g(x)=5/(2x-3), c=-3 Find a power series for the function, centered at c and determine the interval of convergence.

If x is a variable, then an infinite series of the form sum_(n=0)^ooa_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+......+a_n(x-c)^n+..... is a power series centered at x=c, where c is a constant.
Given g(x)=5/(2x-3) , c=-3
Let's write g(x) in the form a/(1-r)
g(x)=5/(2x-3)
=(5/2)/(x-3/2)
=(5/2)/(x+3-3/2-3)
=(5/2)/(x+3-9/2)
=(5/2)/(-9/2(1-2/9(x+3)))
=((5/2)(-2/9))/(1-2/9(x+3))
=(-5/9)/(1-2/9(x+3))
So a=-5/9 and r=2/9(x+3)
So, the power series for g(x) is sum_(n=0)^ooar^n
=sum_(n=0)^oo(-5/9)(2/9(x+3))^n
=-5sum_(n=0)^oo(2^n(x+3)^n)/9^(n+1)
This power series is a geometric series and it converges if |r|<1
|2/9(x+3)|<1
-1<2/9(x+3)<1
-9<(2x+6)<9
-9-6<2x<9-6
-15<2x<3
-15/2Interval of convergence is (-15/2,3/2)