$\sum_{n=1}^{\infty}p_{n}<\infty \iff X_{n} \to 0$ a.s.

Question

Let $(p_{n})_{n}\subset [0,1]$ and $(X_{n})_{n}$ independent random variables, so that $X_{n}$~ $Ber(p_{n})$

Prove that:

$\sum_{n=1}^{\infty}p_{n}<\infty \iff X_{n} \to 0$ a.s

Ideas:

"$\Rightarrow$"

$\sum_{n=1}^{\infty}P(X_{n}=1)=\sum_{n=1}^{\infty}p_{n}<\infty$

I would attempt to use the Lemma of Borel-Cantelli, but have no idea how to use the independence aspect.

"$\Leftarrow$" $P(X_{n} \to 0)=P(\omega\in \Omega: \exists N \in \mathbb N, X_{n}(\omega)=0, \forall n \geq N)=1$. Let fix this $N$ accordingly, this then means $\sum_{n=1}^{\infty}P(X_{n}=1)=\sum_{n=1}^{N-1}P(X_{n}=1)+\sum_{n=N}^{\infty}P(X_{n}=1)=\sum_{n=1}^{N-1}P(X_{n}=1)+0=\sum_{n=1}^{N-1}p_{n}<\infty$

da $p_{n} \in [0,1], \forall n \in \mathbb N$

Answer 1

Note that $ X_n\stackrel{\text{a.s}}{\to} 0 $ iff $P(|X_n|>\varepsilon \; \text{i.o})=0$ for every $\varepsilon>0$. Since we are dealing with bernoulli random variables we only care about $0<\varepsilon<1$. So fix $0<\varepsilon<1$.

In this case, $\sum_n P(|X_n|>\varepsilon)=\sum_n p_n$. By the first Borel Cantelli lemma, if $\sum_n p_n<\infty$, then $X_n\to 0$ almost surely. By the second Borel cantelli lemma (the $X_n$ are independent), if $\sum_n p_n=\infty$, then $P(|X_n|>\varepsilon \; \text{i.o})=1$, so $X_n\not\to0$ almost surely.

Answer 2

Since $X_n$'s are $0,1$ valued we see that $X_n \to 0$ a.s iff $P\{X_n=1 \, \text {i.o.}\}=0$. By independence and Borel Cantelli this is so iff $\sum P\{X_n=1\} <\infty$ or $\sum p_n<\infty$