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Assume that $u_n$ is a real sequence such that $u_n = O(1/n)$ as $n$ tends to infinity. It means that there exists a positive $M$ and integer $n_0$ such that $ \left|\frac{u_n}{n^{-1}}\right| \leq M \text { for all } n \geq n_0.$

So, $\left|u_n \cdot n \right| \leq M \text { for all } n \geq n_0.$

However, $n$ can become very large and therefore in order to stay bounded $u_n$ should get closer and closer to $0$.

Therefore, $u_n$ is $o(1)$?

amWhy
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Eryna
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    $u_n=\mathcal{O}(1/n)\implies \exists M>0 ,, \lvert u_n \rvert \leqslant M/n ,, \forall n \implies \lvert u_n \rvert \to 0$ as $n\to \infty$ since $M/n \to 0$. – jcneek Jan 24 '23 at 10:49
  • @eryna You're right: specifically, $\lvert u_n\rvert\le\varepsilon$ for all $n\ge \max{n_0,M/\varepsilon}$. – Sassatelli Giulio Jan 24 '23 at 10:53
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    More generally, if $f\in o(g)$ then $O(f)\subseteq o(g)$. – J.G. Jan 24 '23 at 10:59

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