Question

How do you find the sum of the infinite geometric series \[\dfrac{1}{2} + \dfrac{1}{4} +
\dfrac{1}{8} + \dfrac{1}{{16}}...\]?

Answer 1

Hint:In the given question, the ratio of each term is less than one. Therefore, we can say that the series is convergent. Therefore, to find the answer to this question, we will first find the common ratio. After that we will use the formula for the sum of a convergent infinite series and get the final answer.

Formula used:
$S = \dfrac{a}{{1 - r}}$, where, $S$is the sum of the infinite geometric series, $a$is the first term of the infinite geometric series and $r$is the common ratio

Complete step by step answer:
We are given the infinite geometric series \[\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +
\dfrac{1}{{16}}...\]
Here, the first term is: $a = \dfrac{1}{2}$
Now, we will find the common ratio which is the ratio of two consecutive terms of a geometric series.
We will take the ratio of the first two terms.
$r = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{2}}} = \dfrac{2}{4} = \dfrac{1}{2}$
Now, to find the sum of this infinite geometric series by using the formula $S = \dfrac{a}{{1 - r}}$.
$ \Rightarrow S = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}} = 1$
Thus, the sum of the given infinite geometric series \[\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +
\dfrac{1}{{16}}...\] is 1.

Note:
Here, we have used the concept of the convergent series. If an infinite series has a sum, the series is said to be convergent. On the contrary, an infinite series is said to be divergent; it has no sum. The infinite geometric series has a sum when the value of the common ratio belongs to the range $\left( { -
1,1} \right)$which means that $\left| r \right| < 1$. But if the value of the common ratio is less than minus one or greater than one, then the series will be the divergent series. Also if the value of the common ratio is greater than or equal to one, then the sum of the infinite geometric series will tend to infinity.