Does the convergence in probability imply the following limit is $1?$

Question

Let $X_n \in \mathbb{R}$ be a sequence of non-constant random variable with continuous PDF converging in probability to $c,$ but not necessarily convergence almost surely, i.e.

$$\lim\limits_{n \to \infty}P[|X_n - c| \le \epsilon]=1 \forall \epsilon > 0.$$

What can we then say about:

$$\lim\limits_{n \to \infty}P[|X_n - c| \le \frac{1}{n}]?$$

In general, choose a sequence $s_n \to 0$ as $n \to \infty.$ Then what can we then say about:

$$\lim\limits_{n \to \infty}P[|X_n - c| \le s_n]?$$

Is it always $1?$ If not, can we have a counter example?

Answer 1

Take $P(X_n = 1) = 1- P(X_n = \frac{2}{n}) = p_n$ with $(p_n)\subset (0,1)$ your favorite sequence satisfying $\lim_{n\to \infty} p_n = 0$. Then $X_n \to 0$ in probability but $P(|X_n|\leq \frac{1}{n}) = 0$.

Edit: if you want the $X_n$ to have a density, let $X \sim \mathcal{U}([0,1])$ be uniformly distributed and set $X_n = \frac{2}{n}X$. Then $X_n \to 0$ in probability (and a.s.) but $$P\left(|X_n|\leq \frac{1}{n}\right) = P\left(X\leq \frac{1}{2}\right) = \frac{1}{2}.$$