How to develop $\frac{1}{1-x}$ into $1+\frac{x}{1+x}+\frac{1\cdot2x^2}{(1+x)(1+2x)}+\dots$ I know that, by long division, or binomial formula or Taylor formula that this function can be developed into the geometric series:
$1/(1-x)=1+x+x^2+x^3+x^4+\ldots $ and I thought that this is the only series that represent the function mentioned above. However, recently, I read "Traite Elementaire des Series" by Eugene Catalan and he said, on page 60, that "The same function can admit multiple expansions". He gives the series below:
$$\frac{1}{1-x}=1+\frac{x}{1+x}+\frac{1\cdot2x^2}{(1+x)(1+2x)}+\frac{1\cdot2\cdot3x^3}{(1+x)(1+2x)(1+3x)}+\dots$$
I am just curious how can we derive this second series. So a function may admit multiple series representation? This is something new that I don't know. Maybe a better way to think about this is that two or multiple series can converge to the same function, which is the same as multiple numerical series may converge to the same sum.
 A: There are two questions we need to separate here. One is how you prove this series if presented with it; the other is how to invent it in the first place. For question 1, let's exploit the fact we've been told the sum is $\frac{1}{1-x}$, subtract partial sums from this target, then see whether the remainders are informative. In particular$$\frac{1}{1-x}-1=\frac{x}{1-x},\,\frac{1}{1-x}-1-\frac{x}{1+x}=\frac{x}{1-x}-\frac{x}{1+x}=\frac{2x^2}{(1-x)(1+x)},$$and so on. Sooner or later, you'll conjecture$$\frac{1}{1-x}-\sum_{k=0}^n\frac{k!x^k}{\prod_{j=1}^k(1+jx)}=\frac{(n+1)!x^{n+1}}{(1-x)\prod_{j=1}^n(1+jx)},$$which you can prove by induction. For question 2, I suspect a mathematician decided to solve$$\frac{1}{1-x}-\sum_{k=0}^na_k=\frac{(n+1)!x^{n+1}}{(1-x)\prod_{j=1}^n(1+jx)},$$giving$$a_n=\left(\frac{1}{1-x}-\sum_{k=0}^{n-1}a_k\right)-\left(\frac{1}{1-x}-\sum_{k=0}^na_k\right)\\=\frac{n!x^n}{(1-x)\prod_{j=1}^{n-1}(1+jx)}-\frac{(n+1)!x^{n+1}}{(1-x)\prod_{j=1}^n(1+jx)}.$$
A: Partial answer: Lets begin with the second term. Then $n$th term of this series satisfies the recurrence relation $$f(n)=\dfrac{nx}{(1+nx)}f(n-1),\qquad f(1)=\dfrac{x}{1+x}.$$ For summing series, my favorite method is telescoping. So, let me find another sequence $(a_n)$ such that $f(n)=a_{n-1}-a_n$ with the additional relation $a_{n}=\dfrac{nx}{(1+nx)}a_{n-1}.$ Then we have $$f(n)=a_{n-1}-\dfrac{nx}{(1+nx)}a_{n-1}=\dfrac{a_{n-1}}{1+nx},\qquad  a_{n-1}=(1+nx)f(n),$$ and it determines $(a_n)$ uniquely.  After telescoping we get $$\sum_{n=1}^mf(n)=a_0-a_n$$ and substituting for $(a_n)$ simplifies this to $x-(n+1)xf(n).$
