# How do you calculate sum_(n=1)^(∞)(3/2)^(1-2n)?

## How do you calculate sum_(n=1)^(∞)(3/2)^(1-2n)?

Jan 24, 2018

${\sum}_{n = 1}^{\infty} {\left(\frac{3}{2}\right)}^{1 - 2 n} = \frac{6}{5}$

#### Explanation:

Note that:

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

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} {\left(\frac{3}{2}\right)}^{1 - 2 n}} = {\sum}_{n = 1}^{\infty} \left(\frac{2}{3}\right) {\left(\frac{2}{3}\right)}^{2 n - 2}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} {\left(\frac{3}{2}\right)}^{1 - 2 n}} = {\sum}_{n = 1}^{\infty} \left(\frac{2}{3}\right) {\left(\frac{4}{9}\right)}^{n - 1}$

This is now in the standard form of a geometric series with initial term $a = \frac{2}{3}$ and common ratio $\frac{4}{9}$.

Given any geometric series, the general term can be written:

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

where $a$ is the initial term and $r$ the common ratio.

Then we find:

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

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - {\sum}_{n = 2}^{N + 1} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a \left(1 - {r}^{N}\right)$

Dividing both ends by $\left(1 - r\right)$ we find:

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

So if $\left\mid r \right\mid < 1$ then:

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

Applying this in our example with $a = \frac{2}{3}$ and $r = \frac{4}{9}$ we find:

${\sum}_{n = 1}^{\infty} {\left(\frac{3}{2}\right)}^{1 - 2 n} = {\sum}_{n = 1}^{\infty} \left(\frac{2}{3}\right) {\left(\frac{4}{9}\right)}^{n - 1} = \frac{\frac{2}{3}}{1 - \frac{4}{9}} = \frac{\frac{2}{3}}{\frac{5}{9}} = \frac{6}{5}$