Find the sum of the infinite geometric series 8 + 4 + 2 + 1 +…

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