<s>[INST] Find sum of the infinite series $$\frac{4}{20}+\frac{4 \cdot 7}{20 \cdot 30}+\frac{4 \cdot 7 \cdot 10}{20 \cdot 30 \cdot 40}+\dots$$
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.
 [/INST] 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.
</s>