Convergence Almost Surely Exercise We have to check convergence almost surely of $X_n$ to $0$ 
$$P(X_n=e^n)=\frac{1}{n+1}, \quad P(X_n=0)=1-\frac{1}{n+1}$$
I'm not sure I understand in practice the convergence Almost Surely and its difference from convergence in Probability. 
 A: For 2, here are two realizations which give different answers:


*

*Assume in addition that $(X_n)$ are independent. Then by the 2nd Borel-Cantelli's lemma together with the estimate
$$ \sum_{n=0}^{\infty} \Bbb{P}(X_n = e^n) = \infty, $$
it follows that $\Bbb{P}(X_n = e^n \text{ i.o.}) = 1$.Therefore in this case, $(X_n)$ diverges almost surely.

*Now assume that $(X_n)$ is constructed in the following way: $(Z_n)_{n\geq1}$ are independent and
$$ \Bbb{P}(Z_n = e) = \frac{n}{n+1}, \qquad \Bbb{P}(Z_n = 0) = \frac{1}{n+1}. $$
Now let $X_0 = 0$ and $X_n = Z_1 \cdots Z_n$. Then
$$ \Bbb{P}(X_n = e^n) = \prod_{i=1}^{n} \Bbb{P}(Z_i = e) = \prod_{i=1}^{n}\frac{i}{i+1} = \frac{1}{n+1}, \qquad \Bbb{P}(X_n = 0) = 1 - \frac{1}{n+1}. $$
Notice that $\Bbb{P}(Z_n \neq 0 \text{ for all }n) = \prod_{n=1}^{\infty}\frac{n}{n+1} = 0$. Thus $Z_n = 0$ for some $n$ almost surely, which means that $X_n \to 0$ almost surely.
A: This does not converges Almost Surely, though it converges in probability.
For as converge to hold (in rather informal terms) we need that "almost all realizations" of the sequence $\{X_n \}$ tend to zero. In our case, an individual sequence tends to zero only if there exists some $M$ such that $X_n=0$ for all $n\ge M$. Then, we need to compute this probability.
Let's call $A$ the set of such sequences, and for each $\{X_n \}\in A$ define $M$ as the greatest index such that $X_M\ne 0$. Lets call $A_M$ the sets grouped by this value, so $A= \cup_{M=0}^\infty A_M$
Then $$P(\{X_n\} \in A_M)= P(X_M\ne 0 \wedge X_{M+1}=0 \wedge X_{M+2}=0 \dots)=\frac{1}{M+1} \prod_{k=1}^ {\infty} \frac{M+k}{M+k+1} =0$$
This implies that $P(\{X_n\} \in A) =0$ . But for as convergence we wanted $P(\{X_n\} \in A) =1$
