Prove $\frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$ converges to $\frac{3}{2}$ by definition Let $a_{n+1} = \frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$, $l = \frac{3}{2}$. Prove that $a_n$ converges to $l$.
I used the absolute value of $a_{n} - l$, to obtain $\frac{40n^3 + 17}{4n^5-2}$. I then obtained these inequalities:
$\frac{40n^3 + 17}{4n^5-2} < \frac{40n^3 + 17}{4n^5-2n^5} = \frac{40n^3 + 17}{2n^5} < \frac{40n^3 + 17n^3}{2n^5} = \frac{57n^3}{2n^5} < \frac{57}{2n^2} < \frac{57}{2n}$, which holds for all natural numbers $n\gt1$.
I then did the usual steps letting $N$ be any natural number greater than $\frac{57}{2\epsilon}$. Does this seem correct  ?
Any help ???

I tried a different way...
Let $\epsilon>0$ and consider the formal definition of limit for sequences. Then, we want to find an $N(\epsilon)$ such that for all $n>N(\epsilon)$,
$$\left|\frac{3n^5+20n^3+7}{2n^5-1}-\frac{3}{2}\right|<\epsilon\tag{*}$$
i.e.,
$$\left|\frac{40n^3+17}{2(2n^5-1)}\right|<\epsilon$$
is satisfied. Assuming, $N(\epsilon)\geq 1$, then $2n^5-1>0$ and then,
$$\frac{40n^3+17}{2(2n^5-1)}<\epsilon\implies \frac{10}{\epsilon n^2}+\frac{17}{4\epsilon n^5}+\frac{1}{2n^5}<1.$$
We may assume that $\frac{10}{\epsilon n^2}<\frac{1}{3}$, $\frac{17}{4\epsilon n^5}<\frac{1}{3}$, $\frac{1}{2n^5}<\frac{1}{3}$ or $n>\sqrt\frac{30}{\epsilon}$, $n>\sqrt[5]\frac{51}{4\epsilon}$, $n>\sqrt[5]\frac{3}{2}$.
Now, let $$N(\epsilon)=\max\left\{1, \sqrt\frac{30}{\epsilon}, \sqrt[5]\frac{51}{4\epsilon}, \sqrt[5]\frac{3}{2} \right\}.$$
Then, $(*)$ is satisfied for all $N>N(\epsilon)$.
