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# ?

Question

2882 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-06T21:30:26

The series converges when:

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

If #x = 1.2# then:

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

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#