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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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    Can't you just set $x$ equal to $(1-x)$ in your formula? – 2017-01-20

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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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    He didn't say he was looking for an expansion around 0. – 2017-01-20
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    @Vik78, Actually, I thought about Taylor series. – 2017-01-20
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    Then you can only get power-series around $x=a$ for some non-zero $a$. – 2017-01-20
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    @Vik78 How do you know that the OP is male? – 2017-01-20
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    Idk, it was a verbal slip I guess. Apologies to OP if I was mistaken. – 2017-01-20
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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 $(0, 2)$.