Suppose $X_1,X_2,...$ is a sequence of discrete random variables and $X_n$ has a probability function
$$p_n(x)=\begin{cases} \frac{n-1}{2n} & \text{for } x=-1/n \text{ or } x=1/n\\ \frac1n &\text{for }x=n \\ 0 &\text{otherwise} \end{cases}$$
Show that $X_n$ converges in probability to $0$.
After using the formula I take it it becomes $\lim\limits_{n\to \infty} p(|X_n-0|\geq \epsilon)=0$
But I’m not sure where to go from here
For any $\epsilon>0$ you choose, I can always find $N$ such that for all $n\geq N$, $\frac 1n < \epsilon$. This means that past that $N$, $X_n=\pm \frac 1n<\epsilon$ with probability $\frac{n-1}{2n}$. That means that only when $x=n$ (with probability $\frac 1n$) will $|X_n|$ be greater or equal to $\epsilon$, and thus $p(|X_n|\geq \epsilon)=\frac 1n$, which goes to $0$ as $n\to \infty$. This argument holds for all $\epsilon$, so $X_n\to_p 0$.