# How do you find the sum of the infinite geometric series 1+1/5+1/25+....?

Nov 9, 2015

Divide terms to find $r$ and then apply the geometric series formula to find the sum to be $\frac{5}{4}$

#### Explanation:

A geometric series is a series of the form
$a + a r + a {r}^{2} + a {r}^{3} + \ldots$
where $a$ is an initial value and $r$ is a common ratio between terms.

In general, the sum of a geometric series is
$a + a r + a {r}^{2} + \ldots + a {r}^{n - 1} = a \frac{1 - {r}^{n}}{1 - r}$ (see below for derivation)

If $| r | < 1$ then as $n \to \infty , {r}^{n} \to 0$ so the formula reduces to

$a + a r + a {r}^{2} + \ldots = \frac{a}{1 - r}$

Now, in order to find $r$, it suffices to divide any term in the series after the first by the term prior to it, as
$\frac{a {r}^{n}}{a {r}^{n - 1}} = r$

So, picking the second and first terms, for this series we get
$r = \frac{\frac{1}{5}}{1} = \frac{1}{5}$

As the first term in the series gives us $a = 1$ we get the final sum
$\frac{1}{1 - \left(\frac{1}{5}\right)} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$

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What follows is a short derivation for the geometric series formula, and is not required for understanding the above solution.

Let ${S}_{n}$ be the $n$th partial sum of a geometric series with ratio $r$ and first term $a$, that is,
${S}_{n} = a + a r + a {r}^{2} + \ldots + a {r}^{n - 1}$

$\implies r {S}_{n} = a r + a {r}^{2} + \ldots + a {r}^{n}$

$\implies {S}_{n} - r {S}_{n} = a - a {r}^{n}$

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

$\implies {S}_{n} = a \frac{1 - {r}^{n}}{1 - r}$