Give the series $$\sum_{n=0}^{\infty} \dfrac{(x + 10)^n}{3^n (n+1)},$$ find the intervals which result in point-wise and absolute convergence.
Applying the root test we have,
$$L(x) = \lim\limits_{n \to \infty} \sqrt[n]{\left| \dfrac{(x + 10)^n}{3^n (n+1)} \right|} = \left| \dfrac{x+10}{3} \right| \dfrac{1}{\lim\limits_{n \to \infty} n^{1/n} \cdot \lim\limits_{n \to \infty} (1 + 1/n)^{1/n}} = \left| \dfrac{x+10}{3} \right|.$$
The series is absolutely convergent when $L(x) < 1$ so, $$\left| \dfrac{x+10}{3} \right| < 1 \implies I_{\rm abs} = (-13, -7).$$
Where $I_{\rm abs}$ is the interval for which the series converges absolutely.
Given that absolute convergence $\implies$ point-wise convergence, is it sufficient to just check if the series converges on the boundary for the point-wise case? i.e. taking $x = -13, -7$ in the above we find $I_{\rm pntw} = [-13, -7]$ for the point-wise interval. Is it correct?
You are correct in finding the radius of convergence and the statement that is is absolutely convergent for $$ \left| \dfrac{x+10}{3} \right| < 1 \implies I_{\rm abs} = (-13, -7).$$
When it comes to the endpoint, notice that the series $$\sum_{n=0}^{\infty} \dfrac{(x + 10)^n}{3^n (n+1)}$$ does not converge at $x=-7$ since the harmonic series diverges. It does converge at $x=-13$ since the alternating harmonic series converges.
Please modify your interval of convergence accordingly.