For this proof, I'm having trouble minimizing the denominator. Here is what I have so far. Can anyone please help me out?
Let $a_n = \frac{2-2n}{(n^2+n)(n^2-3)}$. By definition, prove that $\{a_n\}$ converges. Make sure to fully justify your work.
Suppose $n>N$
Let $\epsilon > 0$ be arbitrary.
$\lim{n \to \infty} \frac{2-2n}{(n^2+n)(n^2-3)} = 0$
$\frac{2-2n}{n^4-3n^2+n^3-3n}$
$\le \frac{2-2n}{n^3-3n}$
$ = \frac{2(1-n)}{n(n^2-3)}$
On
Given $\epsilon > 0$, fix some $0< \eta <1$, $$ |a_n| = \left|\frac{2-2n}{(n^2+n)(n^2-3)}\right| \overset{(*)}{<} \frac{2|1-n|}{(n^2+0)(n^2-\eta n^2)} < \frac{2n}{n^4(1-\eta)} = \frac{2}{1-\eta} \cdot \frac{1}{n^3} $$
where we use $n^2-3>n^2-\eta n^2$ at $(*)$ which holds for $n>\sqrt{\frac{3}{\eta}}$. Now we want $$\frac{2}{1-\eta} \frac{1}{n^3} < \epsilon,\quad \text{so require}\quad n >\sqrt[3]{\frac{2}{\epsilon(1-\eta)}}$$
Hence if $n > \max\left\{\sqrt{\frac{3}{\eta}},\sqrt[3]{\frac{2}{\epsilon(1-\eta)}}\right\}$ then $|a_n|<\epsilon$.
For $n\geq 3$, \begin{align*} \left|\dfrac{2-2n}{(n^{2}+n)(n^{2}-3)}\right|&\leq\dfrac{2+2n}{(n+n)(n^{2}-(1/2)n^{2})}\\ &=\dfrac{2(n+1)}{2n(1/2)n^{2}}\\ &=2\left(1+\dfrac{1}{n}\right)\dfrac{1}{n^{2}}\\ &\leq\dfrac{4}{n^{2}}, \end{align*} and $4/n^{2}<\epsilon$ for all large $n$.