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Show that if $x_n=O(\alpha_n)$, then $x_n/\ln n=o(\alpha_n)$

I know that $x_n=O(α_n)$ means there exists a constant $C$ and integer $n_0$ such that $|x_n|≤C|α_n|$ for all $n≥n_0$ so $|\frac{x_n}{\alpha_n\ln n}|\leq|\frac{Cx_n}{x_n \ln n}|=|\frac{C}{\ln n}|\to 0$ for all $n≥n_0$ since $|x_n|≤C|α_n|\Leftrightarrow \frac{1}{|\alpha_n|}\leq\frac{C}{|x_n|}$ and so $\lim_{n\to \infty}\frac{x_n}{\ln n}/\alpha_n=\lim_{n\to \infty}\frac{x_n}{\alpha_n\ln n}=0$, with which $x_n/\ln n=o(\alpha_n)$.

Is this reasoning right? Thank you very much.

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    $\begingroup$ Your arguments work, but it is more direct to use $|x_n|\leq C|\alpha_n|$, getting: $$\left|\frac{x_n}{\alpha_n \ln n}\right|\leq \left|\frac{C\alpha_n}{\alpha_n \ln n}\right|=\left|\frac{C}{\ln n}\right|.$$ $\endgroup$ – Thomas Andrews Dec 14 '17 at 18:35
  • $\begingroup$ @ThomasAndrews Why is this not true in general? $x_n=O(α_n)$ means there exists a constant $C$ and integer $n_0$ such that $|x_n|≤C|α_n|$ for all $n≥n_0$ and $x_n=o(\alpha_n)$ means that $\lim_{n\to 0}x_n/\alpha_n=0$ $\endgroup$ – user402543 Dec 14 '17 at 19:23
  • $\begingroup$ I deleted that comment - jumped the gun. $\endgroup$ – Thomas Andrews Dec 14 '17 at 19:55

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