Any known conclusions for $\sum_{k=0}^\infty k^n x^k$ where $n$ is a nonnegative integer and $0<x<1$?
You may have a look at the following interesting reference, due to Khristo N. Boyadzhiev, which adresses your issue: A series transformation formula and related polynomials.
The author proved that
$$ \sum_{k=0}^\infty k^n x^k=\frac1{1-x}\: \omega_n\!\left(\frac{x}{1-x}\! \right),\quad |x|<1, $$
where $\omega_n(\cdot)$ are the polynomials $$ \omega_n(x)=\sum_0^n {\small{n\brace k}}k!\:x^k $$ and where $\displaystyle {\small{n\brace k}}$ are the Stirling numbers of the second kind. One may notice that $\displaystyle {\small{n\brace k}}$ is the number of partitions of a set of $n$ elements into $k$ disjoint nonempty subsets.
One can remark that $$\frac{d}{dx}\left(\sum_{k=0}^\infty k^n x^k\right) = \sum_{k=1}^\infty k^{n+1} x^{k-1} = \frac{1}{x}\left(\sum_{k=0}^\infty k^{n+1} x^{k}-1\right).$$ Hence, if we note $f_n(x) = \sum_{k=0}^\infty k^n x^k$, we find a relation $$xf'_n(x) = f_{n+1}(x)-1.$$ For example, we know that $f_0(x) = \frac{1}{1-x}$ and hence we find that $$f_1(x) = x \left(\frac{1}{1-x}\right)' +1 =\frac{x}{(1-x)^2}+1,$$ and so on.
Doesn't answer convergence, but a quick wolfram search gives:
http://mathworld.wolfram.com/LerchTranscendent.html
$\sum_{k=0}^\infty k^n x^k = \Phi(x,-n,0)$
Define for nonnegative integers $n$ the function $$f_n(z) = \sum_{k=0}^\infty a_k(z) = \sum_{k=0}^\infty k^n z^k.$$ The ratio test shows that $$\left|\frac{a_{k+1}(z)}{a_k(z)}\right| = |z| \left(1 + \frac{1}{k}\right)^n,$$ and the limit as $k \to \infty$ for a fixed $n$ clearly shows that the function is absolutely convergent whenever $|z| < 1$, and divergent for $|z| > 1$. The behavior for $|z| = 1$ is left to the reader to explore.
As for the sum itself, we observe $$f_{n+1}(z) = z \frac{d}{dz}[f_n(z)], \quad f_0(z) = (1-z)^{-1}$$ by formal differentiation of the power series. This gives, for example $$\begin{align*} f_1(z) &= z(1-z)^{-2} \\ f_2(z) &= z(1+z)(1-z)^{-3} \\ f_3(z) &= z(1+4z+z^2)(1-z)^{-4} \\ f_4(z) &= z(1+z)(1+10z+z^2)(1-z)^{-5} \\ f_5(z) &= z(1+26z+66z^2+26z^3+z^4)(1-z)^{-6}. \end{align*}$$ This clearly suggests that $$f_{n+1}(z) = P_n(z) (1-z)^{-n}$$ for some suitable monic polynomial $P_n(z)$ with degree $n$ and integer coefficients, and that we should be interested in the coefficients of $P$; e.g., define $$P_n(z) = \sum_{m=0}^n c_{n,m} z^m.$$ Then we have the triangle of coefficients $$1 \\ 0, 1 \\ 0, 1, 1 \\ 0, 1, 4, 1 \\ 0, 1, 11, 11, 1 \\ 0, 1, 26, 66, 26, 1 $$ and so forth. These happen to be Eulerian numbers. It is left as an exercise for the reader to prove various recursion relationships among these coefficients and that they satisfy the desired properties.
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.
