
How do you find the sum of the infinite geometric series $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \dfrac{1}{{81}} + ...$ ?
Answer
530.1k+ views
Hint: In a geometric progression, each term is multiplied by a common ratio to get the next term. We can use the formula to find the sum of the series up to $n$ terms given by, ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$. Using this formula, we can find the sum of the series by limiting $n \to \infty $
Formula used:
${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$
${S_\infty } = \dfrac{a}{{(1 - r)}}$
Complete step by step solution:
We are given a series $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \dfrac{1}{{81}} + ...$
We are given that this series is a Geometric Progression.
We can find the common ratio of the series by dividing the consecutive terms.
We can calculate the common ratio as $r = \dfrac{{1/9}}{{1/3}} = \dfrac{{1/27}}{{1/9}} = \dfrac{{1/81}}{{1/27}} = \dfrac{1}{3}$.
Thus, the given series is a Geometric Progression (GP) with common ratio $r = \dfrac{1}{3}$.
Now we can use the formula to find the sum of the series up to $n$ terms given by, ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$.
We have to find the sum of the series up to infinite terms.
Since $r < 1$, as $n \to \infty $ we can say that ${r^n} \to 0$. Therefore, $(1 - {r^n}) \to 1$.
Thus, the formula for sum of the infinite geometric series becomes,
${S_\infty } = \dfrac{a}{{(1 - r)}}$
where, ${S_\infty }$ is the sum of the infinite series
$a$ is the first term of the series
$r$ is the common ratio
In the given series, $a = \dfrac{1}{3}$ and $r = \dfrac{1}{3}$.
Putting all the values in the above formula, we get,
\[
{S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(1 - \dfrac{1}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(\dfrac{{3 - 1}}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(\dfrac{2}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{1}{2} \\
\]
Thus, the sum of the given infinite series is $\dfrac{1}{2}$.
Note: We can find the sum of the geometric series up to $\infty $ terms using the formula only when the absolute value of the common ratio is less than $1$, i.e. $\left| r \right| < 1$. For $\left| r \right| > 1$, we can only calculate sums up to $n$ terms where $n$ is a finite natural number. We can calculate the sum of the series without knowing all the terms, we only need at most three terms to calculate the common ratio.
Formula used:
${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$
${S_\infty } = \dfrac{a}{{(1 - r)}}$
Complete step by step solution:
We are given a series $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \dfrac{1}{{81}} + ...$
We are given that this series is a Geometric Progression.
We can find the common ratio of the series by dividing the consecutive terms.
We can calculate the common ratio as $r = \dfrac{{1/9}}{{1/3}} = \dfrac{{1/27}}{{1/9}} = \dfrac{{1/81}}{{1/27}} = \dfrac{1}{3}$.
Thus, the given series is a Geometric Progression (GP) with common ratio $r = \dfrac{1}{3}$.
Now we can use the formula to find the sum of the series up to $n$ terms given by, ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$.
We have to find the sum of the series up to infinite terms.
Since $r < 1$, as $n \to \infty $ we can say that ${r^n} \to 0$. Therefore, $(1 - {r^n}) \to 1$.
Thus, the formula for sum of the infinite geometric series becomes,
${S_\infty } = \dfrac{a}{{(1 - r)}}$
where, ${S_\infty }$ is the sum of the infinite series
$a$ is the first term of the series
$r$ is the common ratio
In the given series, $a = \dfrac{1}{3}$ and $r = \dfrac{1}{3}$.
Putting all the values in the above formula, we get,
\[
{S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(1 - \dfrac{1}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(\dfrac{{3 - 1}}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{{(\dfrac{1}{3})}}{{(\dfrac{2}{3})}} \\
\Rightarrow {S_\infty } = \dfrac{1}{2} \\
\]
Thus, the sum of the given infinite series is $\dfrac{1}{2}$.
Note: We can find the sum of the geometric series up to $\infty $ terms using the formula only when the absolute value of the common ratio is less than $1$, i.e. $\left| r \right| < 1$. For $\left| r \right| > 1$, we can only calculate sums up to $n$ terms where $n$ is a finite natural number. We can calculate the sum of the series without knowing all the terms, we only need at most three terms to calculate the common ratio.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

