Find the sum of the series: lim n → ∞ sin ⁡ 1 1 + sin ⁡ 2 2 + sin ⁡ 3 3 + . . . + sin ⁡ n n I have no clue how to find this.

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Find the sum of the series:      lim          n      →      ∞                  sin      ⁡      1        1    +            sin      ⁡      2        2    +            sin      ⁡      3        3    +  .  .  .  +            sin      ⁡      n        n  I have no clue how to find this.

Sawyer Day · Accepted Answer

consider the Fourier series of the   2  π-periodic function that equals             π      −      x        2   over   (  0  ,  2  π  )Yet another way is to use the limit case of the Euler-Maclaurin summation formula: since   sinc  (  x  ) is a very well-behaved even analytic function, we simply have:      ∑          n      ≥      1                  sin      ⁡      n        n    =      ∫          0              +      ∞                  sin      ⁡      x        x      d  x  −      1    2    .

Semaj Christian · Accepted Answer

Here&#039;s a step-by-step breakdown using the fact that   sin  ⁡  x  =  Im  (      e          i      x        )      lim          n      →      ∞            ∑          k      =      1        n              sin      ⁡      k        k    =      lim          n      →      ∞            ∑          k      =      1        n              Im      (              e                  i          k                    )        k    =  Im      (          ∑              k        =        1            ∞                      e                  i          k                    k        )    =  Im      (          ∑              k        =        1            ∞                      (                  e          i                          )          k                    k        )  Now recall that the Taylor Series for   −  log  ⁡  (  1  −  x  ) is       ∑          k      =      1        ∞              x      k        k    =  Im  (  −  log  ⁡  (  1  −      e    i    )  )  =  Im  (  Arg  (  1  −      e    i    )  )We now express   1  −      e    i   as   1  −  cos  ⁡  (  1  )  −  i  sin  ⁡  (  1  ). Noting that   1  −  cos  ⁡  (  x  )  &amp;gt;  0    (  ∀  x  ∈      R    ) we can simply let   Arg  (  1  −      e    i    )  =  arctan  ⁡      (                  sin        ⁡        (        1        )                    1        −        cos        ⁡        (        1        )              )    =  arctan  ⁡      (    cot    ⁡    (          1      2        )    )

Find the sum of the series: <munder> <mo movablelimits="true" form="prefix">lim <mro

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