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I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. 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 $xin (0,infty)$? I can't find any. Thanks.

The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.

On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.

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.

The geometric series $sum_{ngeq0}(1-x)^n$ converges for

$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$

and we have

$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$

The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:

• 
  ${a}$;


• 
  $(a-r,a+r)$, for some $rin(0,+infty)$;


• 
  $(a-r,a+r]$, for some $rin(0,+infty)$;


• 
  $[a-r,a+r)$, for some $rin(0,+infty)$;


• 
  $[a-r,a+r]$, for some $rin(0,+infty)$;


• 
  $mathbb R$.

Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.