# Under what circumstances does the infinite series 1+(2x/3)+(2x/3)^2+(2x/3)^3+... converge and what is the sum when x=1.2 ?

Sep 6, 2016

The series converges when:

$- \frac{3}{2} < x < \frac{3}{2}$

If $x = 1.2$ then:

$1 + \left(2 \frac{x}{3}\right) + {\left(2 \frac{x}{3}\right)}^{2} + {\left(2 \frac{x}{3}\right)}^{3} + \ldots = 5$

#### Explanation:

This is a geometric series with initial term $1$ and common ratio $\frac{2 x}{3}$

A non-zero geometric series with common ratio $r$ will converge if and only if $\left\mid r \right\mid < 1$.

In our case, that means:

$\left\mid \frac{2 x}{3} \right\mid < 1$

Multiplying both sides by $\frac{3}{2}$, that becomes:

$\left\mid x \right\mid < \frac{3}{2}$

which expands to mean:

$- \frac{3}{2} < x < \frac{3}{2}$

The sum of an geometric series with initial term $a$ and common ratio $r$ is given by the formula:

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

So given $a = 1$ and $r = \frac{2}{3} x = \frac{2}{3} \cdot 1.2 = 0.8$ we find:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{1}{1 - 0.8} = \frac{1}{0.2} = 5$