Question

How do you find the radius of convergence of the power series $\sum {{2}^{n}}{{n}^{3}}{{x}^{n}}$ from $n=\left[ 0,\infty \right)$?

Answer 1

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.

Complete step-by-step solution:
We have been given a power series $\sum {{2}^{n}}{{n}^{3}}{{x}^{n}}$.
We have to find the radius of convergence for the given interval.
Now, we know that to find the radius of convergence first we will have to find that the given series is convergence or not. For this we have to perform the ratio test.
The given series is $\sum {{2}^{n}}{{n}^{3}}{{x}^{n}}$ so the ${{n}^{th}}$ term of the series will be ${{a}_{n}}={{2}^{n}}{{n}^{3}}{{x}^{n}}$.
Now, we know that the ratio test for the series is given as $L=\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$
Now, substituting the values we will get
$\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{2}^{n+1}}{{\left( n+1 \right)}^{3}}{{x}^{n+1}}}{{{2}^{n}}{{n}^{3}}{{x}^{n}}} \right|$
Now, we know that $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$
So, by simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left| {{2}^{n+1-n}}\dfrac{{{\left( n+1 \right)}^{3}}}{{{n}^{3}}}{{x}^{n+1-n}} \right| \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }2\dfrac{{{\left( n+1 \right)}^{3}}}{{{n}^{3}}}\left| x \right| \\
\end{align}\]
Now, if n tends to infinity we get the value of L as $\Rightarrow L=2\left| x \right|$
Now, the series will be convergent for $L<1$
Now, substituting the value we will get
$\begin{align}
  & \Rightarrow 2\left| x \right|<1 \\
 & \Rightarrow \left| x \right|<\dfrac{1}{2} \\
\end{align}$
Hence we get the radius of convergence as $\dfrac{1}{2}$.

Note: The point to be noted is 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 that the test is inconclusive. Before finding the radius we have to perform the ratio test to confirm that the series is convergent or divergent.