# How do you determine if a_n=1+3/7+9/49+...+(3/7)^n+... converge and find the sums when they exist?

Mar 18, 2018

Convergent.

Sum to infinity $= \frac{7}{4}$

#### Explanation:

This is a geometric series:

The $\boldsymbol{n t h}$ term in a geometric series is given by:

$a {r}^{n - 1}$

Where:

$\boldsymbol{a}$ is the first term, $\boldsymbol{r}$ is the common ratio and $\boldsymbol{n}$ is the $\boldsymbol{n t h}$ term.

If:

$a , b , c$ are in geometric sequence, then:

$\frac{b}{a} = \frac{c}{b}$ This is known as the common ratio.

So we have:

$\frac{\frac{3}{7}}{1} = \frac{\frac{9}{49}}{\frac{3}{7}} = \frac{3}{7}$

Common ratio $\boldsymbol{\frac{3}{7}}$

First term is $\boldsymbol{1}$

The sum of a geometric series is given by:

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

From this we can see that if:

$| r | < 1$ then:

${\lim}_{n \to \infty} a \left(\frac{1 - {r}^{n}}{1 - r}\right) = \frac{a}{1 - r}$

This is a finite value, and the sum is said to converge.

If:

$| r | > 1$ then:

The sum increases without bound, and is said to diverge.

We have $r = \frac{3}{7}$

$| \frac{3}{7} | < 1$ so the series converges.

${\sum}_{n = 1}^{\infty} {\left(\frac{3}{7}\right)}^{n - 1} = \frac{1}{1 - \frac{3}{7}} = \textcolor{b l u e}{\frac{7}{4}}$