Using Dirichlet series test I proved that the series ∑ n = 2 ∞ sin ⁡ n x n log ⁡ n converges for all x ∈ R

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Using Dirichlet series test I proved that the series             ∑              n        =        2            ∞                      sin        ⁡        n        x                    n        log        ⁡        n             converges for all   x  ∈      R

engaliar0l · Accepted Answer

Let       {          a      n        }   be a decreasing sequence of real numbers such that   n  ⋅      a    n    →  0. Then the series       ∑          n      ⩾      2            a    n    sin  ⁡  (  n  x  ) is uniformly convergent on       R  Thanks to Abel transform, we can show that the convergence is uniform on   [  δ  ,  2  π  −  δ  ] for all   δ  &amp;gt;  0. Since the functions are odd, we only have to prove the uniform convergence on   [  0  ,  δ  ]. Put       M    n    :=      sup          k      ⩾      n        k      a    k  , and       R    n    (  x  )  =      ∑          k      =      n              ∞            a    k    sin  ⁡  (  k  x  ). Fix   x  ≠  0 and N such that       1    N    ⩽  x  &amp;lt;      1          N      −      1      . Put for   n  &amp;lt;  N      A    n    (  x  )  =      ∑          k      =      n              N      −      1            a    k    sin  ⁡  k  x       and         B    n    (  x  )  :=      ∑          k      =      n              +      ∞            a    k    sin  ⁡  (  k  x  )  ,and for       A    n    (  x  )  :=  0Since       |    sin  ⁡  t      |    ⩽  t for   t  ≥  0 we have      |        A    n    (  x  )      |    ⩽      ∑          k      =      n              N      −      1            a    k    k  x  ⩽      M    n    x  (  N  −  n  )  ⩽            N      −      n              N      −      1            M    n    ,so       |        A    n    (  x  )      |    ⩽      M    n  If   N  &amp;gt;  n, we have after writing       D    k    =      ∑          j      =      0        k    sin  ⁡  j  x on   (  0  ,  δ  ] for some constant c. Indeed, we have       |        D    k    (  x  )      |    ⩽      1          2      (      1      −      cos      ⁡      x      )       and   cos  ⁡  x  =  1  −            x      2        2    (  1  +  ξ  ) where       |    ξ      |    ⩽      1    2   so   2  (  1  −  cos  ⁡  x  )  ⩾            x      2        2   and       |        D    k    (  x  )  ⩽            2        x  . Therefore      |        B    n    (  x  )      |    ⩽            2        x        ∑          k      =      N              +      ∞        (      a    k    −      a          k      +      1        )  +      a    N              2        x    =            2              2              x        a    N    ⩽  2      2    N      a    N    ⩽  2      2        M    n    .

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