If a sequence of random variables converges in probability to $0$, will the sequence still converge to $0$ when multiplied by $\sqrt n$

Question

Assume we have a sequence of random variables $X_1,X_2...$ such that the sequence converges to $0$ in probability. Under what general conditions will the sequence still converge in probability to $0$ when multiplied by $\sqrt n $?

Answer 1

The sequence $X_1, X_2, ...$ converging in probability to $0$ means that the following holds: \begin{align} \lim_{n \rightarrow \infty} P(|X_n - 0| > \varepsilon) = 0. \end{align} To check wether the sequence $\sqrt{n} X_1, \sqrt{n} X_2, ...$ converges in probability to $0$ you would have to check if the following holds: \begin{align} \lim_{n \rightarrow \infty} P(|\sqrt{n} X_n - 0| > \varepsilon) = 0. \end{align} Now observe that \begin{align} \lim_{n \rightarrow \infty} P(|\sqrt{n} X_n - 0| > \varepsilon) =& \lim_{n \rightarrow \infty} P(\sqrt{n} |X_n| > \varepsilon) \\ =& \lim_{n \rightarrow \infty} P(|X_n| > \frac{\varepsilon}{\sqrt{n}}) \end{align} Using Markovs inequality we can find an upper bound as \begin{align} \lim_{n \rightarrow \infty} P(|X_n| > \frac{\varepsilon}{\sqrt{n}}) \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n} E(|X_n|)}{\varepsilon} \end{align} So one general condition would be that the expectation of the sequence $|X_1|,|X_2|,...$ (mind the absolute value!) converges faster to zero than $\frac{1}{n}$. Note: You could also use a quadratic function in Markovs inequality and obtain a condition for the second moment.
Another possibility is to use that convergence in $d$ and convergence in $p$ coincide if the sequence converges to a constant. So if you have more knowledge about the cdf and are able to show that \begin{align} F_{\sqrt{n}X_n}(t) = P(\sqrt{n}X_n \leq t) = P(X_n \leq \frac{t}{\sqrt{n}}) \rightarrow \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases} \end{align} holds, you can also go this way.