Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 13

f(x)=(4x)/(x^2+2x-3), c=0
Let's first factorize the denominator of the function,
x^2+2x-3=x^2+3x-x-3
=x(x+3)-1(x+3)
=(x+3)(x-1)
Now let, (4x)/(x^2+2x-3)=A/(x+3)+B/(x-1)
4x=A(x-1)+B(x+3)
4x=Ax-A+Bx+3B
4x=(A+B)x-A+3B
equating the coefficients of the like terms,
A+B=4 ----------------(1)
-A+3B=0 ------------(2)
From equation 2,
A=3B
Substitute A in equation 1,
3B+B=4
4B=4
B=1
Plug in the value of B in equation 2,
-A+3(1)=0
A=3
The partial fraction decomposition is thus,
(4x)/(x^2+2x-3)=3/(x+3)+1/(x-1)
=3/(3(1+x/3))+1/(-1(1-x))
=1/(1-(-x/3))+(-1)/(1-x)
Since both fractions are in the form of a/(1-r)
Power series is a geometric series,
=sum_(n=0)^oo(-x/3)^n+sum_(n=0)^oo(-1)x^n
=sum_(n=0)^oox^n/(-3)^n+sum_(n=0)^oo(-1)x^n
=sum_(n=0)^oo(1/(-3)^n-1)x^n
Interval of convergence |-x/3|<1,|x|<1
|x/3|<1 and |x|<1
-3Interval of convergence is the smaller of the intervals of convergence of the two individual fractions,
So, Interval of convergence is (-1,1)