Suppose a random sequence is defined by $X_n := n B_n$, where $B_n$ is a Bernoulli sequence such that $\mathbb{P}(B_n = 1) = 1/n$. I am interested in the convergence properties of this random process and am not sure how to interpret the results.
To show pointwise (sure) convergence, we need to show that $\lim_{n\rightarrow\infty} X_n(\omega) = X(\omega), \ \forall \omega \in \Omega$. In this case, $\Omega = \{0,1\}$, i.e., success or fail of the Bernoulli trial. Further, $X_n(\omega = 0) = 0$ and $X_n(\omega = 1) = n$, from which we see that the cae of $\omega = 1$ does not yields convergence as clearly $n$ diverges.
For convergence in probability, we need to show that $\lim_{n\rightarrow\infty} \mathbb{P}(|X_n - X| > \epsilon) = 0$, however I am not sure what $X$ should be here. If I assume $X = 0$, then
$$ \mathbb{P}(|X_n - 0| > \epsilon) = \mathbb{P}(X_n > \epsilon) = \mathbb{P}\bigg(B_n > \frac{\epsilon}{n}\bigg) = \frac{1}{n}, $$ since $n,\epsilon > 0$, so the above expression reduces to $\mathbb{P}(B_n = 1)$. Plugging this back in gives $\lim_{n\rightarrow\infty} 1/n = 0$, so $X_n$ does converge in probability.
The same procedure can be done for mean-square (MS) convergence as well, from which I find that $X_n$ diverges in that sense as well.
Am I doing something wrong here when trying to calculate the convergence properties? And if not, what is the intuition behind why $X_n$ does not converge to any random variable?
Edit 1: Cumulative Distribution Functions of $X_n$