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