If $X\in L^2$, how do I show that $nP(|X|> \varepsilon \sqrt{n}) \to 0$? Consider a random variable $X$ with $E(X^2)<\infty$. How do I show that $nP(|X|> \varepsilon \sqrt{n}) \to 0$ for each $\varepsilon>0$? I don't even know where to start, and would like any hint.
As $X \in L^2$, $\{X\}$ is an absolutely continuous collection, wich means that there is an $N \in \mathbb{N}$ such that
$$\int_{|X|>N}X\,dP < \varepsilon$$. Where can I go from there?
 A: $$nP(|X|> \varepsilon \sqrt{n}) = \frac{1}{\epsilon^2}\cdot (\epsilon^2 n) \cdot P(X^2> \epsilon^2 n)\le \frac{1}{\epsilon^2}\int_{X^2\ge\epsilon^2n} X^2 dP\rightarrow 0$$
A: Unfortunatelly, Markov's inequality shows only that $nP(|X|>\varepsilon \sqrt{n}) \le \frac{EX^2}{\varepsilon^2}$, so we should use another way.
Lemma:  if $a_k \ge 0$, $a_k$ is nonincreasing and $\sum_{k \ge 1} a_k < \infty$ then $a_k = o(k^{-1})$.
Proof (sketch):
$$[\frac{n}2]a_n = \sum_{k = \frac{n}2}^n a_n \le \sum_{k = [\frac{n}2]}^n a_k \le \sum_{k \ge [\frac{n}2]} a_k \to 0, \quad n \to \infty.$$
Consider $\eta = \frac{X^2}{\varepsilon^2}$. Hence $E \eta < \infty$ and we want to prove that $P(\eta > n) = o(n^{-1})$. Put $\xi = [\eta]$. Then $E\xi \le E \eta < \infty$. As $\xi$ takes values in $\{0, 1, 2, \ldots, ... \}$ we have $E \xi = \sum_{k \ge 1} P(\xi \ge k) < \infty$. Hence by lemma we have $P(\xi \ge k) = o(k^{-1})$. Now it's sufficient to notice that $P(\xi > k)  = P([\eta] > k) \le  P(\eta > k) \le P([\eta] + 1 > k) = P([\eta] > k-1) = P(\xi > k-1)$.
As $P(\xi \ge k) = o(k^{-1})$ we have $P(\eta > k) = o(k^{-1})$, q.e.d.
