# How do you find the sum of the infinite geometric series 1, 5, -25, 125,…?

Jan 23, 2016

This is not a geometric series.

If the first term is replaced with $- 1$ then the sum of the resulting geometric series does not converge.

#### Explanation:

If the sequence of numbers in the question is accurate then this is not a geometric series.

The ratios of successive pairs of terms are $5$, $- 5$, $- 5$.

It would be a geometric series if the first term was $- 1$, with common ratio $- 5$.

$- 1 , 5 , - 25 , 125 , \ldots$

The general term of this sequence is ${a}_{n} = \left(- 1\right) {\left(- 5\right)}^{n - 1}$

Then assuming that was the intention, the sum of this geometric series does not converge.

The sum of an infinite geometric sequence with general term ${a}_{n} = a {r}^{n - 1}$ converges when $\left\mid r \right\mid < 1$ and not otherwise.

In general:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}$

provided $\left\mid r \right\mid < 1$

We find:

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

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

$= a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

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

Dividing both ends by $\left(1 - r\right)$ we get:

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

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

Hence:

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