# Convergence of sum of Bernoulli random variables $X_k\sim \text{Bernoulli}\left(\frac{p}{k}\right)$

Let $X_k\sim\text{Bernoulli}\left(\frac{p}{k}\right)$ for some $0<p<1$ and let $S_n:=\dfrac{\sum_{k=1}^nX_k}{\sum_{k=1}^n\frac1k}$. Then

1. does $S_n\to p$ in probabibility?
2. does $S_n\to p$ almost surely?

We have that $\mathbb E\left(S_n\right)=p$ and $\text{Var}\left(S_n\right)=\dfrac{1}{H_n^2}\sum_{k=1}^n\frac{p}{k}\left(1-\frac{p}{k}\right)$, where $H_n$ is the $n-$th harmonic number. Since $\text{Var}\left(S_n\right)<p/H_n$, the variances are bounded and hence by Chebyshev's Weak Law, we get convergence in probability. But I am not sure how to proceed with almost sure convergence. I think that it does not converge almost surely to $p$ but how can I show this? Do I have to use the second Borel-Cantelli lemma?

• @Lonidard Sorry but the quantities $P(S_n=p)$ are totally irrelevant. – Did Mar 28 '18 at 7:23