Answer
The radius of convergence of the series is $1$.
Work Step by Step
Here, we have $a_n=(-1)^n \dfrac{x^n}{n+1}$
Ratio Test states that when $\Sigma a_n$ is an infinite series with positive terms and, then $r=|\lim\limits_{n \to \infty}\dfrac{a_{n+1}}{a_n}|$
a) When $0 \leq r \lt 1$, the series converges. (b) When $r \gt 1$, or, $\infty$, so the series diverges. (c) When $r=1$, the ratio test is inconclusive.
Now, $r=\lim\limits_{n \to \infty}|\dfrac{\dfrac{x^{n+1}}{n+2}}{\dfrac{x^n}{n+1}}|\\=\lim\limits_{n \to \infty}|\dfrac{x(n+1)}{n+2}|\\=|x|$
Therefore, the series converges for $|x| \lt 1$ by the ratio test. This implies that the radius of convergence of the series is $1$.
