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?
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.
The series you are asking about is $$ S(x) \!:=\! 1 \!+\! \frac{x}{1\!+\!x} \!+\! \frac{1\cdot 2\cdot x^2}{(1\!+\!x)(1\!+\!2x)} \!+\! \frac{1\cdot 2\cdot 3\cdot x^3}{(1\!+\!x)(1\!+\!2x)(1\!+\!3x)} \!+\! \cdots. \tag{1} $$ One of the first things to do in such a series is to find the ratio of consecutive terms which gives the sequence $$ \frac{x}{1+x},\;\; \frac{2x}{1+2x},\;\; \frac{3x}{1+3x}\;\; \dots,\;\; \frac{nx}{1+nx},\;\; \dots $$ which is a rational function in $\,n\,$ and this is the characteristic property of a Hypergeometric series.
Assuming $\,x\ne 0\,$, let $\, y := 1/x.\,$ Then $$ S(x) \!=\! 1 \!+\! \frac{1!}{(1\!+\!y)} \!+\! \frac{2!}{(1\!+\!y)(2\!+\!y)} \!+\! \frac{3!}{(1\!+\!y)(2\!+\!y)(3\!+\!y)} \!+\! \cdots. \tag{2} $$ This is a simple Hypergeometric series $$ S(x) = {}_2F_1(1,1;1+1/x;1) = 1/(1-x) \tag{3} $$ where the left side series has a complicated domain of convergence and the right side has a simple pole at $\,x=1.\,$
Your question was
How does one obtain this series?
Quoting from the Wikipedia article:
Many of the common mathematical functions can be expressed in terms of the hypergeometric function, or as limiting cases of it.
In this particular case, assume the Ansatz $$ f(x) \!:=\! a_0 \!+\! \frac{a_1\,x}{1\!+\!x} \!+\! \frac{a_2\,x^2}{(1\!+\!x)(1\!+\!2x)} \!+\! \frac{a_3\,x^3}{(1\!+\!x)(1\!+\!2x)(1\!+\!3x)} \!+\! \cdots. \tag{4} $$ Then by expanding into power series in $\,x\,$ we have the result $$ f(x) \!=\! a_0 \!+\! a_1\,x \!+\! (a_2\!-\!a_1)x^2 \!+\! (a_3\!-\!3a_2\!+\!a_1)x^3 \!+\! (a_4\!-\!6a_3\!+\!7a_2\!-\!a_1)x^4 \!+\! \cdots \tag{5} $$ which gets the power series coefficients of $\,f(x)\,$ from those of the series in equation $(4)$.
For this particular hypergeometric series, there is another simple method to try. Define the partial sums $$ S_n := \sum_{k=0}^n k!/(1+1/x)_k. \tag{6} $$ Then we can observe that $$ S_n = P_n x^n/(1+1/x)_n \tag{7} $$ where $\,P_n\,$ is a polynomial of degree $\,n\,$ with positive integer coefficients appearing in OEIS sequence A109822. For example $$ P_1\!=\! 1\!+\!2x, P_2\!=\! 1\!+\!4x\!+\!6x^2, P_3 = 1\!+\!7x\!+\!18x^2\!+\!24x^3. \tag{8} $$
But notice that the same coefficients appear in OEIS sequence A096747 which has an extra $\,(n+1)!\,$ for each row. This suggests looking at $$ 1/(1-x) - S_n = (n+1)! \frac{x}{(1-x)(1+1/x)_{n+1}}. \tag{9} $$ This equality of two rational functions can be proved by induction using telescoping sums.
Is there a way to find the limiting function of this series?
– James Warthington Jan 04 '20 at 01:34Proof of the Formula
Below, we show inductively $$ \frac1{1-x}=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag1 $$ where the empty sum is $0$ and the empty product is $1$.
Gautschi's Inequality says that $$ \begin{align} \frac{n!\,x^n}{\prod_{j=1}^n(1+jx)} &=\frac{\Gamma(n+1)\,\Gamma\!\left(1+\frac1x\right)}{\Gamma\!\left(n+1+\frac1x\right)}\\ &\sim\frac{\Gamma\!\left(1+\frac1x\right)}{(n+1)^{1/x}}\tag2 \end{align} $$ Thus, for $0\lt x\lt1$, the series $$ \sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}\tag3 $$ converges and the remainder term $$ \frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag4 $$ vanishes as $n\to\infty$. Therefore, for $0\lt x\lt1$, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}=\frac1{1-x}}\tag5 $$
Inductive Proof of $\bf{(1)}$
Trivially, we have that $(1)$ is true for $n=0$.
Suppose that we have $(1)$ for some $n$. Then $$ \begin{align} \frac1{1-x} &=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\frac{1+nx}{1-x}\\ &=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\left(\color{#C00}{1}+\color{#090}{\frac{(n+1)x}{1-x}}\right)\\ &=\sum_{k=0}^{\color{#C00}{n}}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\color{#090}{\frac{(n+1)!\,x^{n+1}}{\prod_{j=1}^{n+1}(1+jx)}\frac{1+(n+1)x}{1-x}}\tag6 \end{align} $$ Thus, $(1)$ holds for $n+1$.
Using @Andrew Chin's comments, we face for the second series $$S_\infty=\sum_{n=0}^\infty\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}$$
Let $$a_n=\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}=\frac{\Gamma (n+1)}{\left(1+\frac{1}{x}\right)_n}$$ where appear Pochhammer symbols. So $$S_p=\sum_{n=0}^p a_n=\frac{\frac{\Gamma (p+2)\, \Gamma \left(\frac{1}{x}\right)}{\Gamma \left(p+1+\frac{1}{x}\right)}-1}{x-1}$$ and, assuming $0< x <1$ $$\lim_{p\to \infty } \, \frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma \left(p+1+\frac{1}{x}\right)}=0$$ since, using expansion for large $p$ $$\frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma \left(p+1+\frac{1}{x}\right)}=\Gamma \left(\frac{1}{x}\right) p^{1-\frac{1}{x}}\left(1+ \frac{(x-1) (2 x+1)}{2 x^2}\frac 1p+O\left(\frac{1}{p^2}\right)\right)$$
