Answer
$\left( { - 1,1} \right)$
Work Step by Step
$$\eqalign{
& \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right){x^n}} \cr
& {\text{Let }}{u_n} = {\left( { - 1} \right)^{n + 1}}\left( {n + 1} \right){x^n} \cr
& {\text{Using the ratio test}} \cr
& \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 2}}\left( {n + 2} \right){x^{n + 1}}}}{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right){x^n}}}} \right| \cr
& {\text{Simplifying}} \cr
& = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( { - 1} \right)\left( {n + 2} \right){x^n}x}}{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right){x^n}}}} \right| \cr
& = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( { - 1} \right)\left( {n + 2} \right)x}}{{\left( {n + 1} \right)}}} \right| \cr
& = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{n + 2}}{{n + 1}}} \right| \cr
& = \left| x \right| \cr
& {\text{By the Ratio Test}},{\text{ the series converges for }} \cr
& \left| x \right| < 1 \cr
& - 1 < x < 1 \cr
& \cr
& {\text{At the endpoints}}{\text{, we have}} \cr
& *x = - 1 \cr
& \to \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right){{\left( { - 1} \right)}^n}} = - \sum\limits_{n = 1}^\infty {\left( {n + 1} \right)} ,{\text{ diverges}} \cr
& *x = 1 \cr
& \to \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)} = - \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)} ,{\text{ diverges}} \cr
& {\text{Therefore}}{\text{, the interval of convergence is}} \cr
& \left( { - 1,1} \right) \cr} $$
