$$\frac{4}{20}+\frac{4 \cdot 7}{20 \cdot 30}+\frac{4 \cdot 7 \cdot 10}{20 \cdot 30 \cdot 40}+\dots$$

How to find the nth term of this series? In particular, how to write the series as a summation in terms of some variable (say n) where n ranges from 0 to infinity? I think finding the sum of series would then be easier (or probably that is the task to be done). Please elaborate how to proceed.

  • 1
    $\begingroup$ See math.stackexchange.com/questions/746388/… $\endgroup$ Jan 20, 2017 at 8:09
  • $\begingroup$ $\sum\limits_{n=0}^{\infty}\prod\limits_{k=0}^{n}\frac{4+3k}{20+10k}$ $\endgroup$ Jan 20, 2017 at 8:12
  • $\begingroup$ Here $n$ term is $$10\dfrac{1\cdot4\cdot7\cdots(3n-2)}{10^n n!}$$ Can you find the missing terms $\endgroup$ Jan 20, 2017 at 8:13
  • $\begingroup$ Thank you @labbhattacharjee ... it was easy to proceed. $\endgroup$
    – Fella
    Jan 20, 2017 at 8:59

2 Answers 2


The denominator involves all multiples of $10$ and can be expressed via factorials, $10^nn!$, with $n$ starting at $n=2$.

The numerator doesn't follow this scheme as the factors are of the form $3n-2$. You need to use a product symbol,


Alternatively, you can define a "triple factorial" notation $n!!!$ such that only every third factor is used (this follows the idea of the double factorial).

Hence your sum is


By the generalized binomial theorem,

$$(1-r)^{-p/q}=1+\frac{p}q r+\frac{p}q\frac{p+q}q\frac{r^2}{2!}+\frac{p}q\frac{p+q}q\frac{p+2q}q\frac{r^3}{3!}+\cdots=\sum_{k=0}^\infty\frac{\prod_{j=0}^{k-1}(p+jq)}{k!}\left(\frac rq\right)^k.$$

In your case, by identification, $q=3$, $p=1$, $r/q=1/10$, the first two terms are omitted and a power of $10$ is missing. Then

$$\frac S{10}=\left(1-\frac3{10}\right)^{-1/3}-1-\frac1{10}.$$

  • $\begingroup$ Using triple factorial, how do I evaluate the sum? $\endgroup$
    – Fella
    Jan 20, 2017 at 8:20
  • $\begingroup$ Read the post referred to by @labbhattacharjee and make the link to the generalized binomial formula. $\endgroup$
    – user65203
    Jan 20, 2017 at 8:49
  • $\begingroup$ yeah! generalized binomial formula! :) $\endgroup$
    – Fella
    Jan 20, 2017 at 9:01

Let $\displaystyle S = 10\bigg[\frac{4}{2!}\bigg(\frac{1}{10}\bigg)^2+\frac{4\cdot 7}{3!}\bigg(\frac{1}{10}\bigg)^3+\frac{4\cdot 7\cdot 10}{4!}\bigg(\frac{1}{10}\bigg)^4+\cdots \cdots \bigg]$

$\displaystyle = 10 \bigg[\frac{\left[-\frac{1}{3}\left(-\frac{1}{3}-1\right)\right]}{2!}\cdot \bigg(\frac{3}{10}\bigg)^2 +\frac{\left[-\frac{1}{3}\left(-\frac{1}{3}-1\right)\left(-\frac{1}{3}-2\right)\right]}{3!}\cdot \bigg(\frac{3}{10}\bigg)^3+\cdots \cdots \bigg]$

$\displaystyle = 10\bigg[\bigg(1-\frac{3}{10}\bigg)^{-\frac{1}{3}}-\bigg(1+\frac{1}{3}\cdot \frac{3}{10}\bigg)\bigg] = 10 \bigg[\bigg(\frac{10}{13}\bigg)^{\frac{1}{3}}-\frac{11}{10}\bigg] = 10\bigg(\frac{10}{7}\bigg)^{\frac{1}{3}}-11$

Sorry I have not noticed that Yves Daoust have already write above.


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