<s>[INST] How to derive this series 
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.

I know the first series and it is easy to obtain it. But the second series is strange. It is not a power series, not a Taylor series. How does one obtain this series?
 [/INST] We can show the  identity
\begin{align*}
\sum_{n=0}^\infty  \frac{n!x^n}{\prod_{j=1}^n(1+jx)}=\frac{1}{1-x}\qquad\qquad0<x<1\tag{1}
\end{align*}
with the help of Gauss' summation formula.

We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty  \frac{n!x^n}{\prod_{j=1}^n(1+jx)}}
&=\sum_{n=0}^{\infty}\frac{n!}{\left(1+\frac{1}{x}\right)^{\overline{n}}}\tag{2}\\
&=\sum_{n=0}^{\infty}\frac{1^{\overline{n}}1^{\overline{n}}}{\left(1+\frac{1}{x}\right)^{\overline{n}}}\,\frac{1}{n!}\tag{3}\\
&={}_2F_1\left(1,1;1+\frac{1}{x};1\right)\tag{4}\\
&=\frac{\Gamma\left(\frac{1}{x}+1\right)\Gamma\left(\frac{1}{x}-1\right)}{\Gamma\left(\frac{1}{x}\right)\Gamma\left(\frac{1}{x}\right)}\tag{5}\\
&=\frac{\frac{1}{x}\Gamma\left(\frac{1}{x}\right)\Gamma\left(\frac{1}{x}-1\right)}
{\Gamma\left(\frac{1}{x}\right)\,\left(\frac{1}{x}-1\right)\Gamma\left(\frac{1}{x}-1\right)}\tag{6}\\
&=\frac{\frac{1}{x}}{\frac{1}{x}-1}\tag{7}\\
&\,\,\color{blue}{=\frac{1}{1-x}}
\end{align*}
and the claim (1) follows.

Comment:

*

*In (2) we expand with $\frac{1}{x^n}$ and use the rising factorial notation $q^{\overline{n}}=q(q+1)\cdots (q+n-1)$.


*In (3) we write $1^{\overline{n}}=n!$ and prepare the representation for use of hypergeometric series.


*In (4) we use the hypergeometric series notation
\begin{align*}
{}_2F_1\left(a,b;c;z\right)=\sum_{n=0}^{\infty}\frac{a^{\overline{n}}b^{\overline{n}}}{c^{\overline{n}}}\,\frac{z^n}{n!}
\end{align*}
with $a=b=z=1$ and $c=1+\frac{1}{x}$.


*In (5) we use Gauss' summation formula
\begin{align*}
{}_2F_1\left(a,b;c;1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}
\end{align*}
with $a=b=1$ and $c=1+\frac{1}{x}$ valid for $\Re\left(\frac{1}{x}\right)>1$.


*In (6) we use the identity $\Gamma(x+1)=x\Gamma(x)$ for all $x\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}$.


*In (7) we finally cancel terms.
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