TITLE: How do I show a random variable converges in probability?
QUESTION [5 upvotes]: I've been given the following problem:

For $n \in \mathbb{N}^{*}$, let $X_{n}$ be a random variable such that $\mathbb{P} \left[ X_{n} = \frac{1}{n} \right] = 1 - \frac{1}{n^{2}}$ and $\mathbb{P} \left[ X_{n} = n \right] = \frac{1}{n^{2}}$. Does $X_{n}$ converge in probability?

And the definition I've been given for convergence in probability of a random variable is:

Let $\left( T_{n} \right)_{n \geq 1}$ be a sequence of r.v. and $T$ a r.v. ($T$ may be deterministic). Then:
$T_{n}\overset{\mathbb{P}}{\underset{n \rightarrow \infty}\rightarrow} T$  if and only if   $\mathbb{P} \left[ \left| T_{n} - T \right| \geq \epsilon \right] {\underset{n \rightarrow \infty}\rightarrow} 0$, for all $\epsilon > 0$.

But I don't understand how to apply this definition to the problem. I'm not even sure what I'm supposed to be seeing if $X_{n}$ is converging to. We have no $X$ as far as I can see. Can anyone give me some insight on this?

REPLY [5 votes]: Fix any number $\epsilon>0$. Then, by density of rationals, $\exists n\in \mathbb{Z}^+$, such that $\epsilon>1/n$. Then, $\forall m\ge n$, $$P(X_m>\epsilon)\le P(X_m> 1/m )=\frac{1}{m^2}\stackrel{m\to \infty}{\to} 0$$ Hence $X_m\stackrel{p}{\to}0 $.