General formula for $\sum_{i=1}^{n} i^pA^i$

Question

I was trying to sum series for fun and then I thought that since I can find the sum of polynomials and that of a geometric series then maybe I can find the sum of their product..I could find $$ \sum_{i=1}^n iA^i $$ Where $A$ is a constant easily but then when I tried higher powers for $i$ like $i^2$ $i^3$ $...$ I found it very difficult and I got curious if there is a formula like Faulhabeer's..and after a lot of tries I reached this formula which actually works $$ \sum_{i=1}^{n} i^pA^i =\left (\sum_{j=1}^{n-1} j^p -(j-1)^p\right) \frac{A(A^n-1)}{A-1} +(n^p-(n-1)^p)A^n- \left(\sum_{k=1}^{n-2}((k+1)^p-k^p) \frac{A(A^k-1)}{A-1}\right) $$ But it's not practical so I want to know if there is a simpler and more practical way to find the sum

Answer 1

Just for your curiosity.

There is a formula (the problem is that it still hides a summation !) $$S_n=\sum_{i=1}^{n} i^p\,A^i =\text{Li}_{-p}(A)-A^{n+1}\, \Phi (A,-p,n+1)$$ where appear the polylogarithm function and Lerch transcendent function.

Fo $p=4$ as in your comment, this would give the ugly $$\frac{(A-1)^5}A S_n=\left((A-1)^4 n^4-4 (A-1)^3 n^3+6 (A-1)^2 (A+1) n^2-4 A (A (A+3)-3) n+A (A (A+11)+11)+4 n+1\right) A^n+(-A-1) (A (A+10)+1)$$

It would look better using $A=B+1$ which would make $$\frac{B^5}{B+1} S_n=-B^3-14 B^2-36 B-24+$$ $$\left(B^4 n^4+B^3 \left(-4 n^3+6 n^2-4 n+1\right)+2 B^2 \left(6 n^2-12 n+7\right)+B (36-24 n)+24\right) (B+1)^n$$

Answer 2

$$\sum_{i=0}^n x^i=(x^{n+1}-1)(x-1)^{-1}.$$

Then

$$\sum_{i=0}^n ix^i=\frac d{dx} (x^{n+1}-1)(x-1)^{-1}=(n+1)x^n(x-1)^{-1}-(x^{n+1}-1)(x-1)^{-2}.$$

More generally, using the Leibniz formula,

$$\sum_{i=0}^n i^mx^i=\frac{d^m}{dx^m}(x^{n+1}-1)(x-1)^{-1} \\=(x^{n+1}-1)(m-n+2)!(x-1)^{m-n+1}+ \\\sum_{k=1}^m(-1)^k\binom mk\frac{(n+1)!}{(n-k+1)!}x^{n-k+1}(m-k+2)!(x-1)^{-(m-k+1)} \\=(x^{n+1}-1)(m-n+2)!(x-1)^{m-n+1}+ \\x^{n-m}\sum_{k=1}^m(-1)^k\binom mk\frac{(n+1)!}{(n-k+1)!}(m-k+2)!\left(\frac x{1-x}\right)^{m-k+1}.$$

As the derivative of $x^{n+1}$ cancels after order $m+1$, the summation is indeed up to $k=\min(m,n+1)$.

Answer 3

I'm going to attempt and give a recursive formula solution.

Let

$$ S_j(x,N)=\sum_{k=1}^N{k^j x^k} \ . $$

Then it follows:

\begin{aligned} S_j(x,N) &= \sum_{k=1}^N x \frac{d}{dx} {k^{j-1} x^k} \\ &= x \frac{d}{dx}\sum_{k=1}^N{k^{j-1} x^k} \\ &= x \frac{d}{dx} S_{j-1}(x,N) \ . \end{aligned}

We know that $S_0(x,N)=\frac{x^{N+1}-1}{x-1}$. So the general term becomes:

\begin{aligned} S_j(x,N) &=\left(x \frac{d}{dx} \right)^j S_0(x,N)\\ &= \left(x \frac{d}{dx} \right)^j \left(\frac{x^{N+1}-1}{x-1}\right) \ . \end{aligned}