Answer
a) The radius of convergence is $\dfrac{1}{2}$ and the interval of convergence is: $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$.
b) The interval of absolute convergence is $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$.
c) We get no other values of $x$ for which the given series converges conditionally.
Work Step by Step
Apply the Ratio Test to compute the interval where the series converges absolutely and the radius of convergence of the given series.
$|\dfrac{u_{n+1}}{u_{n}}|=\lim\limits_{n \to \infty} \dfrac{(4)^{n+1} x^{2(n+1)}}{(n+1)!} \cdot \dfrac{n}{(4)^n x^{2n}}=(4) |x^2| \lim\limits_{n \to \infty} |\dfrac{n}{(n+1)}|=(4) |x^2|\lim\limits_{n \to \infty} \dfrac{1}{1+\dfrac{1}{n}}=4x^2 [\dfrac{1}{(1+0)}]=4x^2$
Thus, the series is absolutely convergent for all $x \in R$ when $4x^2 \lt 1 $; then we have: $|x| \lt \dfrac{1}{2}$
We conclude that the series is convergent for all $x \in R$ and satisfies $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$. So, the radius of convergence is $R=\dfrac{1}{2}$.
In order to find the interval of convergence, firstly we need to check the behaviour of the series at the end points $x=\dfrac{1}{2}$ and $x=\dfrac{-1}{2}$.
Case -1: At $x=\dfrac{1}{2}$
$\sum_{n=1}^{\infty} \dfrac{4^n x^{2n}}{n}=\sum_{n=1}^{\infty} \dfrac{4^n (\dfrac{1}{2^{2n})}}{n}==\sum_{n=1}^{\infty} \dfrac{4^n (\dfrac{1}{4})^{n}}{n}=\sum_{n=1}^{\infty} \dfrac{1}{n}$
The series $\sum_{n=1}^{\infty} \dfrac{1}{n}$ represents a p-series with $p=1 \leq 1$ and thus, the series is divergent by the p-test. The series is divergent at $x=\dfrac{1}{2}$ and this means that it is not absolutely convergent. We know that every absolutely convergent series is convergent as well.
Case -2: At $x=\dfrac{-1}{2}$
$\sum_{n=1}^{\infty} \dfrac{4^n x^{2n}}{n}=\sum_{n=1}^{\infty} \dfrac{4^n (\dfrac{-1}{2})^{2n}}{n}=\sum_{n=1}^{\infty} \dfrac{4^n (\dfrac{1}{4})^{n}}{n}=\sum_{n=1}^{\infty} \dfrac{1}{n}$
The series $\sum_{n=1}^{\infty} \dfrac{1}{n}$ represents a p-series with $p=1 \leq 1$ and thus, the series is divergent by the p-test. The series is divergent at $x=\dfrac{-1}{2}$ and this means that it is not absolutely convergent. We know that every absolutely convergent series is convergent as well.
Thus, we get no other values of $x$ for which the given series converges conditionally.
Therefore, we come to the following conclusions:
a) The radius of convergence is $\dfrac{1}{2}$ and the interval of convergence is: $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$.
b) The interval of absolute convergence is $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$.
c) We get no other values of $x$ for which the given series converges conditionally.
