# Question #d7237

Mar 21, 2017

$\textcolor{b l u e}{\text{Thus the sum must be converging}}$
$\textcolor{b l u e}{{\lim}_{n \to \infty} s = 30 \times \frac{5}{3} = 50}$

$\textcolor{b l u e}{s = {\sum}_{i = 1 \to \infty} 30 {\left(\frac{2}{5}\right)}^{i - 1}}$

#### Explanation:

${a}_{1} = 30 {r}^{0} \to 30 {r}^{1 - 1}$

${a}_{2} = 30 {r}^{1} \to 30 {r}^{2 - 1}$

${a}_{3} = 30 {r}^{2} \to 30 {r}^{3 - 1}$

As $r = \frac{2}{5}$ then ${r}^{i}$ is decreasing as $i$ is increasing

Thus $30 {r}^{i}$ is decreasing as $i$ is increasing

$\textcolor{b l u e}{\text{Thus the sum must be converging}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set the sum to be s

$s = 30 + 30 \left(\frac{2}{5}\right) + 30 {\left(\frac{2}{5}\right)}^{2} + 30 {\left(\frac{2}{5}\right)}^{3} + . . + 30 {\left(\frac{2}{5}\right)}^{n}$

$s \left(\frac{2}{5}\right) = 30 \left(\frac{2}{5}\right) + 30 {\left(\frac{2}{5}\right)}^{2} + . . + 30 {\left(\frac{2}{5}\right)}^{n} + 30 {\left(\frac{2}{5}\right)}^{n + 1}$

$s \left(1 - \frac{2}{5}\right) = 30 - 30 {\left(\frac{2}{5}\right)}^{n + 1} = 30 \left[1 - {\left(\frac{2}{5}\right)}^{n + 1}\right]$

$s = 30 \times \frac{1 - {\left(\frac{2}{5}\right)}^{n + 1}}{1 - \frac{2}{5}}$

As this is an infinite series we have

${\lim}_{n \to \infty} s = 30 \times \frac{1 - 0}{\frac{3}{5}}$

$\textcolor{b l u e}{{\lim}_{n \to \infty} s = 30 \times \frac{5}{3} = 50}$

$\textcolor{b l u e}{s = {\sum}_{i = 1 \to \infty} 30 {\left(\frac{2}{5}\right)}^{i - 1}}$