Answer
The interval of convergence is $\left( { - 1,1} \right)$.
Work Step by Step
Let ${a_n} = n{x^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\left( {n + 1} \right){x^{n + 1}}}}{{n{x^n}}}} \right| = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n + 1}}{n}} \right|$
$ = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {1 + \dfrac{1}{n}} \right| = \left| x \right|$
The series converges if $\rho = \left| x \right| \lt 1$. That is,
$ - 1 \lt x \lt 1$
Next, we check the endpoints: $x=-1$ and $x=1$.
1. $x=-1$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty n{x^n} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}n$
Since the $n$th term does not converge to $0$, by the $n$th Term Divergence Test, this series diverges.
2. $x=1$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty n{x^n} = \mathop \sum \limits_{n = 0}^\infty n{1^n} = \mathop \sum \limits_{n = 0}^\infty n$
Since the $n$th term does not converge to $0$, by the $n$th Term Divergence Test, this series diverges.
Thus, the interval of convergence is $\left( { - 1,1} \right)$.
