I read something like this:
Let $X$ be a non-negative random variable. If $\mathbb E X^2 < \infty$, then for every fixed $\delta > 0$, $\mathbb P(X \ge \delta \sqrt{n}) = o(1/n)$. It follows that there exists $\delta_n \to 0$ such that $$\mathbb P(X \ge \delta_n \sqrt{n}) = o(1/n).$$
I know how to prove the first part:
Let $Y_m = m^2 \mathbb 1_{[X \ge m]}$. Then $Y_n \to 0$ a.s. Since $Y_m \le X^2$, by dominated converge theorem, $\mathbb E Y_m \to 0$. In other words $$\mathbb P(X \ge m) \times m^2 \to 0.$$
But I cannot figure out how "it follows...". Any clue?
There exists $N_1$ such that $n\mathbb P(X\ge\frac 1 2\sqrt n) < 1/2$ for all $n\ge N_1$. Now, there exists $N_2 > N_1$ such that $n\mathbb P(X\ge \frac 1 4\sqrt n) < 1/4$ for all $n\ge N_2$. Going on, there exists $N_3 > N_2$ such that $n\mathbb P(X\ge\frac 1{16}\sqrt n) < 1/16$ for all $n\ge N_3$. So, we get an increasing sequence $(N_k)$ such that $$ n\mathbb P\left(X\ge \frac 1 {2^k}\sqrt n\right) < \frac 1 {2^k} $$ for all $N_{k-1}\le n < N_k$. Now, put $\delta_n := 2^{-k}$ if $N_{k-1}\le n < N_k$. Then $$ n\mathbb P\left(X\ge \delta_n\sqrt n\right) < \delta_n $$ for all $n$.