is it possible to estimate the "large number" in Borel-Cantelli lemma? I have a sequence of events $A_n$.
If i can show that $\sum_{n\to\infty} P(A_n)<\infty$, then Borel-Cantelli lemma guarantees that the events in $A_n$ can happen at most finite times, e.g. $P(\limsup_{n\to\infty}A_n)=0$.
In other word, there exists a large number $N$, such that for all numbers $n>N$, $P(A_n) = 0$, where $n>N$.
If i know exactly the expression of $\sum_{n\to\infty} P(A_n)$, is it possible to estimate this large number $N$?
 A: What have you have written is not true. For example, consider $\Omega =[0,1]$ with Lebesgue measure as probability measure and $A_n = (0, 1/2^n)$. Then $\sum_n P(A_n) = \sum_n 1/2^n < \infty$, but $P(A_n) > 0$ for all $n$.
What is true is that for almost all $\omega \in \Omega$ (where $\Omega$ is your probability space), there is some $N_\omega \in \Bbb{N}$ (depending on $\omega$) with $\omega \notin A_n$ for all $n > N_\omega$.
I don't think there is any way in which one can obtain an explicit bound for $N_\omega$, given just the stated assumptions.
A: As PhoemueX noted, the "In other words, ..." statement is false.  But there is a 
random variable $N = \max(n \; : \; A_n)$ (i.e. $N(\omega) = \max(n \; :\; \omega \in A_n)$) that is almost surely finite.
Thus $$P(N \ge n) = P(\bigcup_{j = n}^\infty A_j) \le \sum_{j = n}^\infty P(A_j)$$
Now
$$ E[N] = \sum_{n=1}^\infty P(N \ge n) \le \sum_{n=1}^\infty \sum_{j =n}^\infty P(A_j) = \sum_{j=1}^\infty j \;P(A_j)$$
(which may or may not be finite)
