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Theorem. We have that $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}=\dfrac{1}{(1-x)^{m+1}}\sum_{k=0}^{m-1} A(m,k) x^{m-k}$ where $A(a,k)$ are the Eulerian numbers and $m$ is a nonnegative integer.

I am looking at the polylogarithm function $\operatorname{Li}_{-s}(z) = \displaystyle \sum_{k=1}^{\infty} \dfrac{z^k}{k^{-s}}$ and trying to find its formula for nonnegative integers $s$. Since it is easily verified that

$$\displaystyle \sum_{k=1}^{\infty}k^m x^k =\underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}},$$ the result follows from this. I have noticed that the change of variables $x = e^t$ gives us $$\underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}} = \dfrac{d^m}{dt^m}\left(\dfrac{1}{1-e^t} \right),$$ but I am wondering how to utilize this to prove the theorem.

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It is known that $Li_{-n}(x)$ is just

$$\sum_{k=0}^nk!S(n+1,k+1)(\frac{x}{1-x})^{k+1}$$

see, L. Lewin, Polylogarithms and associated functions, North Holland,1981

and https://web.williams.edu/Mathematics/sjmiller/public_html/math/papers/PolyLogIdentity01.pdf

Moreover,

Negative polylogarithm correspounds to Poly-Bernoulli numbers at negative integers

see theorem 1 here http://www2.math.kyushu-u.ac.jp/~mkaneko/papers/44poly-bernoulli_numbers.pdf

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  • $\begingroup$ I am looking for a way to prove my theorem and not prove the polylogarithm result directly using poly-bernoulli numbers. In the latter case we would still have the Stirling numbers of the second kind as coefficients, but it might be possible to show it is equal by using the formula you have. $\endgroup$ – user19405892 Apr 29 '16 at 17:38
  • $\begingroup$ see math.stackexchange.com/questions/1130184/… $\endgroup$ – user21574 Apr 29 '16 at 17:40
  • $\begingroup$ I think the result is more easily proved with a intermediate step such as my theorem above. $\endgroup$ – user19405892 Apr 29 '16 at 17:44
  • $\begingroup$ This type of questions are suitable for stack.mathexchange and not for mathoverflow. It is seems to be exersise in BA level $\endgroup$ – user21574 Apr 29 '16 at 17:47

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