Answer
$ a.\quad R=\displaystyle \frac{1}{4}.\ \quad $Interval of convergence:$\quad -\displaystyle \frac{1}{2} \lt x \lt 0$
$b. \quad$ Interval of absolute convergence:$\displaystyle \quad -\frac{1}{2} \lt x \lt 0$
$ c.\quad$ There are no values for which the series converges conditionally
Work Step by Step
(See text: "How to Test a Power Series for Convergence".)
$\text{Step 1.} $
Use the Ratio Test to find the interval where the series converges absolutely.
Ordinarily, $|x-a|\lt R\quad $ or $\quad a-R\lt x\lt a+R.$
$\begin{align*}
\displaystyle \lim_{n\rightarrow\infty}|\frac{u_{n+1}}{u_{n}}|&\lt 1 \\
\displaystyle \lim_{n\rightarrow\infty}|\frac{(-1)^{n+1}(4x+1)^{n+1}}{(-1)^{n}(4x+1)^{n}}|&\lt 1 \\
\displaystyle \lim_{n\rightarrow\infty}|\frac{(4x+1)^{n+1}}{(4x+1)^{n}}|&\lt 1 \\
\end{align*}.$
This is true when:
$\begin{align*}
|4x+1|&\lt 1 \\
-1& \lt 4x+1 \lt 1 \\
-2& \lt 4x \lt 0 \\
-\displaystyle \frac{1}{2}& \lt x \lt 0 \end{align*}.$
Determine the center and radius:
$a -R\lt x\lt a+R$
$-\displaystyle \frac{1}{4} -\frac{1}{4}\lt x\lt -\frac{1}{4} +\frac{1}{4}$
$ a=-\displaystyle \frac{1}{4},\quad$
the radius is $R=\displaystyle \frac{1}{4},$
the interval of convergence is $-6 \lt x \lt -4$
$\text{Step 2.} $
If the interval of absolute convergence is finite,
test for convergence or divergence at each endpoint.
$ x=-\displaystyle \frac{1}{2}\quad \Rightarrow$
$ \displaystyle \sum_{n=1}^{\infty}(-1)^{n}(-1)^{n}=\sum_{n=1}^{\infty}(-1)^{2n}=\sum_{n=1}^{\infty}1 \qquad$ ... a divergent series.
$ x=0\quad \Rightarrow$
$ \displaystyle \sum_{n=1}^{\infty}(-1)^{n}(1)^{n}=\sum_{n=1}^{\infty}(-1)^{n} \qquad$ ... a divergent series.
$\text{Step 3.} $
If the interval of absolute convergence is $a -R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$ a.\quad R=\displaystyle \frac{1}{4}.\ \quad $Interval of convergence:$\quad -\displaystyle \frac{1}{2} \lt x \lt 0$
$b. \quad$ Interval of absolute convergence:$\displaystyle \quad -\frac{1}{2} \lt x \lt 0$
$ c.\quad$ There are no values for which the series converges conditionally
