I'm trying to prove $6n^5 - 4n^3 + 2n \not\in \mathcal{O}(5n^4 - 3n^3 + 1)$. I'm not quite sure if my proof is correct, and was wondering if you guys can look over it.
Proof
We want to prove that $\forall c, n_0 \in \mathbb{R}^+, \exists n \in \mathbb{N}, n \geq n_0 \wedge 6n^5 - 4n^3 + 2n > c(5n^4 - 3n^3 + 1).$ Let $c, n_0 \in \mathbb{R}^+$ and let $n = c + n_0 + 2$. Thus, $n \geq n_0$, proving the first part of the statement. Now for the second part.
We know that $-3n^2 \cdot n = -3n^2$ n times $= -3n_1^2 - 3n_2^2 - \ldots -3n_n^2$. So then, since $n > 1$, we can say that $-3n^2 \cdot n < -n^2 - 3n^2 = -4n^2$.
Also, $6n^4 > 5n^4$ (Since $n>1$). Then $6n^4 - 4n^2 > 5n^4 - 3n^3$ (Since we showed $-4n^2 > -3n^2 \cdot n = -3n^3$ above). So then $6n^4 - 4n^2 + 2 > 5n^4 - 3n^3 + 1$ (Since $2 > 1$). So then $n(6n^4 - 4n^2 + 2) > c(5n^4 - 3n^3 + 1)$ (Since $n > c$). Thus, $n \geq n_0 \wedge 6n^5 - 4n^3 + 2n > c(5n^4 - 3n^3 + 1)$, which completes our proof. $\blacksquare$
Thank you!