Answer
series has radius of convergence is $1$ and the intervals of convergence is $(-1,1)$.
Work Step by Step
$\frac{x^{2}}{1+x}=\Sigma_{n=0}^\infty(-1)^{n}x^{n+2}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty }|\frac{(-1)^{n+1}x^{n+3}}{(-1)^{n}x^{n+2}}|$
$=|-x|$
$=|x|\lt 1$
Thus, the series has radius of convergence is $1$ and the intervals of convergence is $(-1,1)$.
