Borel Cantelli Lemma for non independent random variables Suppose $Y_n$ is a sequence of random variables, not independent, which converges to $0$ in probability. Define a set $A_n = \{|Y_n|>\epsilon\}$. If $\sum_1^\infty P(A_n) = \infty$, can it be concluded that $A_n$ does not converge to zero almost surely?
 A: No. Borel-Cantelli Lemma states that $P(A_n i.o.) = 1$ only when $\sum_{n=1}^{\infty}{P(A_n)} = \infty$ and $A_n$'s are mutually independent. 
Since in your case, $Y_n$'s are not independent, $A_n$'s are also not mutually independent. Hence, you cannot conclude "$A_n$ does not converge to zero almost surely" just using Borel-Cantelli Lemma. You will need additional information.
Edit 1: Adding an example as per suggestion in comments.   
Consider $ \Omega = [0,1]$,  $\sigma$-algebra to be the boral $\sigma$-algebra
$$Y_n = \begin{cases} 1, \ if \ \ \ \ w \in \left[ 0, \frac{1}{n} \right) \\
0 \ \ \ otherwise \end{cases} $$
Let the Probability Measure be defined as length of the interval i.e.
$$ P(Y_n = 1) = \frac{1}{n} \ , P(Y_n=0) = 1 - \frac{1}{n}$$
Here, you can clearly observe that $Y_n$'s are not independent. But if you define $A_n$ as
$$A_n = \{Y_n = 1\}$$ we get
$$P(A_n) = P(Y_n = 1) = \frac{1}{n}$$ so you have
$$\sum_{n=1}^{\infty}{P(A_n)} = \sum_{n=1}^{\infty}{\frac{1}{n} = +\infty}$$
But for this sequence $Y_n$, you can easily see that $Y_n$'s converge almost surely to $0$ as $ n \to \infty$
So, we have a case where $Y_n$'s are not independent and $P(A_n)$ sums upto $\infty$, but still $Y_n$'s converge almost surely to $0$.
Hence, Independence is must to apply Borel-Cantelli Lemma.
Edit 2: Adding reference
Example taken from this Tutorial Notes. Example 12.3.1
