    n                                                  )                                                                                      =                              sin            ⁡            (                          ϕ              2                        )            ⋅            sin            ⁡                                          (                                      (            1            −                          1              n                        )                          ϕ              2                                                      )                                                                        sin              ⁡              (              ϕ                              /                            2              n              )                                      1                              /                            n                                          Doing   n  →  +  ∞ we take            lim              n        →        +        ∞                    1      n              (      sin      ⁡              ϕ        n            +      sin      ⁡                        2          ϕ                n            +      ⋯      +      sin      ⁡                        (          n          −          1          )          ϕ                n            )        =          2      ϕ              sin      2        ⁡    (    ϕ          /        2    )    .  You can also use Riemann sums, for example in this case                                          lim                          n              →              +              ∞                                            1            n                                (            sin            ⁡                          ϕ              n                        +            sin            ⁡                                          2                ϕ                            n                        +            ⋯            +            sin            ⁡                                          (                n                −                1                )                ϕ                            n                        )                                              =                  lim                      n            →            +            ∞                                    ∑                      k            =            1                                n            −            1                          sin        ⁡        (                              k            ϕ                    n                )                  1          n                                                  =                  ∫          0          1                sin        ⁡        (        ϕ        x        )        d        x                                          =                  1          ϕ                (        1        −        cos        ⁡        (        ϕ        )        )                                          =                  2          ϕ                          sin          2                ⁡        (        ϕ                  /                2        )

I tried to evaluate the limit of the following sequence: 1 n </mfrac> ( si

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New Questions in Calculus 1