Consider the sequence of independent random variables $\{X_n\}$ such that $$\begin{align} P(X_n = 1) &= 1/n \\P(X_n = 0) &= 1 - 1/n \end{align}$$
I saw this as an example of convergent in probability but not almost surely (https://courses.cit.cornell.edu/econ620/reviewm6.pdf). But I do think it is convergent a.s.
And I know the most common example of convergent in p but not a.s. is more complex than this...
Any suggestions? Thanks~
The sequence is not convergent a.s. Indeed, is the typical example of convergence in probability but not convergence a.s.
See for example this answer.
If you have trouble understanding the concept of convergence a.s., and wish to show that $X_n$ does not converge a.s. without resorting to Borel-Cantelli, you can think along these lines: imagine the experiment of producing an instance of the complete random sequence $(x_1,x_2 \cdots )$, ask yourself : does this (given, fixed) sequence converge (now not in probability, but in the traditional sense: convergence of a non-random sequence) ? A little reflection should show you that this can only happen if it converges to $1$ or (more probable, it seems) to $0$. Now, for the sequence to converge to zero we need to have, either $x_k=0$ for all $k$, or $ x_m=1$ and $x_k=0$ for all $k>m$, for some (finite) $m \ge 0$. Denote by $E_m$ this event. Show that $P(E_m) = 0$ and hence the probability of the sequence converging to zero, being the countable sum of infinite events of zero probability, is zero. (this is also true for convergence towards $1$, of course). Hence, the probability of producing a sequence that converges is zero. But the (random ) sequence converges a.s. if the probability of (non random) convergence is one. Hence, this does not converge a.s.