# How do you find the sum of the infinite geometric series 8 + 4 + 2 + 1 +...?

Mar 3, 2018

$16$

#### Explanation:

This converging geometric progression is a special one.
$\text{G.P } 8 , 4 , 2 , 1 , \frac{1}{2} , \frac{1}{4.} . .$
Common multiple is $\frac{1}{2}$ .

When you find the sum of this G.P
$\implies 8 + 4 + 2 + 1 + \frac{1}{2} + \frac{1}{4.} \ldots$

$\implies 15 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8.} \ldots .$

$\implies 15 + 1$

Basically, you need to prove $\frac{1}{2} + \frac{1}{4} + \frac{1}{8.} . . = 1$

Proof:
${S}_{n} = \frac{1}{2} + \frac{1}{4} + \ldots \frac{1}{2} ^ \left(n - 1\right)$

Multiply $2$ both sides

$2 {S}_{n} = \frac{2}{2} + \frac{2}{4} + \ldots \frac{1}{2} ^ n$

Subtract ${S}_{n}$ from this.

$2 {S}_{n} - {S}_{n} = 1 + \frac{2}{4} - \frac{1}{2} + \frac{2}{8} - \frac{1}{4.} \ldots$

${S}_{n} = 1 - \frac{1}{2} ^ n$

As you increase the value of $n$ the value of ${S}_{n}$ tends to be $1$.

There's one more proof which is easy to understand and really amazing.

Given below is a square of side 1unit. First cut it half, then ${\left(\frac{1}{4}\right)}^{t h}$ and then ${\left(\frac{1}{8}\right)}^{t h}$ and so on. You'll see that as you increase the value of $n$ the sum tends to be $1$. 