How do you find the fifth partial sum of sum_(i=1)^oo 2(1/3)^i?

Jul 2, 2017

${\sum}_{i = 1}^{5} 2 {\left(\frac{1}{3}\right)}^{i} = \frac{242}{243}$

Explanation:

This is a geometric series with initial term $a = \frac{2}{3}$ and common ratio $r = \frac{1}{3}$.

The general term of the series may be written:

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

We find:

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

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

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

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

So dividing both ends by $\left(1 - r\right)$, we find:

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

So in the given example, with $a = \frac{2}{3}$ and $r = \frac{1}{3}$, we find:

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

$\textcolor{w h i t e}{{\sum}_{i = 1}^{5} 2 {\left(\frac{1}{3}\right)}^{i}} = \frac{\frac{2}{3} \left(1 - \frac{1}{243}\right)}{\frac{2}{3}}$

$\textcolor{w h i t e}{{\sum}_{i = 1}^{5} 2 {\left(\frac{1}{3}\right)}^{i}} = \frac{242}{243}$