Delta epsilon proof of convergence for $a_n = \frac{2-2n}{(n^2+n)(n^2-3)}$

Question

Let $a_n = \frac{2-2n}{(n^2+n)(n^2-3)}$. By definition, prove that $\{a_n\}$ converges.

So far what I have is; Let $\varepsilon > 0$ and if $n>N$, $L = 0$ and $a_n \rightarrow L$

Want to show: For all $\varepsilon > 0$, there exists $N> 0$ such that if $n>N$, $|a_n-L|< \varepsilon$ However, I need help to start the question:

I started with $a_n < (2-2n)/(n^2-3)$ because $(n^2+n) > 0$ but this doesn't hold for some reason? Any help is appreciated.

Answer 1

HINT: For large $n$ $$ \left| \frac{2-2n}{(n^2+n)(n^2-3)} \right|<\frac1{n^2+n}<\frac1{n^2}, $$ leading to a simple calculation of $N$ depending on $\varepsilon$.

Answer 2

Note that for $n\ge2$ we have $$\bigg | \frac {2-2n}{(n^2+n)(n^2-3)}\bigg |<\bigg |\frac {2-2n}{n^2}\bigg |\le\\\bigg |\frac {2}{n^2}\bigg |+\bigg |\frac {2}{n}\bigg |\to 0+0=0$$