I've encountered these two examples (used to show how a.s. convergence doesn't imply convergence in $R$th mean and visa versa). In one case we have a random variable $X_n = n$ with probability $\frac{1}{n}$ and zero otherwise (so with probability $1-\frac{1}{n}$). In another case same deal with only difference being $X_n=1$, not $n$ with probability $\frac{1}{n}$. Assume $X_n$'s are independent in both.
It is clear to me that in the second case, where $X_n = 1$ w.p. $\frac{1}{n}$, the summation is $\sum_{n=1}^\infty P(X_n=1) =\sum_{n=1}^\infty \frac{1}{n} =\infty $, so by Borel Cantelli Lemma 1 we have that "$X_n=1$" occurs infinitely often and hence $X_n$ doesn't converge almost surely to zero.
In the first case however, where $X_n = n$ w.p. $\frac{1}{n}$, the argument goes that $X_n$ converges to zero almost surely. Why is that? Isn't it the same, only here we are summing $P(X_n \ne 0)$ for $n=1$ through $\infty$ and these probabilities are again resulting in a diverging harmonic series?
