Convergence in probability of normal distributions I need some help with the following. I am given that $X_n\sim\mathcal{N}\left(0,\frac{1}{n}\right)$ i.e. $X_n$ has a normal distribution with mean $0$ and variance $1/n$. I need to show that $X_n$ converges in probability to $0$. I started by doing
$$\lim_{n\to\infty}P\left(\lvert X_n-0 \rvert\ge\epsilon\right)=\lim_{n\to\infty}P\left(X_n\ge\epsilon\right)$$
and I don't know where to go from here. I though that I could use the cdf of the normal but this doesn't seem correct.

By Markov's inequality, for each $\varepsilon>0$,
$$
\Pr\{|X_n|\ge\varepsilon\}\le\frac{\operatorname EX_n^2}{\varepsilon^2}=\frac1{\varepsilon^2n}\to0\quad\text{as}\quad n\to\infty.
$$
Hence, $X_n\to0$ in probability as $n\to\infty$
