conversion of a powerseries $-3x+4x^2-5x^3+\ldots $ into $ -2+\frac 1 x - 0 - \frac 1 {x^3} + \ldots $ This is initially a funny question, because I've found this on old notes but I do not find/recover my own derivation... But then the question is more general.
Q1:     
I considered the function
$ f(x) = - \frac {2x^2+3x}{(x+1)^2}  $ 
I expressed this by a powerseries $ f_1(x) = -3x + 4x^2 - 5x^3 + 6x^4 - \ldots $ 
and stated without the derivation that this is also
$  f_2(x) = \frac {-2}{1} -\frac {-1}{x} + 0 - \frac  {1} {x^3} + \frac {2}{x^4} - \ldots + \ldots $ 
and - well: hell, - don't see it now how I did it.
What was interesting to me was, that after looking for the fixpoints $ x_0=0, x_{1,2} =-2 $ the range of convergence in the expression by $f_1$ is obviously $ |x|<1 $ limited to the unit-interval but in that by $f_2$ it is infinity and $ |x|>1 $ .      
Q2:
I would like to be able to translate also other powerseries into an $f_2$-type-expression. (I remember to have read a remark of "expanding a powerseries at infinity" but have never seen an explanation of this - so this might be irrelevant for this case?) So: what is the technique to do this given a function in terms of a usual powerseries, for instance for the geometric series $ g(x)=1+x+x^2+ \ldots $ or some series $ h(x) = K + a*x + b*x^2 + c*x^3 + \ldots $ ?
[edit: minus-sign in f(x) was missing, one numerator in f2 was wrong]

Divide the numerator and denominator of $f(x)$ by $x^2$ and set $y=1/x$ then expand for $y$ and you have your expansion at infinity.
