Absolute and uniform convergence of ∑ n = 1 n = ∞ 2 n sin ⁡ 1 3 n z

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Absolute and uniform convergence of       ∑          n      =      1              n      =      ∞            2          n        sin  ⁡      1                  3                  n                    z

Alberto Duffy · Accepted Answer

To get the absolute convergence, just use the inequality       |    sin  ⁡  t      |    ≤      |    t      |   for any real number t, which be established thanks to the fundamental theorem of analysis:      |    sin  ⁡  t      |    =      |        ∫    0    t    −  cos  ⁡  s  d  s      |    ≤      ∫    0                  |            t              |              cos  ⁡  s  d  s  ≤      |    t      |  Let S a subset of       C   of the form   S  :=  {  z  =  0  &amp;lt;      |    z      |    &amp;lt;  δ  }, we have to show that the convergence is not uniform on S. To see that, note that for n large enough, say larger than       n    0   contains points of the form       3          −      n      , so       sup          z      ∈      S            |          ∑              n        =        N                    +        ∞                    2      n        sin    ⁡          1                        3          n                z              |    ≥      2    N    sin  ⁡  1  , since the terms for   z  =      3          −      n       are non-negative if   n  ≥      n    0

Absolute and uniform convergence of <munderover> ∑ <mrow class="MJX-TeXA

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