Almost sure convergence does not imply complete convergence We say $X_1, X_2, \ldots$ converge completely to $X$ if for every $\epsilon>0$ $\sum_{n=1}^\infty \text{P}\left(|X_n-X|>\epsilon\right) <\infty$. 
With Borel Cantelli's lemma is straight forward to prove that complete convergence implies almost sure convergence. 
I am looking for an example were almost sure convergence cannot be proven with Borel Cantelli. This is, a sequence of random variables that converges almost surely but not completely.
 A: A little late to the party but here's my attempt. Stumbled upon this when doing a similar problem, but came up with a different solution.
Usually, questions of this sort will involve some kind of dependent random variable (since the converse of the Borel-Cantelli lemma fails when there's non-independence).
Consider the martingale $S_n = \sum_{i=0}^n X_i$, which starts at $S_0 = X_0 = 1$, with increments $P(X_k = \pm 1) = \frac{1}{2}$.
Define the stopping time $\tau = \inf \{n: S_n = 0\}$. Since we are dealing with a simple symmetric random walk, $\tau < \infty$ almost surely, and hence, $S_{n \wedge \tau} = 0$ almost surely as $n \rightarrow \infty$.
Before we provide the counter example to show almost sure convergence does not imply complete convergence, lets first calculate $\mathbb{P} (n < \tau)$. The strategy used here is similar to the gambler ruin's problem, namely,
$$
1 = \mathbb{E}[S_{n \wedge \tau}] \leq n (\mathbb{P} (n < \tau)) + 0(\mathbb{P} (n \geq \tau)),
$$
where the inequality is from the observation that $S_n \leq n$ and therefore, $\mathbb{P} (n < \tau) \geq \frac{1}{n}$.
Now, we proceed to demonstrate by counterexample that $S_{n \wedge \tau} \stackrel{a.s.}{\rightarrow} 0 $ does not imply that $S_{n \wedge \tau}$ converges completely to 0. By the definition of complete convergence,
$$
\begin{split}
\sum_{n=1}^\infty \mathbb{P}(|S_{n \wedge \tau}-0|>\epsilon) &= \sum_{n=1}^\infty \mathbb{P}(|S_{n \wedge \tau}|>\epsilon) \\
&= \sum_{n=1}^\infty \mathbb{P}(S_{n \wedge \tau} \neq 0) \\
&=  \sum_{n=1}^\infty \mathbb{P}(n < \tau) \\
&\geq \sum_{n=1}^\infty \frac{1}{n} \\
&= \infty,
\end{split}
$$
and therefore, a.s. convergence does not necessarily imply complete convergence.
A: Let $\Omega=(0,1)$ with the Borel sigma-algebra $\mathfrak{F}$ and uniform measure $\mu$.  Define
$$X_n(\omega) = 2 + (-1)^n \text{ when } \omega \le 1/n$$
and $X_n(\omega)=0$ otherwise.  The $X_n$ are obviously measurable on the probability space $(\Omega, \mathfrak{F}, \mu)$.

For any $\omega\in\Omega$ and all $N \gt 1/\omega$ it is the case that $X_n(\omega)=0$.  Thus, by definition, the sequence $(X_n)$ converges to $0$ (not just almost surely!).
However, whenever $0 \lt \epsilon \lt 1$, $\Pr(X_n\gt \epsilon) = \Pr(X_n \ne 0) = 1/n$, whence
$$\sum_{n=1}^\infty \Pr(X_n \gt \epsilon) = \sum_{n=1}^\infty \frac{1}{n},$$
which diverges to $\infty$.
