<s>[INST] On $\sum_{x=0}^\infty x^n r^x$ I am looking for either a closed form or recursive expression for $$\sum_\limits{x=0}^\infty x^n r^x\ $$ that does not include differential operators, where $n \in \mathbb N_0$. It is clear to me that $$\sum_\limits{x=0}^\infty x^n r^x = r \cfrac{\text d}{\text dr}\sum_\limits{x=0}^\infty x^{n-1} r^x \ $$ and from this I found some patterns in the first few expressions:
$$\sum_{x=0}^\infty xr^x = \cfrac r {(1-r)^2} \\ \sum_{x=0}^\infty x^2 r^x = \cfrac {r+r^2} {(1-r)^3} \\ \sum_{x=0}^\infty x^3 r^x = \cfrac {r^3+4r^2+r} {(1-r)^4} \\ \sum_{x=0}^\infty x^4 r^x = \cfrac {r^4+11r^3+11r^2+r} {(1-r)^5} \\ \sum_{x=0}^\infty x^5 r^x = \cfrac {r^5+26r^4+66r^3+26r^2+r} {(1-r)^6} \\ \sum_{x=0}^\infty x^6 r^x = \cfrac {r^6 + 57r^5+302r^4+302r^3+57r^2+r} {(1-r)^7} \\ \sum_{x=0}^\infty x^7 r^x = \cfrac {r^7+120r^6 + 1191 r^5 + 2416r^4+1191r^3+120r^2+r} {(1-r)^8}$$
It's interesting that the denominator increases by a factor of $1-r$ each time, and another interesting thing is that the coefficients of row $n$ add up to $n!$. I haven't been able, though, to leverage this information in a way that spits out the individual coefficients. Is my goal of finding a closed form/recursive formula for these possible? Thanks.
 [/INST] Terminology ... polylogarithm
$$
\operatorname{Li}_s(z) = \sum_{k=0}^\infty \frac{z^k}{k^s}
$$
So your question asks about the polylogarithm $\operatorname{Li}_{-n}(r)$.
That page shows $\operatorname{Li}_s(z)$ for $s=-1,-2,-3,-4$, the first four
examples you computed.
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