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

S (n) = 16 i think

almost positive