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I have a sum $\sum_{i=0}^{\infty} i^{k} D^{i}$, where $k = \frac{1}{2} n$ for some $n \in \mathbb{N}$ and $1>D>0$. The ratio test proves that this sum converges, but I struggle finding a closed form for the limit. Any suggestions how to get there?

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The explicit expression for the sum is: $$ \sum_{i=0}^{\infty} i^{k} D^{i}=\frac{\sum_{i=0}^{k-1}A_i^{k}D^{i+1}}{(1-D)^{k+1}}, $$ where $A_i^k$ are the Eulerian numbers. Can be easily proved by induction.

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  • $\begingroup$ Thanks for the help! I am afraid you might have overlooked that $k$ is not a natural number, but a fraction. Does your result still hold then? $\endgroup$
    – Jeremias K
    Mar 27, 2018 at 10:11
  • $\begingroup$ Yes I have overlooked it. Do you look for the solution only for odd $n$? $\endgroup$
    – user
    Mar 27, 2018 at 10:16
  • $\begingroup$ No I am afraid not, though your answer led me to look at the polylogarithmic function, and the way I see it, that might be what I am looking at $\endgroup$
    – Jeremias K
    Mar 27, 2018 at 10:18
  • $\begingroup$ Yes the general expression for the sum is $\text{Li}_{-k}(D)$. $\endgroup$
    – user
    Mar 27, 2018 at 10:26

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