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

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

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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 $n\ge n_0$.

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  • $\begingroup$ I have a very similar question. Namely, what is $\lim_{n\to\infty}(n\cdot\exp(\mathcal{O}(nz_n^2)))$. $\endgroup$
    – Rhjg
    Commented Jan 22, 2018 at 19:00

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