Answer
$R=1$ ; interval of convergence $(-1,1)$
Work Step by Step
Let $a_{n}=(-1)^{n}nx^{n}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(-1)^{n+1}(n+1)x^{n+1}}{(-1)^{n}nx^{n}}|$
$=|x|$
converges if $|x|\lt 1$, so $R=1$ and the interval of convergence is from $-1$ to $+1$
When $x = -1$; $a_{n}=(-1)^{n}n(-1)^{n} = (-1)^{2n}n = n$
$\Sigma^{\infty}_{n=1}n $ is divergent, Therefore -1 is not included in interval of convergence
When $x = -1$; $a_{n}=(-1)^{n}n(1)^{n} = (-1)^{n}n$
$\Sigma^{\infty}_{n=1}(-1)^{n}n $ is divergent, Therefore 1 is not included in interval of convergence
Hence, $R=1$ ; interval of convergence is $(-1,1)$
