# Show $\lim_n f(n)=0$

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 $$\lim_{n\to\infty}f(n)=0.$$

• What if $z_n=n^{-1/3}$? Commented Jan 22, 2018 at 18:26
• Yes, $f$ goes to $0$. Just use the fact that $n/(n-1)\to1$ and $z_n^2\to0$. Commented Jan 22, 2018 at 18:28

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 $n\ge n_0$.
• I have a very similar question. Namely, what is $\lim_{n\to\infty}(n\cdot\exp(\mathcal{O}(nz_n^2)))$.