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

Question

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.

Answer 1

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

Answer 2

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.