I have question about Slutsky's theorem, specifically regarding the property that $X_n\xrightarrow{p}X$, $Y_n\xrightarrow{p}c$, then $\frac{X_n}{Y_n}\xrightarrow{p}\frac{X}{c}$ (provided $c\neq0$ and $X_n, X ,$ and $Y_n$ are all random variables).
I'm wondering if this will hold in the case where $X_n\xrightarrow{p}0$ and $Y_n\xrightarrow{p} 0$, but $X_n$ converges to zero at a "faster rate" than $Y_n$.
Specifically, assume that $\mathbb{E}[X_n] = a_n$ and $\mathbb{E}[Y_n] = b_n$ for each $n$ with $(a_n),(b_n)$ being a sequence of real numbers such that
$$a_n\xrightarrow{p.w.} 0\quad \quad b_n\xrightarrow{p.w.} 0\quad \quad\frac{a_n}{b_n}\xrightarrow{p.w.} 0$$
Also assume that $X_n,Y_n,a_n,b_n\geq0$
Want: $$\frac{X_n}{Y_n}\xrightarrow{p}0$$
My attempt:
Writing, $$\frac{X_n}{Y_n} =\frac{b_nX_n}{b_nY_n} = \frac{b_n}{Y_n}\frac{X_n}{b_n}, $$ we see that
$$\mathbb{P}\big[\big\vert\frac{X_n}{b_n}\big\vert>\epsilon\big]\leq\frac{\mathbb{E}[X_n]}{\epsilon b_n} = \epsilon^{-1}\frac{a_n}{b_n}\rightarrow 0$$
Which shows that $\frac{X_n}{b_n}\xrightarrow{p} 0$
Now since $\mathbb{E}[Y_n]=b_n$ implies $\mathbb{E}[\frac{Y_n}{b_n}]=1$ for all $n$ trivially, we know that $\mathbb{E}[\frac{Y_n}{b_n}]\rightarrow 1$ which shows that $\frac{Y_n}{b_n}\xrightarrow{p}1$
Then by the continuous mapping theorem we see that for $g(x) = x^{-1}$ $$g\bigg[\frac{Y_n}{b_n}\bigg] = \frac{b_n}{Y_n}\xrightarrow{p}g(1) = 1$$
By Slutksy's theorem, we then have
$$\frac{X_n}{Y_n} = \frac{b_n}{Y_n}\frac{X_n}{b_n}\xrightarrow{p}0\cdot 1 = 0$$
Does anything look incorrect? Is there an easier way?