Question

How do you find the sum of the infinite series ${\sum {\left( {\dfrac{1}{{10}}} \right)} ^k}$ from $k = 1$ to $\infty $?

Answer 1

Hint: This problem deals with the sum of the infinite series of a geometric progression. A geometric progression is a series, with which there is a common ratio associated with it, which means that every consecutive term in the geometric progression have the same common ratio, throughout the geometric progression, here the common ratio is $r$, and the initial term is $a$. Here $r$ is less than 1, hence the sum of the infinite terms is given by:
$ \Rightarrow \dfrac{a}{{1 - r}}$

Complete step-by-step answer:
Given that the expression of the infinite terms is given by ${\sum {\left( {\dfrac{1}{{10}}} \right)} ^k}$, where $k$ is from 1 to $\infty $.
\[ \Rightarrow {\sum\limits_{k = 1}^\infty {\left( {\dfrac{1}{{10}}} \right)} ^k}\]
Here the common ratio, $r$ is $\dfrac{1}{{10}}$, which is less than 1.
$\therefore r = \dfrac{1}{{10}}$
The first term of the G.P is obtained when $k = 1$, which is given by:
$ \Rightarrow a = {\left( {\dfrac{1}{{10}}} \right)^1}$
$\therefore a = \dfrac{1}{{10}}$
Now substituting these values in the formula of the sum of the infinite terms in a G.P:
$ \Rightarrow \dfrac{a}{{1 - r}}$
$ \Rightarrow \dfrac{{\dfrac{1}{{10}}}}{{1 - \dfrac{1}{{10}}}} = \dfrac{1}{9}$
$\therefore $The sum of the given infinite series ${\sum {\left( {\dfrac{1}{{10}}} \right)} ^k}$ , where k varies from 0 to $\infty $, is equal to $\dfrac{1}{9}$.

Final Answer: The sum of the infinite series is \[{\sum\limits_{k = 1}^\infty {\left( {\dfrac{1}{{10}}} \right)} ^k} = \dfrac{1}{9}\]

Note:
Please note that the formula used for finding the sum of $n$ terms in the G.P varies accordingly, that is if the common ratio of the G.P is greater than 1, then the formula applied in the problem is used.
But the sum of the$n$ terms in G.P if the common ratio is less than 1 which is $r < 1$, is given by:
$ \Rightarrow \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$
The sum of the infinite terms of a G.P when $\left| r \right| < 1$, is given by:
$ \Rightarrow \dfrac{a}{{1 - r}}$