Answer
The interval of convergence is $[ - 1,1)$.
Work Step by Step
Let ${a_n} = \dfrac{{{x^n}}}{{n + 1}}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x^{n + 1}}}}{{n + 2}}\cdot\dfrac{{n + 1}}{{{x^n}}}} \right| = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n + 1}}{{n + 2}}} \right|$
$ = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{1 + \dfrac{1}{n}}}{{1 + \dfrac{2}{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. $x=-1$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n + 1}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{n + 1}}$
Since the sequence $\left\{ {\dfrac{1}{{n + 1}}} \right\}$ is positive, decreasing and converge to $0$, by the Alternating Series Test, this series converges.
2. $x=1$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n + 1}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{1}{{n + 1}}$
We apply the Integral test:
$\mathop \smallint \limits_0^\infty \frac{1}{{x + 1}}{\rm{d}}x = \left[ {\ln \left( {x + 1} \right)} \right]_0^\infty = \infty $
Since $\mathop \smallint \limits_0^\infty \frac{1}{{x + 1}}{\rm{d}}x$ diverges, by the Integral Test, $\mathop \sum \limits_{n = 0}^\infty \frac{1}{{n + 1}}$ also diverges.
Thus, the interval of convergence is $[ - 1,1)$.
