Recall that $g \in \mathcal{O}(f)$ means $$\exists c \in \mathbb{R^+}, \exists n_0 \in \mathbb{R^+}, \forall n \in \mathbb{N}, n \geq n_0 \implies g(n) \leq cf(n).$$
I thought of a stronger Big-O definition that we can call big-Zeta, which is $$g \in \zeta(f) : \forall c \in \mathbb{R^+}, \exists n_0 \in \mathbb{R^+}, \forall n \in \mathbb{N}, n \geq n_0 \implies g(n) \leq cf(n).$$ What this says instead is that: no matter how much we scale $f$ by, $f$ will eventually dominate $g(n)$. Now, I've thought of an example problem I tried to prove which I believe is true:
For all positive reals $a,b$ with $a < b$ prove that $n^a \in \zeta(n^b)$. Try not to use limits to solve this so the goal is to find an $n_0$.
Bonus problem*: Prove that for $f,g$ that map naturals to positive reals, if $g \in \zeta(f)$ then $f \not \in \mathcal{O}(g)$.
(*) This is harder but a clue is that $f$ and $g$ map to positive reals.
This is exactly what $g\in \mathcal{o}(f)$ means! Note the lower-case $\mathcal{o}$: this is called the little-$\mathcal{o}$ notation.