<s>[INST] 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 ???
 [/INST] Yes you are on the right track. To elaborate: So let $\epsilon$ be any positive number. We show that there is an $n_0$ such that the equation $|a_n - l| \le \epsilon$ for all integers $n \ge n_0$, holds. That suffices to prove convergence of the sequence $\{a_n\}; n \in \mathbb{N} \ $, to $l=\frac{3}{2}$.
Indeed, let $n_0 = \left\lceil \frac{57}{2\epsilon} \right\rceil$. Then on the one hand, As noted in the OP,
$$|a_n- l| < \frac{57}{2n} \quad \forall n \in \mathbb{N}.$$
On the other hand, for any $n \ge n_0$:
$$\frac{57}{2 \epsilon} \le n \implies \frac{2 \epsilon}{57} \ge \frac{1}{n} \implies  \frac{57}{2n} \le \epsilon .$$
Thus in particular, both statements
$$|a_n- l| < \frac{57}{2n} \quad \forall n \in \mathbb{N},$$ and $$\frac{57}{2n} \le \epsilon \quad \forall n \ge n_0 = \left \lceil \frac{57}{2 \epsilon} \right\rceil,$$ hold. Putting these statements together gives
$$|a_n- l| \le \epsilon \quad \forall {\text{ integers }} n \ge n_0=\left\lceil \frac{57}{2\epsilon} \right\rceil.$$ Thus indeed, we've shown that there is an $n_0$ such that the equation $|a_n - l| \le \epsilon$ for all integers $n \ge n_0$ holds, namely $n_0 = \left \lceil \frac{57}{2 \epsilon} \right \rceil$. As noted above, this suffices to establish convergence of the sequence $\{a_n\}; n \in \mathbb{N} \ $,to $l=\frac{3}{2}$.
</s>