Convergence in probability for a given distribution $X_n$ assumes the values $1 , \frac{1}{2} , \frac{1}{3}, \ldots, \frac{1}{n}$. with equal probability . Examine whether  $X_n$ converges in probability to some constant?
This is what I tried :
$\mathrm{E}(X_n) = \frac{1}{n}\sum_{k=1}^{n}\left(\frac{1}{k}\right)$, here I am confused how to solve limit $n \to \infty$ on RHS.
 A: This is an interesting question because it helps expose the meaning of convergence in probability and thereby develops our intuition.
When probabilities are involved and the question is general, the most basic tool to consider is the cumulative probability function, or CDF.  For any number $x$ its value at $x$ is
$$F_X(x)= \Pr(X \le x).$$
This is useful because convergence in probability can be established by showing that the sequence of CDFs converges to some limiting CDF except possibly at points where the limiting CDF has jumps.
Take a look at the CDFs of some of the random variables $X_n$ in this sequence. 

To reflect the probabilities of $1/n$ at the values $1/n, 1/(n-1), \ldots, 1/2, 1/1,$  the CDF of $X_n$ has a vertical jump of $1/n$ at each of those numbers. As $n$ grows, those jumps get smaller but they also concentrate close to $0,$ since $1/n\approx 0$ for large $n.$ 
For reference, I have drawn the CDF for a constant jump of $1$ at the value $0$ with a dashed blue line.  This is the CDF for the constant, or "atomic," random variable that is always zero.  Let's call this CDF $F_0.$ The CDFs of the $X_n$ appear to be approaching $F_0.$

To prove that the CDFs of the $X_n$ do approach $F_0,$ first consider any positive number $x.$  The value of $F_{X_n}(x)$ is $1/n$ times the number of jumps made at values less than or equal to $x.$  These would be the integers between $1/x$ and $n,$ inclusive.  Writing $[1/x]$ for the smallest integer exceeding $1/x,$ we have shown that
$$F_{X_n}(x) = \frac{n - [1/x] + 1}{n}\tag{1}$$
provided $n \gt x,$ which eventually will occur no matter what value $x$ might have.  As $n$ grows large, the fraction in $(1)$ approaches $1$--just as the figure suggests.
Next, if $x \lt 0,$ $F_{X_n}(x)=0$ is a constant sequence with the limiting value $0.$
Putting these results together establishes two limits:


*

*$\lim_{n\to\infty}F_{X_n}(\epsilon) = 1$ for positive $\epsilon.$

*$\lim_{n\to\infty}F_{X_n}(\epsilon) = 0$ for negative $\epsilon.$
We don't care about what happens at $x=0$ because $F_0$ has a jump there.
Consequently, we have established that

$F_0$ has a jump at $0.$  For all $x \ne 0,$ $\lim_{n\to\infty}F_{X_n}(x) = F_0(x).$ Therefore $\operatorname{plim}_{n\to\infty} X_n = 0.$

