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

The series converges when:

-3/2 < x < 3/2

If x = 1.2 then:

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

Explanation:

This is a geometric series with initial term 1 and common ratio (2x)/3

A non-zero geometric series with common ratio r will converge if and only if abs(r) < 1.

In our case, that means:

abs((2x)/3) < 1

Multiplying both sides by 3/2, that becomes:

abs(x) < 3/2

which expands to mean:

-3/2 < x < 3/2

The sum of an geometric series with initial term a and common ratio r is given by the formula:

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

So given a = 1 and r = 2/3x = 2/3*1.2 = 0.8 we find:

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