How do you find the sum of the infinite geometric series 1/3 + 4/9 + 16/27 + 64/81?

1 Answer
Dec 14, 2015

The series diverges, as

#sum_(n=0)^oo1/3(4/3)^n = oo#

Explanation:

An infinite geometric series is a series of the form

#sum_(n=0)^ooa_0r^n#

where #a_0# is the initial term and #r# is the ratio between terms.

To find #r#, then, we can divide any term after the first by the term prior, as
#(a_0r^k)/(a_0r^(k-1)) = r#

Doing so for the given series, we find that

#r = (4/9)/(1/3) = 4/3#

As #|r| >= 1#, that means the terms do not tend towards #0# as #n->oo#, meaning the series diverges.