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Summation of series

  1. Mar 9, 2010 #1
    what is the sum of this series ??

    [tex] \sum_{n=0}^{\infty}n^{b}e^{ian} [/tex] for every a and b to be Real numbers

    from the definition of POlylogarithm i would say [tex] \sum_{n=0}^{\infty}n^{b}e^{ian}= Li_{-b}(e^{ia})[/tex]

    however i would like to know if the sum is Cesaro summable and what it would be its Cesaro sum [tex] C(k,a,b) [/tex] for k bigger than 'b' , and a=0 , thanks.
  2. jcsd
  3. Mar 9, 2010 #2

    I'm not really familiar with the Cesaro summability, so I'm not possibly answering your question.. at least not the last one.

    Anyway - here's my suggestion:

    First, let c:=ia

    The series you typed is a Dirichlet series. It converges locally unimormly towards a holomorphic function on some right half plain of C. Now the local uniformity allows us to interchange summation and differentiation/integration. (another way of proving the latter is by using measure theory and the counting measure on R or C - limits can be interchanged since the 'e-to-the-i*n' term is bounded for every n by the constant function 1 and the e-function is monotone, so the partial sum would also be bounded (dominated convergence thm.))

    This particular series is obtainable from the series containing only the e-term by differentiating it b times wrt. c . So we could pull out a differential operator to the b-th power in front of the sum by linearity. Then realising the series contains only a 'e-to-the-i' term I could try using the geometric series and do some algebra...

    I hope this could be somehow helpful

    best regards,
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