I was deriving closed forms for some 'quasi-geometric' (for lack of a better name) series, for example $\Sigma^{\infty}_{n=1}n r^n = \frac{r}{(1-r)^2}$ and $\Sigma^{\infty}_{n=1}n^2 r^n = \frac{r(r+1)}{(1-r)^3}$, and naturally I considered the following general case:
$\Sigma^{\infty}_{n=1}n^k r^n$ for $k \in \mathbb{N}$, $|r|<1$
After applying the same techniques I used for the aforementioned special cases, I found that:
$\Sigma^{\infty}_{n=1}n^k r^n = \frac{1}{1-r}\Sigma^{\infty}_{n=1}[n^k-(n-1)^k]r^n$ for $k \in \mathbb{N}$, $|r|<1$
For special cases of k, this result means $\Sigma^{\infty}_{n=1}n^k r^n$ can be found in terms of $\Sigma^{\infty}_{n=1}n^{k-1} r^n$, $\Sigma^{\infty}_{n=1}n^{k-2} r^n$, $...$,$\Sigma^{\infty}_{n=1}r^n$ as $n^k-(n-1)^k$ can be expressed as a linear combination of each power of n from $0$ to $k-1$ as per the Binomial Theorem. By applying the same result to each of these series, $\Sigma^{\infty}_{n=1}n^k r^n$ can (after much tedious computation) be found purely in terms of r and $\Sigma^{\infty}_{n=1}r^n$ and hence r alone (as $\Sigma^{\infty}_{n=1} r^n = \frac{r}{1-r}$).
Would it be possible to find a closed form for this general result? If not, how could one streamline the computations for special cases of k?
