Why is ∑ n = 1 ∞ 1 n sin ⁡ ( x n n ) uniformly convergent?

Question

Why is       ∑          n      =      1              ∞            1    n    sin  ⁡      (                  x        n                    n              )   uniformly convergent?

Tianna Deleon · Accepted Answer

Use Weirestrass M test. We have       |    sin  ⁡  (  x  )      |    ≤      |    x      |    ,    ∀  x  ∈      R              |              1      n        sin    ⁡          (                        x          n                          n                    )              |        ≤          |              1      n              |              |                      x                  n                            n                    |        ≤          1      n              1              n            . This is because   x  ∈  [  −  1  ,  1  ]Since       ∑          n      =      1              ∞            1          n              n             converges you have uniform convergence by Weirestrass-M Test .

Llubanipo · Accepted Answer

It&#039;s a serie of definitely alternating sign. So, we must study the absolute convergence. Let:      a    n    =      |          1      n        sin    ⁡          (                        x          n                          n                    )        |  We can apply the asympotic criteria because       a    n   is always non-negative. We have:      lim          n      →      +      ∞            a    n    =      lim          n      →      +      ∞            |          1      n        sin    ⁡          (                        x          n                          n                    )        |    ∼      lim          n      →      +      ∞            |                  x        n                    n        ⋅                  n                      |    ≤      1          n      ⋅              n              =      1          n                        3          2                    Here, we have use the fact that                     |                    x        n                    |                    n        →  0 if   x  ∈  [  −  1  ,  1  ]. The series       ∑          n      =      1              +      ∞            1          n                        3          2                     converges, thus by comparison test, converges also your series.

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