Question

How do you find the sum of the infinite geometric series $- 3{\text{, }} - \dfrac{3}{2}{\text{, }} - \dfrac{3}{4}{\text{, }} - \dfrac{3}{8}{\text{,}}......$?

Answer 1

Since this is a geometric series, the ratios of any two consecutive terms will be the same throughout the series. Find this ratio which is also called the common ratio by considering any two consecutive terms. Then apply the formula of the sum of infinite geometric series which is ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio.

Complete step by step answer:
According to the question, we have to show how to calculate the sum of the infinite geometric series $- 3{\text{, }} - \dfrac{3}{2}{\text{, }} - \dfrac{3}{4}{\text{, }} - \dfrac{3}{8}{\text{,}}......$
Let the sum of terms is denoted by a letter $S$, then we have:
$\Rightarrow S = - 3 + \left( { - \dfrac{3}{2}} \right) + \left( { - \dfrac{3}{4}} \right) + \left( { - \dfrac{3}{8}} \right) + ....$
This can be simplified as:
$\Rightarrow S = - 3 - \dfrac{3}{2} - \dfrac{3}{4} - \dfrac{3}{8} - ....$
If we take -3 outside of the entire expression, we’ll get:
$\Rightarrow S = - 3\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} - ....} \right){\text{ }}.....{\text{(1)}}$
The expression $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} - ...$ is the sum of another geometric series. Its first term is 1(i.e. $a = 1$). And we can determine its common ratio by taking the ratio of any two consecutive terms. If we consider second and third terms, we have:
$\Rightarrow r = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}} = \dfrac{1}{2}$
Now we know that the formula to determine the sum of infinite geometric series is given as:
$\Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$
Putting this formula in equation (1), we’ll get:
$\Rightarrow S = - 3\left( {{S_\infty }} \right) \\\ \Rightarrow S = - 3\left( {\dfrac{a}{{1 - r}}} \right)$
Putting the values of $a$ and $r$ as calculated above, we’ll get:
$\Rightarrow S = - 3\left( {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right) \\\ \Rightarrow S = - 3\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) \\\ \Rightarrow S = - 3\left( 2 \right) = - 6$

Therefore, the sum of the infinite geometric series $- 3{\text{, }} - \dfrac{3}{2}{\text{, }} - \dfrac{3}{4}{\text{, }} - \dfrac{3}{8}{\text{,}}......$ is -6.
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Note: The formula to calculate the sum of finite geometric series is given as:
$\Rightarrow S = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$
And the formula to calculate the sum of infinite geometric series is given as:
$\Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$
The last formula we have already used in the above problem. If the number of terms in the geometric series is finite then we can use the first formula to calculate the sum.