
How do you find the sum of the infinite geometric series $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \dfrac{1}{{81}} + ...$ ?
Answer
476.4k+ 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.
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