# Proving expectation to be infinite with an inequality

Let $$X$$ be a non-negative random variable, and suppose that $$P(X \geq n) \geq 1/n$$ for each $$n \in \mathbb{N}$$. Prove that $$E(X) = \infty$$.

I have been stuck with this problem for a few days now. I guess it can make some sense intuitively because you have some probability mass everywhere, and we're looking at probability of it being greater than some value. I tried to use inequalities like Markov's and Chebyshev's with no luck. I was hoping if someone can please explain to me how to answer this problem. It is coming from an introductory probability with measure theory book, and I am trying my best to get better at these kind of problems.

Suppose we have a discrete random variable $$Y$$ with $$\mathbb P(Y=n) = \frac{1}{n(n+1)}$$ for all positive integers $$n$$

then $$\mathbb P(Y\ge n) = \frac1n \le \mathbb P(X \ge n)$$ for all positive integers $$n$$

and $$\mathbb P(Y\ge x) \le \mathbb P(X \ge x)$$ for all real $$x$$

so $$X$$ has weak first-order stochastic dominance over $$Y$$

making $$\mathbb E[Y]\le \mathbb E[X]$$

but $$\mathbb E[Y] = \sum\limits_{n=1}^\infty \frac{1}{n+1}= +\infty$$

showing $$\mathbb E[X] = +\infty$$

I am giving a proof for the continuous case. Integrating by parts in the definition of expected value and observing that $$P(X<-t)=0$$ for any $$t>0$$ you have $$\mathbb{E}[X]=\int\limits_0^{\infty}P(X>t)\,dt\ge \int\limits_0^{\infty}\frac{1}{t+1}\,dt=\infty$$ (since $$P(x>t)\ge P(x>{\rm{ceiling\ of}}\, t)$$ and $${\rm{ceiling\ of}}\, t\le t+1$$