Show that $E(X)=\sum_{n\in\mathbb{N}}P(X\ge n)$

Question

Let $X$ be a discrete random variable over $\mathbb{N}$, $E$ its expected value.

I want to show that $\displaystyle E(X)=\sum_{n\in\mathbb{N}}P(X\ge n)$

I thought of an "Euler sum" approach :

$\displaystyle E(X)=\sum_{n\in\mathbb{N}}nP(X=n)=P(1)+2P(2)+3P(3)+\cdots=\\ [P(1)+P(2)+P(3)+\cdots]+[P(2)+2P(3)+\cdots]+\dots=[P(1)+P(2)+P(3)+\cdots]\\ +[P(2)+P(3)+\cdots]+[P(3)+\cdots]+\cdots=\sum_{n\in\mathbb{N}}P(X\ge n)$

but that approach seems "too easy" and not rigorous enough to me (I do know that changing the order of the terms in the expected value's sum does not affect its final value, though).

What would be a rigorous way to prove it ?

Answer 1

$\displaystyle \sum_{n\geq 1}P(X\geq n)=\sum_{n\geq 1}\sum_{k\geq n}P(X=k)=\sum_{k\geq 1}\sum_{n\leq k}P(X=k)=\sum_{k\geq 1}kP(X=k)=E(X)$