Convergence / divergence examination ∑ k = 1 ∞ sin k ⁡ ( a ) k

Question

Convergence / divergence examination       ∑          k      =      1              ∞                          sin                  k                    ⁡                        (          a          )                      k

talhekh · Accepted Answer

If   sin  ⁡  (  a  )  =  0, then the series obviously converges.If   sin  ⁡  (  a  )  =  1, then the series rewrites             ∑              1        k             which diverges.If   sin  ⁡  (  a  )  =  −  1, then the series rewrites             ∑                        (          −          1                      )            k                          k             which converges by Alternating Test.If       |    sin  ⁡  (  a  )      |    &amp;lt;  1 and   sin  ⁡  (  a  )  ≠  0, then one has      |                            sin                      k            +            1                          ⁡        (        a        )                  /                (        k        +        1        )                              sin          k                ⁡        (        a        )                  /                k              |    =      |    sin    ⁡    (    a    )    |    ×            k      +      1        k    ⟶      |    sin    ⁡    (    a    )    |  Because       |    sin  ⁡  (  a  )      |    &amp;lt;  1, then by the ratio test, the series converges.In conclusion, the series converges iff   sin  ⁡  (  a  )  ≠  1, i.e.                              the series converges iff           a          ≠                                    π              2                                 mod           2          π

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