# How do you calculate sum_(n=1)^(∞)(4/3)^(2-n)?

## How do you calculate sum_(n=1)^(∞)(4/3)^(2-n)?

Jan 24, 2018

$S = \frac{16}{3}$

#### Explanation:

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First term $\left(n = 1\right)$, is:

${\left(\frac{4}{3}\right)}^{2 - 1} = {\left(\frac{4}{3}\right)}^{1} = \frac{4}{3}$

Second Term ($n = 2$), is:

${\left(\frac{4}{3}\right)}^{2 - 2} = {\left(\frac{4}{3}\right)}^{0} = 1$

Third term $\left(n = 3\right)$, is:

${\left(\frac{4}{3}\right)}^{2 - 3} = {\left(\frac{4}{3}\right)}^{- 1} = \frac{1}{\frac{4}{3}} ^ 1 = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

Fourth Term $\left(n = 4\right)$, is:

${\left(\frac{4}{3}\right)}^{2 - 4} = {\left(\frac{4}{3}\right)}^{- 2} = \frac{1}{\frac{4}{3}} ^ 2 = \frac{1}{\frac{16}{9}} = \frac{9}{16}$

Fifth term $\left(n = 5\right)$, is:

${\left(\frac{4}{3}\right)}^{2 - 5} = {\left(\frac{4}{3}\right)}^{- 3} = \frac{1}{\frac{4}{3}} ^ 3 = \frac{1}{\frac{64}{27}} = \frac{27}{64}$

As can be seen, the first five terms of this series are:

$\frac{4}{3} , 1 , \frac{3}{4} , \frac{9}{16} , \frac{27}{64}$

This is a geometric series with a common ratio of $\left({a}_{n + 1} / {a}_{n}\right) = {a}_{2} / {a}_{1} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$:

$r = \frac{3}{4}$

If $\left\mid r \right\mid < 1$ there is a formula for the sum of infinite geometric series:

$S = {a}_{1} / \left(1 - r\right)$

$S = \frac{\frac{4}{3}}{1 - \left(\frac{3}{4}\right)} = \frac{\frac{4}{3}}{\frac{1}{4}} = \frac{4}{3} \cdot \frac{4}{1} = \frac{16}{3}$

Jan 24, 2018

${\sum}_{n = 1}^{\infty} {\left(\frac{4}{3}\right)}^{2 - n} = \frac{16}{3}$

#### Explanation:

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

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

=${\sum}_{n = 0}^{\infty} {\left(\frac{3}{4}\right)}^{n - 1}$

=${\left(\frac{3}{4}\right)}^{- 1} \cdot \frac{1}{1 - \frac{3}{4}}$

=$\frac{\frac{4}{3}}{\frac{1}{4}}$

=$\frac{16}{3}$