f(x) = x/(x^2 - 3 x + 2) = x/((x-1)(x-2)) = A/(x-1)+B/(x-2)
(A(x-2)+B(x-1))/((x-1)(x-2))=((A+B)x-(2A+B))/((x-1)(x-2))
Solving for A,B
{
(A+B=1),
(2A+B=0)
:}
we obtain A=-1,B=2 So
f(x) = -1/(x-1)+2/(x-2)
Now considering that (x^{n+1}-1)/(x-1)=1+x+x^2+cdots+x^n
if abs(x) < 1 we know
lim_{n->oo}(x^{n+1}-1)/(x-1) = -1/(x-1) = sum_{n=0}^oo x^n
Using those results and supposing that abs(x) < 1 we propose
S(x) = -1/(x-1)+2/(x-2) = sum_{n=0}^oo x^n-sum_{n=0}^oo (x/2)^n
so S(x) = sum_{n=0}^oo(1-1/2^n)x^n.
This serie is convergent to f(x) for abs(x) < 1
Supposing you need a series representation for f(x) in the surroundings of x=3 you can proceed as follows.
Making y = x-3 then
f(y) = -1/(y+2)+2/(y+1)
so with abs(y) < 1
f(y)=lim_{n->oo}{-(1-(-y/2)^n)/(2(1-(-y/2)))+2(1-(-y)^n)/(1-(-y))}
or
f(y) = -1/2 sum_{n=0}^oo(-y/2)^n+2sum_{n=0}^n(-y)^n
Finally
S(x) = sum_{n=0}^oo (-1)^n(2-1/2^{n+1})(x-3)^n
This series converges to f(x) for 2 < x < 4
