Real random variable $X$ such that $ \lim_{n\to \infty}{nP(|X|>n)}=0$ what about $ E|X|<\infty$? Let $X\colon (\Omega,\mathcal F,\mathbb P)\to (\mathbb R,\mathcal B(\mathbb R))$ be a random variable from a probability space to the real numbers with the Borel sets.
I proved that if $\mathbb  E|X|<\infty$ then $ \lim_{n\to \infty}{n\mathbb P(|X|>n)}=0$.
I want to know if the reciprocal is true or not. If not, I'd like to know if there are some extra hypothesis that can assure me that it's true.
 A: The short answer is NO.
There exists a probability distribution on $\mathbb{R}^+$ such that $\lim_{n \to \infty} n \mathbb{P}[X > n] = 0$ but $\mathbb{E}[X] = \infty$.
(The basic observation is simply the fact that $\int_{n_0}^{\infty} \frac{1}{x \ln(x)}$ is divergent.)
Define the distribution such that $\mathbb{P}[X \leq n_0] = 0$ for some $n_0  > 0$, and $\mathbb{P}[X > n] = \frac{C}{n \ln(n)}$ for all $n > n_0$.
For such $\mathbb{P}$ it holds that $\mathbb{E}[X] = \int_{n_0}^{\infty} \mathbb{P}[X > n] dn = \int_{n_0}^{\infty}  \frac{C}{n \ln(n)} = \ln\ln(n)|_{n_0}^\infty = \infty$.
A: In general, you cannot do better than: $\Bbb E |X| < \infty$ if and only if $\sum P(|X| > n) < \infty$. The proof is that these two quantities are actually equal,
$$
\Bbb E(|X|) = \sum_n \Bbb P(|X| > n).
$$


*

*Consequence: if there is some $\alpha > 1$ such that $\lim_{n \to \infty} n^\alpha\, \Bbb P(|X| > n) = 0$, then $\Bbb E|X| < \infty$.

*It is easy to find counter-examples if $\alpha = 1$. For instance, consider a random variable $X$  with distribution
$$
\Bbb P(X = n) = \frac{2\ln 2}{n\ln n} - \frac{2\ln 2}{(n+1)\ln(n+1)}, \qquad n \geq 2
$$
and conclude with the fact that $\sum_{n=2}^\infty \frac{1}{n\ln n} = +\infty$.
