I came across the "summation by parts" formula, and wondered how it could be useful in any way: $$ \sum_{k=n}^m\ a_k \Delta b_k=\left[a_k b_k\right]_{n}^{m+1}-\sum_{k=n}^{m}\ b_{k+1} \Delta a_k $$ Where $\Delta a_k$ is the Forward difference operator. I tried to do the following: with $|\alpha|<1$: $$ \sum_{k=0}^m \ k \alpha^k=\frac{1}{\alpha-1}\sum_{k=0}^m \ k \alpha^k(\alpha-1)=\frac{1}{\alpha-1}\left(\left[ k\alpha^k\right]_0^{m+1}-\sum_{k=0}^m\ \alpha^{k+1}\right) $$ And then I took the limit as $m\to \infty$, getting: $$ \sum_{k=0}^\infty\ k\alpha^k=\frac{\alpha}{(1-\alpha)^2} $$ Which is a result I'm sure many of you know, but then I realized that, with this formula, i could do the same for $\sum_{k=0}^\infty\ k^2 \alpha^k$, being $\Delta k^2=2k+1$, arriving at: $$ \sum_{k=0}^\infty\ k^2 \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(2\cdot\frac{\alpha}{1-\alpha}+1\right) $$ But I could do the same for $\sum_{k=0}^{\infty}\ k^3\alpha^k$, and so on, obtaining this formula: $$ \sum_{k=0}^\infty\ k^3\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(6\cdot \left(\frac{\alpha}{1-\alpha}\right)^2+6\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right) $$ And so on with the fourth degree and the fifth: $$ \sum_{k=0}^{\infty}\ k^4\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(24\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+36\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+14\cdot\left(\frac{\alpha}{1-\alpha}\right)+1\right) $$
$$ \sum_{k=0}^{\infty}\ k^5\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(120\cdot\left(\frac{\alpha}{1-\alpha}\right)^4+240\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+150\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+30\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right) $$ At this point I got bored and wanted to find a general formula, seeing a polynomial pattern, but all i could figure out was this: (I didn't prove any of the following, I just tried to figure out the patterns, which could be misleading)
Given $|\alpha|<1$, and $n\in \mathbb{N}$ $$ \sum_{k=0}^\infty k^n \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(\sum_{k=0}^{n-1} P_{n,k} \left(\frac{\alpha}{1-\alpha}\right)^k\right) $$ Where $P_{n,k}$ is a coefficient defined for $n\in \mathbb{N},\ k\in \{0,1,2,...,n-1\}$, with these defining properties: $$ \forall n \in \mathbb{N}, P_{n,0}=1\\ P_{n,k}=\sum_{h=1}^{n-k}\binom{n}{h}P_{n-h,k-1} $$ I'm having some troubles on how to get to a closed form of theese coefficients, all i know is that the coefficient of the highest term is $n!$, the known term is $1$, and that the linear term is $2^n-2$, the formulas for the other coefficients are a mess beyond my comprehension, with nested sums of binomial coefficients. If you recognize these polynomials by any chance, or if you can find a closed form for them, or any kind of insight on this, please let me know.
Post Scriptum: I didn't prove any of the results i showed about $P_{n,k}$, so if you got a correction to make, please point that out.
