If $\alpha_n\to 0$, $x_n=O(\alpha_n)$, and $y_n=O(\alpha_n)$, then $x_ny_n=o(\alpha_n)$.

I know that $x_n=O(\alpha_n)$ means there exists a constant $C$ and integer $n_0$ such that $|x_n|\leq C|\alpha_n|$ for all $n\geq n_0$. Any solutions or hints are greatly appreciated.

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    $\begingroup$ Try to prove this by definition. You already wrote definition of $O$, write that of $o$ and see why does it hold. $\endgroup$ – zhoraster Sep 15 '15 at 3:41
  • $\begingroup$ For little oh I need to find a positive sequence that converges to zero. $\endgroup$ – Happy Sep 15 '15 at 3:53
  • $\begingroup$ Huh, let me guess. $C|\alpha_n|$? $\endgroup$ – zhoraster Sep 15 '15 at 3:55
  • $\begingroup$ It's as simple as using one of the $\alpha_n$ sequences? $\endgroup$ – Happy Sep 15 '15 at 3:56
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    $\begingroup$ I told you it's by definition. So it's simple, yes. $\endgroup$ – zhoraster Sep 15 '15 at 4:09

By assumption there are $M_{1},M_{2} \geq 0$ such that $|x_{n}| \leq M_{1}|\alpha_{n}|$ and $|y_{n}| \leq M_{2}|\alpha_{n}|$ for large $n$, so $|x_{n}y_{n}| \leq M_{1}M_{2}|\alpha_{n}|^{2}$ for large $n$. Since $\alpha_{n} = o(1)$ by assumption, we are done.

  • $\begingroup$ So you choose the sequence to be $\alpha_n$? $\endgroup$ – Happy Sep 15 '15 at 4:00
  • $\begingroup$ By definition $x_{n} = O(a_{n})$ means that there is some $M \geq 0$ such that $|x_{n}| \leq M|a_{n}|$ for large $n$. No matter what definition you are using, it must be equivalent to this one. $\endgroup$ – Megadeth Sep 15 '15 at 4:04
  • $\begingroup$ Moreover, $\alpha_{n} \to 0$ as $n \to \infty$ is equivalent to $\alpha_{n} = o(1)$.as $n \to \infty$. $\endgroup$ – Megadeth Sep 15 '15 at 4:04

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