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I came across these series while solving a probability question.

enter link description here

let |r| < 1 ,

$$S_n=\sum_{k=0}^{\infty}k^n.r^k$$

For n=0 ,it's a GP. $S_0=\frac{1}{1-r}$

For n=1 ,it's a AGP , $S_1=\frac{-1}{1-r}+\frac{1}{(1-r)^2}$

for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=\frac{-r}{(1-r)^2}+\frac{2r}{(1-r)^3}$.

Is it possible to find sum further in this series . Is there any pattern.

Rishi
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3 Answers3

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If you take the (formal) dereivative of $S_n$ with respect to $r$, then $$ \frac{\mathrm d}{\mathrm dr}S_n=\sum_{k=1}^\infty k^n\frac{\mathrm d}{\mathrm dr}r^k=\sum_{k=1}^\infty k^n\cdot k r^{k-1}=\frac 1r\sum_{k=1}^ \infty k^{n+1}r^k=\frac1rS_{n+1}$$ so you obtain a recursion formula, $$S_{n+1}=r\frac{\mathrm d}{\mathrm dr}S_n. $$

Hagen von Eitzen
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Using the $n$th derivative of $f(e^x)$, we have

$$S_n(e^x)=\frac{\mathrm d^n}{\mathrm dx^n}\frac1{1-e^x}=\sum_{k=0}^n{n\brace k}\frac{k!e^{kx}}{(1-e^x)^{k+1}}$$

and by extension,

$$S_n(r)=\sum_{k=0}^n{n\brace k}\frac{k!r^k}{(1-r)^{k+1}}$$

where $\displaystyle{n\brace k}$ are the Stirling numbers of the second kind.

Simply Beautiful Art
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Given $S_n=\sum_{k=0}^{\infty}k^n\cdot r^k$, it must be: $$\begin{align}S_0&=\sum_{k=0}^{\infty}r^k=\frac1{1-r}; \\ S_0'&=\frac{1}{(1-r)^2}=\color{red}1S_0^2;\\ S_1&=\sum_{k=0}^{\infty}k\cdot r^k=rS_0'=rS_0^2; \\ S_1'&=\color{red}1S_0^2+\color{red}2rS_0^3\\ S_2&=\sum_{k=0}^{\infty}k^2\cdot r^k=rS_1'=rS_0^2+2r^2S_0^3; \\ S_2'&=\color{red}1S_0^2+\color{red}6rS_0^3+\color{red}6r^2S_0^4\\ S_3&=\sum_{k=0}^{\infty}k^3\cdot r^k=rS_2'; \\ S_3'&=\color{red}1S_0^2+\color{red}{14}rS_0^3+\color{red}{36}r^2S_0^4+\color{red}{24}r^3S_0^5\end{align}$$ where the coefficients in red are the number sequence A019538.

Simply Beautiful Art
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farruhota
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  • +1. The main idea is that it is simpler to consider $T_n=\sum_{k=1}^\infty k(k-1)(k-2)\dots (k-n+1)r^k$ first. – Taladris Sep 03 '19 at 02:38
  • Hmm, curious as to why does this have so many upvotes when it's just Hagen's answer and my answer with examples. And the link at the end just furthers that the coefficients are given by my answer. – Simply Beautiful Art Sep 03 '19 at 12:45
  • @Taladris I don't see how your $T_n$ is of any relevance to this answer. – Simply Beautiful Art Sep 03 '19 at 12:46
  • @SimplyBeautifulArt, my answer overlaps with both your and HagenvonEitzen's answers, indeed, yet this method is more accessible and detailed, I assume. After all, let me quote: In science the credit goes to the man who convinces the world, not to the man to whom the idea first occurs. Sir Francis Darwin – farruhota Sep 03 '19 at 12:50
  • Could you elaborate on what this "method" is and what about it is "accessible"? Because it's still not clear to me what this answer adds, given the two other answers, even though 5 other people surely think so. – Simply Beautiful Art Sep 03 '19 at 12:52
  • @SimplyBeautifulArt, you took nth derivative of $f(e^x)$, while this method takes derivatives directly; Hagen's answer does not show the coefficients, while this method does with reference. Also, when I wrote "it must be", I also indicated to the mistake in calculation of $S_0$ in the original post, before the index was changed from $1$ to $0$. And thank you for updating. – farruhota Sep 03 '19 at 13:07
  • Unfortunately I fail to see how that is anything more than simply taking Hagen's answer, writing a few examples, and pointing out the coefficients are in my answer. That is to say, without Hagen's answer or my answer, this answer would seem to be only some examples with unconfirmed coefficients (nothing to back up that these are indeed factorials times Stirling numbers) using $S_{n+1}=rS_n'$ without explicitly mentioning it. – Simply Beautiful Art Sep 03 '19 at 13:17
  • Well, this method shows to look for patterns. For the coefficients, I looked up at the number sequences site. For the derivatives, it is not that difficult to see the pattern. – farruhota Sep 03 '19 at 13:34