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`

`-15/2<x<3/2`

Interval of convergence is `(-15/2,3/2)`

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