I came across these series while solving a probability question.

enter link description here

let |r| < 1 ,


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.

  • $\begingroup$ I assume you meant $S_n$ to start at $k=0$? $\endgroup$ Sep 3, 2019 at 12:40
  • $\begingroup$ It is same since $0^n =0$ $\endgroup$
    – Rishi
    Sep 3, 2019 at 12:47
  • $\begingroup$ If so then your $S_0$ is wrong. $\endgroup$ Sep 3, 2019 at 12:48
  • $\begingroup$ Could we please have it start with $k=0$ then? It makes the results much cleaner... $\endgroup$ Sep 3, 2019 at 12:54
  • $\begingroup$ we can't start from $0$ , It will be $0^0$ for $n=0$ , now it's fine i edited r. $\endgroup$
    – Rishi
    Sep 3, 2019 at 12:55

3 Answers 3


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, $$S_{n+1}=r\frac{\mathrm d}{\mathrm dr}S_n. $$

  • $\begingroup$ fantastic observation! any connection with riemann zeta? $\endgroup$
    – vidyarthi
    Sep 2, 2019 at 14:13
  • 1
    $\begingroup$ @vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $r\to-1$. $\endgroup$ Sep 2, 2019 at 14:39

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.


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.

  • $\begingroup$ +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. $\endgroup$
    – Taladris
    Sep 3, 2019 at 2:38
  • $\begingroup$ 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. $\endgroup$ Sep 3, 2019 at 12:45
  • $\begingroup$ @Taladris I don't see how your $T_n$ is of any relevance to this answer. $\endgroup$ Sep 3, 2019 at 12:46
  • $\begingroup$ @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 $\endgroup$
    – farruhota
    Sep 3, 2019 at 12:50
  • $\begingroup$ 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. $\endgroup$ Sep 3, 2019 at 12:52

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