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I am considering the expression $$ n\cdot e^{\mathcal{O}(n\cdot z_n^2)}$$

where the big-O-Notation is meant for $n\to\infty$; $(z_n)$ is a Zero sequence, i.e.$z_n\to 0$ as $n\to\infty$.

I am wondering what happens with this Expression as $n$ is large, i.e. what is $$ \lim_{n\to\infty} n\cdot e^{\mathcal{O}(n\cdot z_n^2)}. $$


Let $f(n)\in\mathcal{O}(n\cdot z_n^2)$ for $n\to\infty$

by definition, this means that there exists some positive real number $M$ and some $n_0$ such that $$ \lvert f(n)\rvert\leq M\lvert nz_n^2\rvert~\forall n\geq n_0. $$

So I think we have to use this in order to compute $$ \lim_{n\to\infty}n e^{f(n)} $$

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The limit could be anything $\geq 0$. Consider $n e^{-\sqrt{n}}$, $n e^{c-\log n}$, and $n e^{\sqrt{n}}$.

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  • $\begingroup$ What I am really aiming it: I have some function $f$ which has a helpful asymptotic expansion for large arguments. Hence, in order to expand $f(n\cdot e^{\mathcal{O}(nz_n^2)})$ with this large argument expansion it would be interesting for me, whether the argument $n\cdot e^{\mathcal{O}(nz_n^2)}$ is large as $n\to\infty$. Is this the case? $\endgroup$
    – Rhjg
    Jan 23, 2018 at 12:54
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    $\begingroup$ @Rhjg, My answer does address "whether the argument is large as $n \to \infty$". Please consider the three examples of functions of the form $ne^{O(nz_n^2)}$ that I gave. $\endgroup$ Jan 23, 2018 at 20:08
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    $\begingroup$ @Rhjg, You are mistaken. Take $z_n = n^{-1/4}$ to get $nz_n^2 = \sqrt n$. $\endgroup$ Jan 24, 2018 at 0:03
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    $\begingroup$ @Rhjg in the case of Mark's answer, the value represented by $O(n z_n^2)$ is eventually $\geq 0$ for large enough $n$, so $ne^{O(nz_n^2)} \geq n \to \infty$. The extra context makes the claim true in his answer, whereas my examples show that it's false in general. $\endgroup$ Jan 24, 2018 at 19:42
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    $\begingroup$ I thought good mathematics are timeless. (; $\endgroup$
    – Rhjg
    Sep 13, 2018 at 7:29

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