Cesaro sum of ∑n^b e^(ian) for real a, b

[zetafunction]

what is the sum of this series ??

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

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

however i would like to know if the sum is Cesaro summable and what it would be its Cesaro sum $$C(k,a,b)$$ for k bigger than 'b' , and a=0 , thanks.

 

[Marin]

Hi!

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,
marin