Gingan7mhd

2023-03-23

How to find the sum of the infinite geometric series given $1+\frac{2}{3}+\frac{4}{9}+...$?

### Answer & Explanation

Lola Rocha

Any geometric series' general term can be expressed in the following way:

${a}_{n}=a\cdot {r}^{n-1}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ for $n=1,2,3,...$

where $a$ is the initial term and $r$ the common ratio
In our case we have:

${a}_{n}=1\cdot {\left(\frac{2}{3}\right)}^{n-1}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ for $n=1,2,3,...$

with initial term $a=1$ and common ratio $r=\frac{2}{3}$
The general formula for the infinite sum (proved below) is:

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

So in our case:

$\sum _{n=1}^{\infty }1\cdot {\left(\frac{2}{3}\right)}^{n-1}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3$

Background
The general term of a geometric series can be written:

${a}_{n}=a\cdot {r}^{n-1}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ for $n=1,2,3,...$

where $a$ is the initial term and $r$ is the common ratio.
Given such a series, we find:

$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=\sum _{n=1}^{N}a{r}^{n-1}-r\sum _{n=1}^{N}a{r}^{n-1}$
$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=\sum _{n=1}^{N}a{r}^{n-1}-r\sum _{n=1}^{N}a{r}^{n-1}$
$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=\sum _{n=1}^{N}a{r}^{n-1}-\sum _{n=2}^{N+1}a{r}^{n-1}$
$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=a+\overline{)\sum _{n=2}^{N}a{r}^{n-1}}-\overline{)\sum _{n=2}^{N}a{r}^{n-1}}-a{r}^{N}$
$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=a-a{r}^{N}$
$\left(1-r\right)\sum _{n=1}^{N}a{r}^{n-1}=a\left(1-{r}^{N}\right)$

Dividing both ends by $\left(1-r\right)$ we get the general finite sum formula:

$\sum _{n=1}^{N}a{r}^{n-1}=\frac{a\left(1-{r}^{N}\right)}{1-r}$

If $|r|<1$ then $\underset{N\to \infty }{lim}{r}^{N}=0$ and we find:

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

So we have the general formula for the infinite sum:

$\sum _{n=1}^{\infty }a{r}^{n-1}=\frac{a}{1-r}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ when $|r|<1$

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