a We want to find the sum of the given infinite geometric series.

8 + 4 + 2 + \dots

We can find the sum of an infinite geometric series if and only if the absolute value of the common ratio is smaller than

1 .

∣ r ∣ < 1 ∣ r ∣ \geq 1 \Leftrightarrow converges \Leftrightarrow diverges

Let's take a look at the terms of our series!

To get the next term, we multiply by

\frac{1}{2} .

This is the common ratio of the series.

∣ ∣ ∣ ∣ ∣ \frac{1}{2} ∣ ∣ ∣ ∣ ∣ = \frac{1}{2} < 1

Since the absolute value of the common ratio is less than

1,

the series converges, so we can calculate its sum. Recall the formula for the sum of an infinite geometric series.

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

In this formula,

r

is the common ratio and

a

is the first term in the series. Let's substitute

r = \frac{1}{2}

and

a = 8

to evaluate the sum.

\frac{a}{1 - r}

$r = \frac{1}{2}$ , $a = 8$

\frac{8}{1 - \frac{1}{2}}

Rewrite $1$ as $\frac{2}{2}$

\frac{8}{\frac{2}{2} - \frac{1}{2}}

Subtract fractions

\frac{8}{\frac{2 - 1}{2}}

Subtract term

\frac{8}{\frac{1}{2}}

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{8 \cdot 2}{1}

\frac{1 6}{1}

$\frac{a}{1} = a$

16

The sum of the given series is

16 .

Exercise