How do you find the radius of convergence \[\sum{\dfrac{{{x}^{n}}}{{{3}^{n}}}}~\] from $n=\left[ 0,\infty \right)$?
Answer
554.7k+ views
Hint: To find the radius of convergence first we will do the ratio test of the given series and then calculate the radius of convergence in three cases i.e., if limit is zero, if limit is equal to $N\times \left| x-a \right|$ and if limit is $\infty $. . The ratio test is given as $L=\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|$, if the value of $L$ is less than $1$ then the series converges else it diverges.
Complete step by step solution:
From our above problem, we have our power series as \[\sum{\dfrac{{{x}^{n}}}{{{3}^{n}}}}~\].
We have to find the radius of convergence for the given interval.
Now, we know that to find the radius of convergence, we will have to find that the given series is convergence or not.
For this we have to perform the ratio test and ratio test is given as,
$L=\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|$
The given series is \[\sum{\dfrac{{{x}^{n}}}{{{3}^{n}}}}~\] so the ${{n}^{th}}$ term of the series will be ${{a}_{n}}=\dfrac{{{x}^{n}}}{{{3}^{n}}}$.
Now, on applying ratio test on our given series, we get:
$\Rightarrow L=\displaystyle \lim_{n\to \infty }\left| \dfrac{\dfrac{{{x}^{n+1}}}{{{3}^{n+1}}}}{\dfrac{{{x}^{n}}}{{{3}^{n}}}} \right|$
On simplifying, we get our equation as:
\[\Rightarrow \displaystyle \lim_{n\to \infty }\left| \dfrac{{{x}^{n+1}}}{{{x}^{n}}}\cdot \dfrac{{{3}^{n}}}{{{3}^{n+1}}} \right|\]
On further simplification, we get:
$=\left| x \right|\cdot \dfrac{1}{3}$
Now, For the series to converge, we need that, $L<1$, that is $\left| x \right|\cdot \dfrac{1}{3}<1$.
Now, on substituting the value we will get,
$\Rightarrow \left| x \right|\cdot \dfrac{1}{3}<1$
On simplifying, we get that the radius of convergence is,
$\Rightarrow \left| x \right|<3$
Hence, we get the radius of convergence as $3$.
Note: While doing this problem, we have to keep in mind that if the value of $L$ is greater than $1$, then the series is divergent and if the value of $L$ is equal to $1$, then the test is inconclusive. Therefore, before finding the radius we have to perform the ratio test to confirm that the series is convergent or divergent.
Complete step by step solution:
From our above problem, we have our power series as \[\sum{\dfrac{{{x}^{n}}}{{{3}^{n}}}}~\].
We have to find the radius of convergence for the given interval.
Now, we know that to find the radius of convergence, we will have to find that the given series is convergence or not.
For this we have to perform the ratio test and ratio test is given as,
$L=\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|$
The given series is \[\sum{\dfrac{{{x}^{n}}}{{{3}^{n}}}}~\] so the ${{n}^{th}}$ term of the series will be ${{a}_{n}}=\dfrac{{{x}^{n}}}{{{3}^{n}}}$.
Now, on applying ratio test on our given series, we get:
$\Rightarrow L=\displaystyle \lim_{n\to \infty }\left| \dfrac{\dfrac{{{x}^{n+1}}}{{{3}^{n+1}}}}{\dfrac{{{x}^{n}}}{{{3}^{n}}}} \right|$
On simplifying, we get our equation as:
\[\Rightarrow \displaystyle \lim_{n\to \infty }\left| \dfrac{{{x}^{n+1}}}{{{x}^{n}}}\cdot \dfrac{{{3}^{n}}}{{{3}^{n+1}}} \right|\]
On further simplification, we get:
$=\left| x \right|\cdot \dfrac{1}{3}$
Now, For the series to converge, we need that, $L<1$, that is $\left| x \right|\cdot \dfrac{1}{3}<1$.
Now, on substituting the value we will get,
$\Rightarrow \left| x \right|\cdot \dfrac{1}{3}<1$
On simplifying, we get that the radius of convergence is,
$\Rightarrow \left| x \right|<3$
Hence, we get the radius of convergence as $3$.
Note: While doing this problem, we have to keep in mind that if the value of $L$ is greater than $1$, then the series is divergent and if the value of $L$ is equal to $1$, then the test is inconclusive. Therefore, before finding the radius we have to perform the ratio test to confirm that the series is convergent or divergent.
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