Find the sum of the infinite geometric series: 9 - 6 + 4 - ...

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