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

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.

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

• So you choose the sequence to be $\alpha_n$? – Happy Sep 15 '15 at 4:00
• 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. – Megadeth Sep 15 '15 at 4:04
• Moreover, $\alpha_{n} \to 0$ as $n \to \infty$ is equivalent to $\alpha_{n} = o(1)$.as $n \to \infty$. – Megadeth Sep 15 '15 at 4:04