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Prove: If $x_n=O(\alpha_n)$ and $y_n=O(\alpha_n)$, then $x_n+y_n=O(\alpha_n)$.

I know that $x_n=O(α_n)$ means there exists a constant $C_0$ and integer $n_0$ such that $|x_n|≤C_0|α_n|$ for all $n≥n_0$ and $y_n=O(α_n)$ means there exists a constant $C_1$ and integer $n_1$ such that $|x_n|≤C_1|α_n|$ for all $n≥n_0$. Let $n_2:=\max\{n_0,n_1\}$ and so $|x_n+y_n|\leq |x_n|+|y_n|\leq C_0|\alpha_n|+C_1|\alpha_n|=C|\alpha_n|$ where $C=C_0+C_1$, with which $|x_n+y_n|\leq C|\alpha_n|$ for all $n≥n_2$, then $x_n+y_n=O(\alpha_n)$.

Is this reasoning right? Thank you very much.

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    $\begingroup$ Looks good to me. $\endgroup$ – angryavian Dec 19 '17 at 21:36
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    $\begingroup$ Perfect............... $\endgroup$ – DanielWainfleet Dec 20 '17 at 3:55

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