The sum of a geometric sequence can be expressed as:
s(n)=a(1-r^n)/(1-r), a=initial value, r=common ration, n=number of terms.
a=8 and r=4/8=2/4=1/2=1/2 so
s(n)=8(1-0.5^n)/(1-0.5)
However a neat thing happens when r^2<1 and n approaches infinity.
(1-0.5^n) becomes just 1. So what to remember is that when r^2<1 the sum of the infinite series is:
s(n)=a/(1-r), so in this case:
s(n)=8/(1-0.5)
s(n)=8/0.5
s(n)=16
