a.s. convergence of Bernoulli sequence

Question

Let $X_1, \ldots, X_n, \ldots$ be a sequence of Bernoulli random variables, $X_k \sim Bern(p_k)$. Prove that $$ X_n \xrightarrow{a.s.} 0 $$ if and only if $$ \sum_{k = 0}^{+\infty} p_k < +\infty. $$ The "if" part is an easy implication from the Borel-Cantelli theorem, but I do not have any ideas about the second part of the problem (why convergence a.s. implies the convergence of the series)

Answer 1

Let $(X_n)_{n \in \mathbb{N}}$ be also independent. In such case, assume $X_n\to 0$ a.s. (i.e. $P(X_n\to 0)=1)$. Suppose $\sum_np_n=\infty$. Then B-C II implies that $P(X_n=1 \textrm{ i.o.})=1$. This, however, contradicts $X_n \to 0$ a.s.; to see this: $$\{X_n\to 0\}=\bigcap_{\ell \in \mathbb{N}}\bigcup_{n \in \mathbb{N}}\bigcap_{k\geq n}\{X_k\leq 1/\ell\}=\bigcap_{\ell\in \mathbb{N}}\{X_n>1/\ell \textrm{ i.o.}\}^c\subseteq \{X_n>1/2 \textrm{ i.o.}\}^c=\{X_n=1\textrm{ i.o.}\}^c$$ and we would have $P(X_n\to 0)\leq P(\{X_n=1\textrm{ i.o.}\}^c)=0\neq 1$. So $\sum_np_n<\infty$.

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To see that the $X_n$ in @Surb's correct example are not independent, note $X_{n+1}\leq X_{n}$ a.s. $\forall n$

Answer 2

The converse is not true. Take $(\Omega,\mathcal F,\mathbb P)=\big((0,1),\mathcal B(0,1),m\big)$ where $m$ is the Lebesgue measure on $(0,1)$. Let $$X_n(\omega ):=\boldsymbol 1_{\left(0,\frac{1}{n}\right)}(\omega ).$$ Set $$p_n:=\mathbb P\{X_n=1\}=\frac{1}{n}.$$ Then, $X_n\sim \text{Bern}(p_n)$ and $X_n\to 0$ a.s. However, $$\sum_{n\in\mathbb N}p_n=\sum_{n\in\mathbb N}\frac{1}{n}=\infty .$$