Showing $X_n \to 0$ in distribution where $X_n = n^2$ with probability $\frac{1}{n}$ and $X_n = \frac{1}{n}$ with probability $1- \frac{1}{n}$? Suppose that $X_n$ is a sequence of random variables where $X_n = n^2$ with probability $\frac{1}{n}$ and $X_n = \frac{1}{n}$ with probability $1- \frac{1}{n}$. I am wondering how to show that the sequence $X_n$ converges in distribution to $0$. I have started with $P(X_n \leq t)$ but when I apply the limit, I have trouble specifying the $t$ ranges. Does anyone have any ideas?
Is it valid that $X_n \to \mathbb{1}_{t \leq 0}$ in distribution?
 A: Since $P(|X_n|>\epsilon)=1/n$, as $n>1/\epsilon$, so $X_n\to 0$ in probability and $X_n\to 0$ in distribution too.
A: For any fixed $t<0$ you have that $P(X_n\le t)=0$ $($since $X_n\ge 0$ with probability $1)$ and therefore $$\lim_{n\to \infty} P(X_n\le t)=0\tag1$$ For any fixed $t>0$ you have that \begin{align}P(X_n\le t)&=P\left(X_n=\frac1n\right)\cdot \mathbf 1_{ \left\{\frac1n \le t\right\}}+P\left(X_n=n^2\right)\cdot \mathbf 1_{\left\{n^2 \le t\right\}}\\[0.2cm]&=\left(1-\frac1n\right)\mathbf 1_{\left\{\frac1n \le t\right\}}-\frac1n \mathbf 1_{\{n^2\le t\}}\end{align} Hence $$\lim_{n\to \infty }P(X_n\le t)=\lim_{n\to \infty}\left[\left(1-\frac1n\right)\mathbf1_{\left\{\frac1n<t\right\}}+\frac1n\mathbf 1_{\{n^2\le t\}}\right]=(1-0)\cdot 1+0\cdot0=1\tag2$$ Now, $(1)$ and $(2)$ together imply that $$\lim_{n\to\infty}P(X_n\le t)=\begin{cases}0, & t<0\\1, &t>0\end{cases}$$ On the other hand, the distribution function $F$ of the degenerate random variable $X\equiv 0$ is $$F_X(t)=\begin{cases}0, & t<0\\1, &t= 0\\1, &t>0\end{cases}$$ However, according to the definition of convergence in distribution we do not need to consider $t=0$ since it is not a continuity point of $F$ and the conclusion follows.
