$P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$ "Let $X_1, X_2, ...$  independent random variables where $X_n\sim B(p_n)$ and $p_n = \frac{1}{n}$. Calculate $P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$"
I suppose that i can use the lemma of Borel-Cantelli, but I don't know how interpret that limsup...
Thank you very much! 
 A: Hint:
Define the event 
$$
E_n = \lbrace X_n = 0, X_{n+1}=1, X_{n+2}=0 \rbrace
$$
Then find $\mathbb P(E_n)$ and look at $$\sum^\infty \mathbb P(E_n)$$
and use Borel-cantelli Lemma
A: Is that really $X_n = 0$, $X_{n+1} = \color{red}{1}$ and $X_{n+2} = 0$ and not $X_n = 0$, $X_{n+1} = \color{red}{0}$ and $X_{n+2} = 0$?
If so, since the $X_n$'s are independent, the events $X_n = 0$, $X_{n+1} = 1$ and $X_{n+2} = 0$ are independent. Thus, we have
$$P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = P(X_n = 0)P(X_{n+1} = 1)P(X_{n+2} = 0) = \frac{n-1}{(n)(n+2)}$$
By limit comparison test with $\sum_{n=1}^{\infty} \frac{1}{n^2}$, we have that
$$\sum_{n=1}^{\infty} P(\{X_n = 0, X_{n+1} = 1, X_{n+2} = 0\}) = \sum_{n=1}^{\infty} \frac{n-1}{(n)(n+2)} < \infty$$
Hence, by BCL1, $P(\limsup \{X_n = 0, X_{n+1} = 1, X_{n+2} = 0\}) = 0$

If it is $X_n = 0$, $X_{n+1} = 0$ and $X_{n+2} = 0$:
Since the $X_n$'s are independent, the events $X_n = 0$, $X_{n+1} = 0$ and $X_{n+2} = 0$ are independent. Thus, we have
$$P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = P(X_n = 0)P(X_{n+1} = 0)P(X_{n+2} = 0)$$
Are the $X_n$'s Bernoulli? If $P(X_n = 1) = p_n = 1 - P(X_n = 0)$, then we have
$$= (1-\frac{1}{n})(1-\frac{1}{n+1})(1-\frac{1}{n+2})$$
$$= (\frac{n-1}{n+2})$$
Thus,
$$\sum_{n=1}^{\infty} P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = \sum_{n=1}^{\infty} (\frac{n-1}{n+2})$$
By the test for divergence, the series diverges. Since the summand is nonnegative, the series diverges to infinity.
However, BCL2 requires independence (or pairwise independence) of the events $\{X_1 = 0, X_{2} = 0, X_{3} = 0\}$, $\{X_2 = 0, X_{3} = 0, X_{4} = 0\}$, ...
Consider the first events. We can see that $\{X_1 = 0, \color{red}{X_{2} = 0, X_{3} = 0}\}$ and $\{\color{red}{X_2 = 0, X_{3} = 0}, X_{4} = 0\}$ are NOT independent.
However, consider a subsequence of those events: $\{X_1 = 0, X_{2} = 0, X_{3} = 0\}$, $\{X_4 = 0, X_{5} = 0, X_{6} = 0\}$, $\{X_7 = 0, X_{8} = 0, X_{9} = 0\}$, ...
These events are independent or pairwise independent. So let us consider this series instead:
$$\sum_{n=1}^{\infty} P(\{X_{3n-2} = 0, X_{3n-1} = 0, X_{3n} = 0\})$$
$$ = \sum_{n=1}^{\infty} \frac{3n-3}{3n}$$
BCL2 gives you that $P(\limsup \{X_{3n-2} = 0, X_{3n-1} = 0, X_{3n} = 0\}) = 1$.
Convince yourself that $P(\limsup \{X_{n} = 0, X_{n+1} = 0, X_{n+2} = 0\}) = 1$
