# How do you determine if a_n=6+19+3+4/25+8/125+16/625+...+(2/5)^n+... converge and find the sums when they exist?

Oct 22, 2017

See below.

#### Explanation:

From the fourth term on we have a geometric series.

The sum of this can be found using:

$a \left(\frac{1 - {r}^{n}}{1 - r}\right)$

Where $a$ is the first term, $r$ is the common ratio and $n$ is the nth term.

So from 4th term to infinity we have:

$\frac{4}{25} \left(\frac{1 - {\left(\frac{2}{5}\right)}^{n}}{1 - \left(\frac{2}{5}\right)}\right) = \frac{1}{\frac{3}{5}} = \frac{5}{3} \cdot \frac{4}{25} = \frac{20}{75} = \frac{4}{15}$

${\sum}_{n = 2}^{\infty} {\left(\frac{2}{5}\right)}^{n} = \frac{4}{15}$ ( Convergent )

Then we have:

$\sum \left(6 + 19 + 3\right) = 28$

And:

$\sum \left(6 + 19 + 3\right) + {\sum}_{n = 2}^{\infty} {\left(\frac{2}{5}\right)}^{n} = \frac{424}{15}$ ( Convergent )