Question

How do you find the sum of the infinite geometric series 0.9 + 0.09 + 0.009 ….

Answer 1

Hint: We are asked to find the sum of the geometric series. We will learn about the geometric series and then we will find what are the first term, common ratio and the type of geometric progression. To find the common ratio, we use \[r=\dfrac{{{a}_{2}}}{{{a}_{1}}}\] and then to find the sum we will use \[{{S}_{n}}=\dfrac{a}{1-r}\] as our sequence is infinite.

Complete step by step answer;
We are given the series as 0.9, 0.09, 0.009…. We are said to be geometric series and we first will understand that the series which increases by the same difference is known as the geometric series. For example 2, 4, 8, 16… Here this is increased by a ratio of 2 as each term is double of the other. So this is the geometric series and also 1, 0.1, 0.01…. This is also a series as it is by the ratio of \[\dfrac{1}{10}.\] Now we will check what we have in our series. We have 0.9, 0.009, 0.009…. The first term is 0.9, the second term is 0.09 and the third term is 0.009. The ratio of the second term to the first term is \[\dfrac{0.09}{0.9}=0.1.\] The ratio of the third term to the second term is \[\dfrac{0.009}{0.09}=0.1.\] So, here the ratio is the same as it is 0.1. So we have the ratio, r = 0.1.
Now we can see that our series is an infinite series. For finite and infinite series, the formula for the sum is different. For infinite series, the sum of the geometric series is given as \[{{S}_{n}}=\dfrac{a}{1-r}\] if r < 1, where S is the sum, a is the first term and r is the common ratio. Now, we have the first term as 0.9. So, a = 0.9, r = 0.1. Using these values in \[S=\dfrac{a}{1-r},\] we get
\[\Rightarrow S=\dfrac{0.9}{1-0.1}\]
On solving for S, we get,
\[\Rightarrow S=\dfrac{0.9}{0.9}=1\]
So, the sum is 1.

Note:
For the geometric series, we have these formulas, if the series is finite then the sum of n terms is given as \[{{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\] if r > 1 and \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\] if r < 1. And if the series is infinite, the sum is given as \[S=\dfrac{a}{1-r},\] if r < 1 and \[S=\dfrac{a}{r-1},\] if r > 1. Also, the \[{{n}^{th}}\] term formula is given as \[{{a}_{n}}=a{{r}^{n-1}}.\]