Suppose $X_{n}\sim bern(p_{n})$ and $\sum_{n}p_{n}<\infty$. Then $\frac{\sum_{k=1}^{n}X_{k}}{\sum_{n}p_{n}}$ does not converge to $1$ in probability

Question

This may seem like a weird question. Suppose $X_{n}\sim \operatorname{bern}(p_{n})$, independent and $\sum_{k=1}^{\infty}p_{k}<\infty$. Then does $\dfrac{\sum_{k=1}^{n}X_{k}}{\sum_{k=1}^{n}p_{k}}$ converge to $1$ in probability?

Well, the first thing that one notices is that $\operatorname{Var}(\dfrac{\sum_{k=1}^{n}X_{k}}{\sum_{k=1}^{n}p_{k}})$ does not go to $0$ which is the variance of $1$. Ideally, if $(\sum_{k=1}^{n}X_{k})^{2}$ was uniformly integrable, we can conclude that as $\dfrac{\sum_{k=1}^{n}X_{k}}{\sum_{k=1}^{n}p_{n}}$ does not converge in $L^{2}$ to $1$ and hence it cannot converge in proability to $1$ (by using uniform integrability). The other argument is that $S_{n}$ should converge then to $\sum_{n}p_{n}$ which if we assume that it is an integer, then $S_{n}$ must also converge almost surely to it as $S_{n}$'s are monotone but this is getting me nowhere.

To try and conclude uniform integrability, I look at $\bigg(\sum_{k=1}^{n}X_{k}\bigg)^{4}$ and try and show that $\sup_{n}\mathbb{E}\left(\sum_{k=1}^{n}X_{k}\right)^{4}<\infty$. But after opening the brackets and using multinomial theorem, I am getting complicated expressions.

Well, $\sum_{k=1}^{n}X_{k}$ is uniformly bounded in $L^{2}$ but I also think that $\sum_{k=1}^{n}X_{k}$ will be bounded in $L^{4}$ but I am having trouble to show it rigorously.

Any help? Is there an easier way to do this?

Answer 1

Let $Y_n:=\sum_{k=1}^n X_k$. By assumption, the sequence $\left(Y_n\right)_{n\geqslant 1}$ is Caucchy in $\mathbb L^2$ hence converges in $\mathbb L^2$ to some random variable $Y$. This gives uniform integrability of $\left(Y_n^2\right)_{n\geqslant 1}$.