Almost sure convergence {Xn} is a sequence of independent random variables each with the same 
Sample Space {0,1} and Probability {1-1/$n^2$ ,1/$n^2$}
 Does this sequence converge with probability one (Almost Sure) to the constant 0?
Essentially the {Xn} will look like this (its random,but just to show that the frequency of ones drops)
   010010000100000000100000000000000001......

Yes, but how fast does the frequency drop ? The faster it does the more it is probable that (Xn) converges to 0.
Lemma : If (Xn) is a sequence of elements in {0,1}, Xn converges to 0 if and only if Xn has finitely many 1s :
If a sequence (Xn) converges to 0, then by definition of the limit, there exists some integer N such that for all n>=N, |Xn| <= 1/2. Now since Xn is 0 or 1, |Xn| <= 1/2 implies that Xn = 0.
Thus, for all n >= N, Xn = 0.
This means that the sequence (Xn) has finitely (in fact, less than N) many 1s.
Conversely, if a sequence (Xn) has finitely many 1s, there exist an integer N (the index of the last 1 of the sequence) such that Xn = 0 for all n > N.
Then the sequence (Xn) converges to 0 because for any $\varepsilon$ > 0, we do have that for all n > N, $|X_n - 0| = 0 < \varepsilon$.
Here, define $Y_n = \Sigma_{k=1}^n X_n$ and $Y_\infty = \lim_{n \rightarrow \infty} Y_n$ = the number of 1s in the sequence (Xn). The lemma tells us that (Xn) converges to 0 if and only if $Y_\infty < \infty$.
You'll notice that $E[Y_\infty] = \lim E[Y_n] \lt \infty$ thanks to the fact that $\Sigma \frac{1}{n^2}$ is convergent.
This implies that $Y_\infty < \infty$ almost surely, which in turn shows that (Xn) does converge with probability 1.
If you have the frequency drop too slowly, this won't work. For example if P(Xn = 1) = 1/log(n) for n>=3, it will almost always diverge.
