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If we know the geometric series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$, is it possible to derive the series of $1/x$ from it?

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    $\begingroup$ Can't you just set $x$ equal to $(1-x)$ in your formula? $\endgroup$ – man_in_green_shirt Jan 20 '17 at 20:31
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As $\frac1x$ is not analytic (it isn't even continuous at $0$) it cannot have a Taylor expansion around $x=0$.

On the other hand, it has a Laurant series expansion which is $\frac1x$

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  • $\begingroup$ He didn't say he was looking for an expansion around 0. $\endgroup$ – Vik78 Jan 20 '17 at 20:35
  • $\begingroup$ @Vik78, Actually, I thought about Taylor series. $\endgroup$ – user300045 Jan 20 '17 at 20:37
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    $\begingroup$ Then you can only get power-series around $x=a$ for some non-zero $a$. $\endgroup$ – Dennis Gulko Jan 20 '17 at 20:38
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    $\begingroup$ @Vik78 How do you know that the OP is male? $\endgroup$ – Mark Viola Jan 20 '17 at 20:38
  • $\begingroup$ Idk, it was a verbal slip I guess. Apologies to OP if I was mistaken. $\endgroup$ – Vik78 Jan 20 '17 at 20:41
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Following man_in_green_shirt's suggestion, you can substitute $(1-x)$ into the power series for $\frac{1}{1-x}$ to get a power series for $\frac{1}{x}$. Since the original series converges for $x\in (-1,1)$, this series for $\frac{1}{x}$ converges in $(0, 2)$.

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