# How do you find the sum of the infinite geometric series 4 + 3 + 2.25 + 1.6875 +… ?

Nov 10, 2015

Use the formula ${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}$ to find $4 + 3 + 2.25 + \ldots = 16$

#### Explanation:

The sum of a geometric series with initial term $a$ and common ratio $r$ is $\frac{a}{1 - r}$.

To see this note that:

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

$= {\sum}_{n = 1}^{N} a {r}^{n - 1} - {\sum}_{n = 2}^{N + 1} a {r}^{n - 1}$

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

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

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

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

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

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

In our case $a = 4$ and $r = \frac{3}{4}$ so

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r} = \frac{4}{1 - \frac{3}{4}} = 16$