# Under what circumstances does the infinite series #1+(2x/3)+(2x/3)^2+(2x/3)^3+...# converge and what is the sum when #x=1.2# ?

##### 1 Answer

Sep 6, 2016

#### Answer:

The series converges when:

#-3/2 < x < 3/2#

If

#1+(2x/3)+(2x/3)^2+(2x/3)^3+... = 5#

#### Explanation:

This is a geometric series with initial term

A non-zero geometric series with common ratio

In our case, that means:

#abs((2x)/3) < 1#

Multiplying both sides by

#abs(x) < 3/2#

which expands to mean:

#-3/2 < x < 3/2#

The sum of an geometric series with initial term

#sum_(n=1)^oo ar^(n-1) = a/(1-r)#

So given

#sum_(n=1)^oo ar^(n-1) = 1/(1-0.8) = 1/0.2 = 5#