Second Borel-Cantelli Lemma

Statement
Let the events $E_n$ be independent.

Let the sum of the probabilities of the $E_n$ diverges to infinity.

Then the probability that infinitely many of them occur is $1$.

That is:

If $\displaystyle \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$ and the events $\displaystyle \sequence {E_n}^\infty_{n \mathop = 1}$ are independent, then:
 * $\displaystyle \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 1$

Proof
Let $\displaystyle \sum_{n \mathop = 1}^\infty \map \Pr {E_n} = \infty$.

Let $\sequence {E_n}^\infty_{n \mathop = 1}$ be independent.

It is sufficient to show the event that the $E_n$s did not occur for infinitely many values of $n$ has probability $0$.

This is just to say that it is sufficient to show that:


 * $\displaystyle 1 - \map \Pr {\limsup_{n \mathop \to \infty} E_n} = 0$

Noting that:

it is enough to show:
 * $\displaystyle \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$

Since the $\sequence {E_n}^\infty_{n \mathop = 1}$ are independent:

This completes the proof.

Alternatively, we can see:
 * $\displaystyle \map \Pr {\bigcap_{n \mathop = N}^\infty E_n^c} = 0$

by taking negative the logarithm of both sides to get:

Since $-\map \ln {1 - x} \ge x$ for all $x > 0$, the result similarly follows from our assumption that $\displaystyle \sum^\infty_{n \mathop = 1} \map \Pr {E_n} = \infty$.