# How do you find the sum of the infinite geometric series given 1+2/3+4/9+...?

Oct 22, 2016

$1 + \frac{2}{3} + \frac{4}{9} + \ldots = 3$

#### Explanation:

The general term of any geometric series can be written in the form:

${a}_{n} = a \cdot {r}^{n - 1} \text{ }$ for $n = 1 , 2 , 3 , \ldots$

where $a$ is the initial term and $r$ the common ratio

In our case we have:

${a}_{n} = 1 \cdot {\left(\frac{2}{3}\right)}^{n - 1} \text{ }$ for $n = 1 , 2 , 3 , \ldots$

with initial term $a = 1$ and common ratio $r = \frac{2}{3}$

The general formula for the infinite sum (proved below) is:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}$ when $\left\mid r \right\mid < 1$

So in our case:

${\sum}_{n = 1}^{\infty} \textcolor{b l u e}{1} \cdot {\left(\textcolor{p u r p \le}{\frac{2}{3}}\right)}^{n - 1} = \frac{\textcolor{b l u e}{1}}{1 - \textcolor{p u r p \le}{\frac{2}{3}}} = \frac{1}{\frac{1}{3}} = 3$

$\textcolor{w h i t e}{}$
Background

The general term of a geometric series can be written:

${a}_{n} = a \cdot {r}^{n - 1} \text{ }$ for $n = 1 , 2 , 3 , \ldots$

where $a$ is the initial term and $r$ is the common ratio.

Given such a series, we find:

$\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - r {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - r {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - {\sum}_{n = 2}^{N + 1} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a \left(1 - {r}^{N}\right)$

Dividing both ends by $\left(1 - r\right)$ we get the general finite sum formula:

$\textcolor{b l u e}{{\sum}_{n = 1}^{N} a {r}^{n - 1} = \frac{a \left(1 - {r}^{N}\right)}{1 - r}}$

If $\left\mid r \right\mid < 1$ then ${\lim}_{N \to \infty} {r}^{N} = 0$ and we find:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = {\lim}_{N \to \infty} {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} a {r}^{n - 1}} = {\lim}_{N \to \infty} \frac{a \left(1 - {r}^{N}\right)}{1 - r}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} a {r}^{n - 1}} = \frac{a}{1 - r}$

So we have the general formula for the infinite sum:

$\textcolor{b l u e}{{\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}} \text{ }$ when $\textcolor{b l u e}{\left\mid r \right\mid < 1}$