Question

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

Answer 1

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.