Assume $(X_i)_{i\geq1}$ and $(Y_i)_{i\geq1}$ are two independent sequences of i.i.d. random variables such that $\mathbb E[X^k]<\infty$ and $\mathbb E[Y^k]<\infty$ $\forall k\geq1$.
I am wondering if there is there a straightforward way to prove that \begin{equation} \frac{Y_n}{\sum_{i=1}^nX_i} \stackrel{\mathbb P}{\rightarrow}0. \end{equation}
The only approach that comes to my mind is using some approximation theorem to write $\sum_{i=1}^nX_i = n\mathbb E[X] + \varepsilon_n$, where $\varepsilon_n\stackrel{\mathbb P}{\rightarrow}0$, but this seems unnecessarily complicated. Is there any simpler approach?
You can write \begin{align} \frac{Y_n}{\sum_{i=1}^n X_i} &= \frac{Y_n/n}{(\sum_{i=1}^n X_i)/n} \;. \end{align} If $\mathbb{E}[|X|]<\infty$, we have, by the law of large numbers, $(\sum_{i=1}^n X_i)/n\to\mathbb{E}[X]$ almost surely (or in probability). If $\mathbb{P}(|Y|<\infty)=1$, you also have $Y_n/n\to 0$ in probability. You get the desired convergence in probability provided $\mathbb{E}[X]\neq 0$.
If furthermore, $\mathbb{E}[|Y|]<\infty$, you also have almost sure convergence, because \begin{align} \frac{Y_n}{n} &= \frac{\sum_{i=1}^n Y_i}{n}- \frac{n-1}{n}\cdot\frac{\sum_{i=1}^{n-1} Y_i}{n-1} \to 0 \end{align} almost surely, by the law of large numbers.