Let $(z_n)$ be a zero sequence, $\lim_n z_n=0$ and $$ f(n)\in\mathcal{O}(g(n))\text{ as }n\to\infty, $$ where $$ g(n)=\frac{n\cdot z_n^2}{n-1}. $$
Perhaps a naive question, but does this imply $$ \lim_{n\to\infty}f(n)=0? $$
I think, by definion, $f(n)\in\mathcal{O}(g(n))\text{ as }n\to\infty$ implies there exist some positive real number $M$ and some $n_0$ such that $$ \lvert f(n)\rvert\leq M\left\lvert\frac{n\cdot z_n^2}{n-1}\right\rvert~\forall n\geq n_0. $$
I think that for large $n$, we have $\lvert n\cdot z_n^2\rvert\leq 1$, i.e. $$ \lvert f(n)\rvert\leq \frac{M}{\lvert n-1\rvert} $$ implying $$ \lim_{n\to\infty}f(n)=0. $$
We can't state that $|nz_n^2|\le 1$ for big enough $n$ (take for example $z_n=n^{-1/3}$), but there is a workaround.
Let $\epsilon >0$. Then there is some $n_0\in\Bbb N$ (pick it big enough to hold the bound for $f$ and $g$) such that $$|z_n|^2<\frac\epsilon{2M}$$
Then $$|f(n)|\le M\left|\frac{nz_n^2}{n-1}\right|<\frac\epsilon2\cdot\frac n{n-1}\le\epsilon$$ for $n\ge n_0$.