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

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

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Answer 1 · 2015-12-14T11:01:08

The series diverges, as

$\infty \sum n = 0 \frac{1}{3} {(\frac{4}{3})}^{n} = \infty$ $sum_(n=0)^oo1/3(4/3)^n = oo$

Explanation:

An infinite geometric series is a series of the form

$\infty \sum n = 0 a_{0} r^{n}$ $sum_(n=0)^ooa_0r^n$

where $a_{0}$ $a_0$ is the initial term and $r$ $r$ is the ratio between terms.

To find $r$ $r$ , then, we can divide any term after the first by the term prior, as
$\frac{a_{0} r^{k}}{a_{0} r^{k - 1}} = r$ $(a_0r^k)/(a_0r^(k-1)) = r$

Doing so for the given series, we find that

$r = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3}$ $r = (4/9)/(1/3) = 4/3$

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

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

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