TITLE: Why does $\sum_{n\geq0}(1-x)^n=\frac1x$ have such a poor radius of convergence?
QUESTION [5 upvotes]: I am confused as to why $$\sum_{n\geq0}(1-x)^n=\frac1x$$
only works for $x\in (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $\frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $x\in (0,\infty)$? I can't find any. Thanks.

REPLY [4 votes]: The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents,  including  Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ ,  which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius 
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for  $1/z$ which has a simple pole at $z=0$ :
 - when developed at $z=1$ will have a convergence radius of $ 1$;
 - to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
 - there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.