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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.

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  • $\begingroup$ One thing that may be a point of confusion here is the difference between a series and a Taylor series; any given function has a uniquely defined Taylor series (that doesn't necessarily converge to the function, but that's a different discussion...) but there can be any number of series for the same function in the same way that there can be any number of series for a given number, yes. $\endgroup$ Nov 8 '19 at 19:06
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    $\begingroup$ @StevenStadnicki The Maclaurin series (Taylor series expanded at $0$) is unique, but it's often possible to expand Taylor series at multiple points. $\endgroup$
    – J.G.
    Nov 8 '19 at 19:07
  • $\begingroup$ @StevenStadnicki "any given function has a uniquely defined Taylor series (that doesn't necessarily converge to the function, but that's a different discussion...)" Do you know the proof for this? By "not necessarily converge to the function", do you mean asymptotic series? $\endgroup$ Nov 8 '19 at 19:08
  • $\begingroup$ @J.G. mea culpa; good point that! $\endgroup$ Nov 8 '19 at 19:14
  • $\begingroup$ @JamesWarthington The Taylor series (as J.G. points out, I should say Maclaurin series since I'm really only considering the ones around 0) is unique if it exists because of the uniqueness of the derivative of a function at a point. But the series for a function can fail to converge to the function if the function isn't analytic; consider the series for $e^{-1/x^2}$ if you haven't seen it, as this is the classic example. $\endgroup$ Nov 8 '19 at 19:19
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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$$a_n=\left(\frac{1}{1-x}-\sum_{k=0}^{n-1}a_k\right)-\left(\frac{1}{1-x}-\sum_{k=0}^na_k\right)\\=\frac{n!x^n}{(1-x)\prod_{j=1}^{n-1}(1+jx)}-\frac{(n+1)!x^{n+1}}{(1-x)\prod_{j=1}^n(1+jx)}.$$

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  • $\begingroup$ I wish to know how to expand it. $\endgroup$ Nov 8 '19 at 19:04
  • $\begingroup$ @JamesWarthington See edit. $\endgroup$
    – J.G.
    Nov 8 '19 at 19:05
  • $\begingroup$ Can you do this in a few more steps so I can learn? I don't quite understand what do you mean by increasing n by 1. How can one derive the series from $\dfrac{1}{1-x}$ without knowing the series beforehand? $\endgroup$ Nov 8 '19 at 19:12
  • $\begingroup$ @JamesWarthington See my latest edit. $\endgroup$
    – J.G.
    Nov 8 '19 at 19:19
  • $\begingroup$ I think there is something wrong with your second term. It is $\dfrac{2x^2}{(1+x)(1+2x)}$ instead of $\dfrac{2x^2}{(1+x)(1-x)}$ $\endgroup$ Nov 9 '19 at 13:14
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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 $x-(n+1)xf(n).$

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