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I came across these series while solving a probability question.
let |r| < 1 ,
$$S_n=\sum_{k=0}^{\infty}k^n.r^k$$
For n=0 ,it's a GP. $S_0=\frac{1}{1-r}$
For n=1 ,it's a AGP , $S_1=\frac{-1}{1-r}+\frac{1}{(1-r)^2}$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=\frac{-r}{(1-r)^2}+\frac{2r}{(1-r)^3}$.
Is it possible to find sum further in this series . Is there any pattern.
If you take the (formal) dereivative of $S_n$ with respect to $r$, then $$ \frac{\mathrm d}{\mathrm dr}S_n=\sum_{k=1}^\infty k^n\frac{\mathrm d}{\mathrm dr}r^k=\sum_{k=1}^\infty k^n\cdot k r^{k-1}=\frac 1r\sum_{k=1}^ \infty k^{n+1}r^k=\frac1rS_{n+1}$$ so you obtain a recursion formula, $$S_{n+1}=r\frac{\mathrm d}{\mathrm dr}S_n. $$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=\frac{\mathrm d^n}{\mathrm dx^n}\frac1{1-e^x}=\sum_{k=0}^n{n\brace k}\frac{k!e^{kx}}{(1-e^x)^{k+1}}$$
and by extension,
$$S_n(r)=\sum_{k=0}^n{n\brace k}\frac{k!r^k}{(1-r)^{k+1}}$$
where $\displaystyle{n\brace k}$ are the Stirling numbers of the second kind.
Given $S_n=\sum_{k=0}^{\infty}k^n\cdot r^k$, it must be: $$\begin{align}S_0&=\sum_{k=0}^{\infty}r^k=\frac1{1-r}; \\ S_0'&=\frac{1}{(1-r)^2}=\color{red}1S_0^2;\\ S_1&=\sum_{k=0}^{\infty}k\cdot r^k=rS_0'=rS_0^2; \\ S_1'&=\color{red}1S_0^2+\color{red}2rS_0^3\\ S_2&=\sum_{k=0}^{\infty}k^2\cdot r^k=rS_1'=rS_0^2+2r^2S_0^3; \\ S_2'&=\color{red}1S_0^2+\color{red}6rS_0^3+\color{red}6r^2S_0^4\\ S_3&=\sum_{k=0}^{\infty}k^3\cdot r^k=rS_2'; \\ S_3'&=\color{red}1S_0^2+\color{red}{14}rS_0^3+\color{red}{36}r^2S_0^4+\color{red}{24}r^3S_0^5\end{align}$$ where the coefficients in red are the number sequence A019538.
