Given: 

\(f(x)=x^3+3x-9\) on \([-2,3]\). 

Since \(f'(x)=3x^2+3>0\), \(f\) is increasing, so the greatest value is at \(x=3\): 

\(f(3)=27\).

Sum to infinity of decreasing GP: 

\(S=\dfrac{a}{1-r}=27\) with 

Also \(f'(0)=3\Rightarrow\) difference of first two terms: 

\(a-ar=a(1-r)=3\).

From \(a=\;27(1-r)\), 

plug into \(a(1-r)=3\): \(27(1-r)^2=3\Rightarrow(1-r)^2=\dfrac{1}{9}\Rightarrow 1-r=\dfrac{1}{3}\) (take positive)

Common ratio: \(r=1-\dfrac{1}{3}=\boxed{\dfrac{2}{3}}\).