The value of $\sum_{k=0}^\infty k^n x^k$ Any known conclusions for $\sum_{k=0}^\infty k^n x^k$ where $n$ is a nonnegative integer and $0<x<1$?


*

*If $n=0$, the sum is simply $\frac{1}{1-x}$.

*If $n=1$, the sum can be computed as shown in this post

*For general $n$, it seems the method by taking derivative is
extremely complicated though I believe the sum would have a
finite value because $x^k$ decreases faster than $k^n$ increases.


I don't know of any simple solutions for the general summation, but I do know a method that will work. Since $0 < x < 1$ let $x = \frac{1}{y}$ where $y > 1$. Consider $n = 2$
Let $S = \sum_{k = 0}^\infty \frac{k^2}{y^k}$. 
\begin{align}
S &= \frac{1}{y} + \frac{4}{y^2} + \frac{9}{y^3} + \frac{16}{y^4} + \cdots\\
\frac{S}{y} &= \frac{1}{y^2} + \frac{4}{y^3} + \frac{9}{y^4} + \cdots\\
S - \frac{S}{y} = \frac{S(y - 1)}{y} &= \frac{1}{y} + \frac{3}{y^2} + \frac{5}{y^3} + \frac{7}{y^4} + \cdots\\
\frac{S(y - 1)}{y^2} &= \frac{1}{y^2} + \frac{3}{y^3} + \frac{5}{y^4} + \cdots\\
\frac{S(y - 1)}{y} - \frac{S(y-1)}{y^2} = \frac{S(y-1)^2}{y^2} &= \frac{1}{y} + \frac{2}{y^2} + \frac{2}{y^3} + \frac{2}{y^4} + \cdots\\
&= \frac{1}{y} + \frac{2}{y^2}(1 + \frac{1}{y} + \frac{1}{y^2} + \cdots)\\
&= \frac{1}{y} + \frac{2}{y^2}(\frac{1}{1-\frac{1}{y}}) = \frac{1}{y} + \frac{2}{y(y-1)}
\end{align}
Therefore $S = \frac{y^2}{(y-1)^2}(\frac{1}{y} + \frac{2}{y(y-1)}) = \frac{y}{(y-1)^2} + \frac{2y}{(y-1)^3}$. Now try this method for $n = 3$.
\begin{align}
S &= \frac{1}{y} + \frac{8}{y^2} + \frac{27}{y^3} + \frac{64}{y^4} + \cdots\\
\frac{S}{y} &= \frac{1}{y^2} + \frac{8}{y^3} + \frac{27}{y^4} + \cdots \\
S - \frac{S}{y} = \frac{S(y-1)}{y} &= \frac{1}{y} + \frac{7}{y^2} + \frac{19}{y^3} + \frac{37}{y^4} + \cdots \\
\frac{S(y-1)}{y^2} &= \frac{1}{y^2} + \frac{7}{y^3} + \frac{19}{y^4} + \cdots \\
\frac{S(y-1)^2}{y^2} &= \frac{1}{y} + \frac{6}{y^2} + \frac{12}{y^3} + \frac{18}{y^4} + \cdots\\
\frac{S(y-1)^2}{y^3} &= \frac{1}{y^2} + \frac{6}{y^3} + \frac{12}{y^4} + \cdots\\
\frac{S(y-1)^3}{y^3} &= \frac{1}{y} + \frac{5}{y^2} + \frac{6}{y^3} + \frac{6}{y^4} + \frac{6}{y^5} + \cdots
\end{align}
Which yields the geometric series. 
Basically the procedure consists of taking the difference between the sum and multiplying it by the common ratio $\frac{1}{y}$. This procedure will be done $n$ times in order to yield a finite amount of terms plus an infinite convergent geometric series. This occurs due to the property of finite differences. Quadratics have second differences, Cubics have third, $x^n$ has $n$ differences. This means the numerators eventually disappear into constants while the exponentials on the bottom remain the same. So just repeat this process $n$ times to compute the sum. This method only requires elementary algebra.
