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WolframAlpha claims: $$\sum_{n=0}^m n x^n = \frac{(m x - m - 1) x^{m + 1} + x}{(1 - x)^2} \tag{1}$$ I know that one can differentiate the geometric series to compute $(1)$ when it is a series, i.e. $m=\infty$. However, I'm wondering how the closed form for the partial sum is obtained. Actually, WolframAlpha gives an explicit formula for $$\sum_{n=0}^m n(n-1)(n-2)\cdots (n-k)x^n \tag{2}$$ where $k$ is an integer between $0$ and $n-1$, so it seems that there are some differentiation involved. But already for $k=5$, the formula becomes very messy.

My question: How to prove $(1)$ and how to build a similar formula for $(2)$ assuming $k$ is given?

idm
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    For $(1)$, differentiate $\sum_{n=1}^m x^n = x\frac{x^m-1}{x-1}$ w.r.t $x$ to get a closed form of $\sum_{n=0}^{m-1} (n+1)x^m$ and write $$\sum_{n=1}^m n x^n = \sum_{n=1}^m (n+1) x^n - \sum_{n=1}^m x^n$$ – Gabriel Romon Mar 11 '17 at 11:26
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    @LeGrandDODOM good point! and for (2) I guess that one can proceed in the same way. – idm Mar 11 '17 at 11:28
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    You can probably find several closely related posts on this site. For example: http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-limits-r-1d-r-cdot-2r and http://math.stackexchange.com/questions/180198/what-is-the-sum-of-sum-limits-i-1nipi – Martin Sleziak Mar 11 '17 at 12:31
  • @MartinSleziak thanks for the references, I could not find anything when I searched for it. – idm Mar 11 '17 at 13:03
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    You could probably find some posts using Approach0. In the case of questions which are likely to appear often, it might be useful to check frequent questions in relevant tags - in this case probably summation and sequences-and-series. More tips on searching here. – Martin Sleziak Mar 11 '17 at 13:34

2 Answers2

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If one knows the geometric series, then

$$\sum_{n=0}^mx^n=\frac{1-x^{m+1}}{1-x}$$

And by differentiating $k$ times,

$$\sum_{n=0}^mn(n-1)(n-2)\dots(n-k+1)x^{n-k}=\frac{d^k}{dx^k}\frac{1-x^{m+1}}{1-x}$$

Multiply both sides by $x^k$ to then get your sum. By using Leibniz's rule,

$$\frac{d^k}{dx^k}\frac{1-x^{m+1}}{1-x}=\frac{k!(1-x^{m+1})}{(1-x)^{k+1}}-\sum_{n=1}^k\binom{m+1}nk!(k+1-n)\frac{x^{m+1-k}}{(1-x)^{k+1-n}}$$

One may evaluate this using the Gamma function.

$$\begin{align}\sum_{n=0}^mn(n-1)(n-2)\dots(n-k)x^n&=\sum_{n=0}^m\frac{n!}{(n-k-1)!}x^n\\&=\frac1{(n-k-1)!}\sum_{n=0}^m\int_0^\infty t^ne^{-t}\ dt\ x^n\\&=\frac1{(n-k-1)!}\int_0^\infty e^{-t}\sum_{n=0}^m(tx)^n\ dt\\&=\frac1{(n-k-1)!}\int_0^\infty\frac{e^{-t}(1-(tx)^{m+1})}{1-tx}\ dt\\&=\frac1{x(n-k-1)!}\int_0^\infty\frac{e^{-t}(1-t^{m+1})}{1-t}\ dt\end{align}$$

whereupon for any given $m\in\mathbb N$, you can evaluate this integral to obtain a closed form.

Simply Beautiful Art
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use that $$\sum_{i=1}^{n}x^i=\frac{x \left(x^n-1\right)}{x-1}$$ and differentiate with respect to $x$

Dr. Sonnhard Graubner
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