# Power series of function $f(x)=1/x$

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?

• Can't you just set $x$ equal to $(1-x)$ in your formula? – man_in_green_shirt Jan 20 '17 at 20:31

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 $\frac1x$
• Then you can only get power-series around $x=a$ for some non-zero $a$. – Dennis Gulko Jan 20 '17 at 20:38
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 $(0, 2)$.