Let there be a sequence of r.v.'s $\{X_{n}\}$ with the following property \begin{align} \mathbb{P}(X_n = n) = 1 - \mathbb{P}\left(X_{n} = \frac{1}{n}\right) = \frac{1}{n^{2}} \end{align} Show that $X_{n} \xrightarrow[]{\mathbb{P}} 0$.
Usually I define $$\lim_{n \xrightarrow[]{}\infty}\mathbb{P}(|X_{n}-X| > \epsilon)\; \forall \epsilon,$$ in this case $X = 0$, and apply Chebyshev's inequality to get a upper bound and see where it leads me, however I'm struggling with this one, I would appreciate a insight on finding the variance. Also the author does not define the distribution support, is this a mistake? Can we solve the problem without this information? Applying the limit to the probability would get me the result, but is it correct? Thanks in advance!
If $1/n<\epsilon<n$, then $P(|X_n|>\epsilon)=1/n^2 \to 0$. This inequality holds for a fixed $\epsilon>0$ for all sufficiently large $n$.
The choice of $n,1/n$ and $1/n^2$ was not essential, you could have had $P(X_n=x_n)=1-P(X_n=y_n)=p_n$ for any sequence $x_n$, any sequence $y_n \to 0$, and any sequence $p_n \to 0$.
Note that the distribution support is already specified because the two given probabilities sum to $1$.