# Infinite sum of indicators almost sure convergence

Let $S_{n}:= \sum_{i=1}^{n} X_{i,n}$ where for each $n, X_{1,n}, X_{2,n},..., X_{n,n}$ are sequences of independents r.v.'s. $$X_{i,n}=\begin{cases}1, & \text{with probability }p_n\\0,& \text{with probability }1-p_n\end{cases}$$ for $i = 1,\dots, n$. Suppose that $p_{n}={\beta}/n$.

The problem is to show that $S_{n}$ doesn't converge almost surely to a limit.

I proved that it converges in distribution to the Poisson, but I'm not sure how to proceed in proving that $S_{n}$ does not converge almost surely to a limit. The hint given is to show that $P(|S_{m}-S_{n}|>1)$ does not converge to zero. I know this is a Cauchy so possible giving me guidance in this directions. (or any other that you see fit)??

Thanks!!

• Also, you know that if it converged to a limit it should be to a Poisson R.V. (if your calculationis right) ,since almost sure convergence implies convergence in distribution. Maybe supposing there is such R.V. you could get a contradiction? – Max Jan 29 '16 at 23:25
• I mean I'm almost sure it converges in distribution unless I missed something. $S_{n}$ is binomial. as n goes to infinity $p_{n}$ goes to zero, and $n*p_{n} = {beta}$. So then I just used the binomial approximation to the Poisson. – repos Jan 29 '16 at 23:38
• ok, contradiction argument. if poisson then it's memoryless so if $S_{n}$ converged to ${\beta}$ the probability of $S_{n}> {\epsilon}$ i.o =0 . But this contradicts the fact that it's restarts at each point?? – repos Jan 30 '16 at 2:12
• "if poisson then it's memoryless" Sorry but what on Earth are you alluding to? – Did Jan 30 '16 at 9:48
• And this process is certainly not Poisson, since $S_{n+1}-S_n$ is always $0$ or $1$ (to be compared to Poisson random variables, which take any nonnegative integer value with positive probability). – Did Jan 30 '16 at 12:59