Question

How do you find the sum of the infinite geometric series $3 - 2 + \dfrac{4}{3} - \dfrac{8}{9} + ......$ ?

Answer 1

Here we find the sum of an infinite geometric series having rations with an absolute value less than one and use the following formula.

Formula used:
If $- 1 < r < 1$ then the infinite geometric series,
${a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........ + {a_1}{r^{n - 1}}$ Converges to a particular value. This value is given by
$S = \dfrac{{{a_1}}}{{1 - r}}$ where ${a_1}$ is the first term and $r$ is the common ratio. The series converges because each term gets smaller and smaller (since$- 1 < r < 1$).

Complete step by step answer:
Given series is $3 - 2 + \dfrac{4}{3} - \dfrac{8}{9} + ......$, it is a geometric series. Because the common ratio of this series are equal in each term.
Let the first term of the series is ${a_1} = 3$ and find the common ratio by using the formula $r = \dfrac{{{a_2}}}{{{a_1}}}$ this means the ration of second term and the first term.
The second term in this series is ${a_2} = - 2$ then, the common ratio is $r = - \dfrac{2}{3}$
Since the common ration has value between $- 1$ and $1$, we know the series will converge to some value. so we can calculate the sum using the formula,
$S = \dfrac{{{a_1}}}{{1 - r}}$where ${a_1}$ is the first term and $r$ is the common ratio.

$S = \dfrac{3}{{1 - \left( { - \dfrac{2}{3}} \right)}}$
Doing simplification,
$S = \dfrac{3}{{1 + \dfrac{2}{3}}}$
Open the brackets in the denominator
$\Rightarrow S = \dfrac{3}{{\dfrac{5}{3}}}$
We just reverse the denominator fraction
$\Rightarrow S = \dfrac{3}{1} \times \dfrac{3}{5}$
Multiply the above terms
$\Rightarrow S = \dfrac{9}{5}$

Note: An infinite geometric series is the sum of an infinite geometric sequence. This series would have no last term. The general form of the infinite geometric series is ${a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........$ where ${a_1}$ is the first term and $r$ is the common ratio.
We can find the sum of all finite geometric series. But in the case of an infinite geometric series when the common ratio is greater than one, the terms in the sequence will get larger and larger and if you add the larger numbers, we won’t get a final answer. The only possible answer would be infinity. So, we don’t deal with the common ratio greater than one for an infinite geometric series.
An infinite series that has a sum is called a convergent series and the sum ${S_n}$ is called the partial sum of the series.