I understand that the serie's limit will be 0 since as the denominator gets bigger it gets smaller, but I do not know how to justify this answer.
I know there are probably a couple ways for this to be justified, and just so you guys know what I understand so far, the chapter name this problem comes out of is "Infinite Sequences and Series." Hopefully, that helps. thank you.
You can compare asymptotically your $a_n$ with $b_n=\frac{1}{n^3}$. In fact, since $\sum_{n\in\mathbb{N}}\frac{1}{n^3}<\infty$ (if you want, by integral criterion) and given that $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=1$ you get that $\sum_{n\in\mathbb{N}}a_n<\infty$
We know that the series $\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{n^2}$ is convergent. Moreover, $n(n+1)(n+2)=n^3+3n^2+2n>n^2$, then, $\displaystyle\frac{1}{n^2}>\displaystyle\frac{1}{n(n+1)(n+2)}$. Then, we can conclude that $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}\geq \displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{n(n+1)(n+2)}$$and the first series is convergent. Thus, $\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{n(n+1)(n+2)}$ is also convergent.
