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If $x_n=o(\alpha_n)$ and $y_n=o(\alpha_n)$, then $x_n+y_n=o(\alpha_n)$.

All I know is the definition of little oh notation being used, that is, $x_n=o(\alpha_n)$ means that there exists a sequences $\epsilon_n\geq 0$ with $\epsilon_n \to 0$ as $n\to \infty$ and $|x_n|\leq \epsilon_n |\alpha_n|$. I'm not sure how to prove this result. Any solutions or hints are greatly appreciated.

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Hint: Note that if $|x_n| \leq \delta_n \alpha_n$ and $|y_n| \leq \epsilon_n |\alpha_n|$, then $$ |x_n + y_n| \leq |x_n| + |y_n| \leq (\delta_n + \epsilon_n)|\alpha_n| $$

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  • $\begingroup$ Can we take $\gamma_n=\delta_n + \epsilon_n$ and then we have that $x_n+y_n=o(\alpha_n)$? $\endgroup$ – Happy Sep 15 '15 at 3:08
  • $\begingroup$ @DerpMagoo that's exactly right. We note that $\gamma_n \to 0$. $\endgroup$ – Omnomnomnom Sep 15 '15 at 3:09
  • $\begingroup$ awesome, thank you : ) $\endgroup$ – Happy Sep 15 '15 at 3:14
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Note that $$ \bigg| \frac{x_{n}+y_{n}}{\alpha_{n}} \bigg| \leq \bigg| \frac{x_{n}}{\alpha_{n}} \bigg| + \bigg| \frac{y_{n}}{\alpha_{n}} \bigg| $$ and that the right side $\to 0$ by assumption.

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  • $\begingroup$ While this is useful for intuition, it requires particular care if $\alpha_n = 0$ for some $n$. $\endgroup$ – Omnomnomnom Sep 15 '15 at 2:38
  • $\begingroup$ That depends on the original requirement as to whether $\alpha_{n} \neq 0$ for all $n$. @Omnomnomnom $\endgroup$ – Megadeth Sep 15 '15 at 2:39

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