How do you find the sum of the infinite geometric series (2/9)^2 - (2/9)^3 + (2/9)^4 - (2/9)^5 + ......?

Aug 14, 2016

${\left(\frac{2}{9}\right)}^{2} - {\left(\frac{2}{9}\right)}^{3} + {\left(\frac{2}{9}\right)}^{4} - {\left(\frac{2}{9}\right)}^{5} + \ldots = \frac{4}{33}$

Explanation:

Suppose:

$S = {\left(\frac{2}{9}\right)}^{2} - {\left(\frac{2}{9}\right)}^{3} + {\left(\frac{2}{9}\right)}^{4} - {\left(\frac{2}{9}\right)}^{5} + \ldots$

Then:

$\left(\frac{2}{9}\right) S = {\left(\frac{2}{9}\right)}^{3} - {\left(\frac{2}{9}\right)}^{4} + {\left(\frac{2}{9}\right)}^{5} - {\left(\frac{2}{9}\right)}^{6} + \ldots$

So we find:

$S + \left(\frac{2}{9}\right) S$

$= {\left(\frac{2}{9}\right)}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{3}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{4}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{5}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{6}}}} - \ldots$
$\textcolor{w h i t e}{= {\left(\frac{2}{9}\right)}^{2}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{3}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{4}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{5}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{2}{9}\right)}^{6}}}} + \ldots$
$= {\left(\frac{2}{9}\right)}^{2}$

That is:

$\left(\frac{11}{9}\right) S = \left(1 + \left(\frac{2}{9}\right)\right) S = S + \left(\frac{2}{9}\right) S = {\left(\frac{2}{9}\right)}^{2} = \frac{4}{81}$

Multiply both ends by $\frac{9}{11}$ to get:

$S = \frac{4}{33}$

$\textcolor{w h i t e}{}$
What is missing?

There is one vital thing missing from the above calculation, namely the fact that we have assumed that:

${\left(\frac{2}{9}\right)}^{2} - {\left(\frac{2}{9}\right)}^{3} + {\left(\frac{2}{9}\right)}^{4} - {\left(\frac{2}{9}\right)}^{5} + \ldots$ converges.

$\textcolor{w h i t e}{}$
A more formal approach

The general term of a geometric series can be written:

${a}_{n} = a \cdot {r}^{n - 1}$

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

Then for any positive integer $N$ we find:

$\left(1 - r\right) {\sum}_{n = 0}^{N} {a}_{n}$

$= {\sum}_{n = 0}^{N} a {r}^{n - 1} - r {\sum}_{n = 0}^{N} a {r}^{n - 1}$

$= a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 1}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 1}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

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

If $r \ne 1$ then we can divide both ends by $\left(1 - r\right)$ to get the general formula:

${\sum}_{n = 0}^{N} {a}_{n} = \frac{a \left(1 - {r}^{N}\right)}{1 - r}$

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

${\sum}_{n = 0}^{\infty} {a}_{n} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} {a}_{n} = {\lim}_{N \to \infty} \frac{a \left(1 - {r}^{N}\right)}{1 - r} = \frac{a}{1 - r}$

On the other hand, if $\left\mid r \right\mid > 1$ then ${\lim}_{N \to \infty} {r}^{N}$ does not converge to a finite limit, so the limit of the sum as $\frac{a}{1 - r}$ does not hold.

In our example:

${a}_{n} = {\left(\frac{2}{9}\right)}^{2} \cdot {\left(- \frac{2}{9}\right)}^{n - 1} \text{ }$ i.e. $a = {\left(\frac{2}{9}\right)}^{2}$ and $r = - \frac{2}{9}$

So $\left\mid r \right\mid < 1$ and we have:

${\sum}_{n = 0}^{\infty} {a}_{n} = \frac{{\left(\frac{2}{9}\right)}^{2}}{1 - \left(- \frac{2}{9}\right)} = \frac{\frac{4}{81}}{\frac{11}{9}} = \frac{4}{33}$