Relation between (non-random) Big O and probability little o Let $X_n = o_p(a_n)$, $Y_n = o_p(b_n)$ where $a_n$ and $b_n$ are non-random sequences of real numbers and $X_n$, $Y_n$ are sequences of random variables. If $a_n = O(b_n)$ as $n \rightarrow \infty$, then is it true that $X_n + Y_n = o_p(b_n)$?
My intuition says this should be true but I am not sure how to prove it formally. My reasoning is that since $a_n$ doesn't grow faster than $b_n$, then the $b_n$ term should dominate. We know that $X_n + Y_n = o_p(a_n+b_n)$, then since $b_n$ dominates, we can simplify it to $o_p(b_n)$. How can I prove this statement (if true) formally? I am using the definitions of little-p o on wikipedia: https://en.wikipedia.org/wiki/Big_O_in_probability_notation

First we show that $X_n = o_p(b_n).$ Since $a_n = O(b_n)$ there exists an $M$ such that $a_n \le Mb_n$ for sufficiently large $n$. Then $$ \frac{X_n}{b_n} \le M\frac{X_n}{a_n}.$$ So for any $\epsilon>0$ (and sufficiently large $n$), $$ P(X_n/b_n > \epsilon) \le P(X_n/a_n > \epsilon/M) \to 0.$$
With that done we can finish with $$ P\left(\frac{X_n+Y_n}{b_n}\ge \epsilon\right)\le P(\{X_n/b_n\ge \epsilon/2\}\cup\{Y_n/b_n\ge \epsilon/2\}) \\\le  P(X_n/b_n\ge \epsilon/2)+P(Y_n/b_n\ge \epsilon/2) \to 0.$$
