Question

How do you find the sum of the infinite geometric series given $1 + \dfrac{2}{3} + \dfrac{4}{9} + ...$ ?

Answer 1

Hint: In the geometric series each the previous term is multiplied by the common ratio which gives the next term. The formula for this is ${a_n} = {a_1}.{r^{n - 1}}$. To calculate the sum of series we use the formula ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$.

Complete step by step solution:
The objective of the problem is to find the sum of the infinite geometric series $1 + \dfrac{2}{3} + \dfrac{4}{9} + ...$
Given series $1 + \dfrac{2}{3} + \dfrac{4}{9} + ...$
The given series is the geometric series since there is a common ratio between each term of the series. In the given series the common ratio is $\dfrac{2}{3}$.
In the given series multiplying the previous term with a common ratio that is $\dfrac{2}{3}$ gives the next term.
The general form can be written as ${a_n} = {a_1}.{r^{n - 1}}$. Where r=$\dfrac{2}{3}$.
Now , the sum of series is calculated using the formula ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$. For the sum of a geometric series ${S_\infty }$ as n reaches $\infty $ $1 - {r^n}$ reaches one. Thus the formula becomes $\dfrac{a}{{1 - r}}$.
Here is the first term and r the common difference.
In the given infinite geometric series $1 + \dfrac{2}{3} + \dfrac{4}{9} + ...$ we have $a = 1,r = \dfrac{2}{3}$
Therefore the infinite series ${S_\infty } = \dfrac{a}{{1 - r}}$
Substitute the values of a and r in the infinite series formula, we get
${S_\infty } = \dfrac{1}{{1 - \dfrac{2}{3}}}$
On solving the denominator we get
${S_\infty } = \dfrac{1}{{\dfrac{1}{3}}}$
Multiply the numerator and denominator with three we get
$
  {S_\infty } = \dfrac{{1 \times 3}}{{\dfrac{1}{3} \times 3}} \\
  {S_\infty } = 3 \\
$

Therefore , the sum of the infinite geometric series $1 + \dfrac{2}{3} + \dfrac{4}{9} + ...$ is 3.

Note:
The general form of the geometric series is $a,ar,a{r^2},.....$ where a is the first term and r is the common difference and r should not be equal to one. The infinite geometric series converges if and only if the absolute value of the common ratio is less than one that is \[\left| r \right| < 1\].