# How do you find the sum of the infinite geometric series given 5/3-10/9+20/27-...?

Jul 20, 2017

$S = 1$

#### Explanation:

We first find the factor :

$r = \frac{- \frac{10}{9}}{\frac{5}{3}} = - \frac{2}{3}$

The sum of an infinite geometric series is given by the formula :

$S = {a}_{1} / \left(1 - r\right)$ where ${a}_{1}$ is the fisrt term and $| r | \le 1$ (which in our case it is).

So the sum is :

$S = {a}_{1} / \left(1 - r\right) = \frac{\frac{5}{3}}{1 - \left(- \frac{2}{3}\right)} = \frac{\frac{5}{3}}{1 + \frac{2}{3}} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$

Jul 20, 2017

Sum of the infinite geometric series is $1$

#### Explanation:

In the series $\frac{5}{3} - \frac{10}{9} + \frac{20}{27} - \ldots . .$

whilr first term ${a}_{1} = \frac{5}{3}$, common ratio is $\frac{- \frac{10}{9}}{\frac{5}{3}} = \frac{\frac{20}{27}}{- \frac{10}{9}} = - \frac{2}{3}$

As for common ratio $r = - \frac{2}{3}$, $| r | = | - \frac{2}{3} | < 1$,

sum of the infinite geometric series is

${a}_{1} / \left(1 - r\right) = \frac{\frac{5}{3}}{1 - \left(- \frac{2}{3}\right)} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$

Jul 20, 2017

${S}_{\infty} = 1$

#### Explanation:

$\text{for a geometric sequence the sum of n terms is}$

S_n=(a(1-r^n))/(1-r);(r!=1)

$\text{where a is the first term and r, the common ratio}$

$\text{as " ntooo,r^nto0" and " S_n" can be expressed as}$

S_oo=a/(1-r);(|r|<1)

$\Rightarrow r = \frac{- \frac{10}{9}}{\frac{5}{3}} = - \frac{2}{3} \rightarrow | r | < 1$

$\Rightarrow {S}_{\infty} = \frac{\frac{5}{3}}{1 + \frac{2}{3}} = 1$