Upper bound of tail of probability distribution assuming only finite first moment. Let $X$ be a random variable with $\mathbb E{X}<\infty$. By Markov's inequality we have
$$
\mathbb P(X > n)\le \mathbb E(X)/n = O(1/n).
$$
I sort of remembering I saw somewhere $\mathbb E{X}<\infty$ actually implies that
$$
\mathbb P(X > n) = o(1/n).
$$
I cannot remember where I saw it and I also cannot find a counter example. Is this actually true?

Suppose $X$ is non-negative and $EX<\infty$. Then $xP(X>x)\to 0$ as $x\to \infty$. Indeed
first note that $$xI(X>x)\leq XI(X>x)$$
where $I$ is the indicator function. Taking expectations yields that
$$
xP(X>x)\leq EXI(X>x)\to 0
$$
by the dominated convergence theorem.
