The defining characteristic of all geometric sequences is a common ratio which is a constant when dividing any term by the term preceding it.
In this case the common ratio is: -6/9=4/-6=r=-2/3
An infinite series will have a sum when r^2<1, so in this case the sum will converge to an actual value because (-2/3)^(+oo) approaches zero.
The sum of any geometric sequence is:
s(n)=a(1-r^n)/(1-r), since we have a common ratio of -2/3 and we want to calculate an infinite series, ie, n approaches infinity, the sum becomes simply:
s(n)=a/(1-r) (because (1-r^+oo) approaches 1 as n approaches +oo)
So our infinite sum is:
s(+oo)=9/(1--2/3)
s(+oo)=9/(1+2/3)
s(+oo)=9/(5/3)
s(+oo)=27/5
s(+oo)=54/10
s(+oo)=5.4