rigorous proof that : $E[X] = \sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k)$ Let $X$ be a random variable : $\mathbb{N} \to \mathbb{N}$. Then if $X$ has an expectation we have : 
  $$E[X] = \sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k)$$


It's quite logical that its true. The problem is that I don't know how to prove this result rigourously just by manipulating sums. For example I can do the following : $\sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k) = \sum_{k \in \mathbb{N}} \sum_{i = k}^\infty \mathbb{P}(X = i)$. It would be nice if I can isolate the $\mathbb{P}(X = i)$ for single $i$, but I don't know how to do this. Maybe I should introduce indicator functions somehow ? and then use Fubini to interchange the summations ? 

Thank you !
For non-negative numbers $a_{nm}$ we always have $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} a_{nm}=\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} a_{nm}$ even without the assumption of convergence. Fubini's Theorem is not required for this. You can just use partial sums and take limits to show that each side is less than or equal to the other. Apply this result with $a_{nm}=P(X= n)$ for $n \geq m$ and $0$ for $n&lt;m$ to get the expression for $EX$.
Here is how you can "discover" the desired result. Begin with
$$P(X\geq 1)=P(X=1)+P(X\geq 2)$$
Add $P(X\geq 2)$ to both sides:
$$P(X\geq 1)+P(X\geq 2)=P(X=1)+2P(X\geq 2)=P(X=1)+2P(X=2)+2P(X\geq 3)$$
in the r.h.s equality the fact that $X$ takes only integer values is used.
Add $P(X\geq 3)$ to both sides:
$$\sum_{k=1}^3P(X\geq k)=\sum_{k=1}^3kP(X=k)+3P(X\geq 4)$$
And by induction, you get that for every natural number $n$
$$\sum_{k=1}^nP(X\geq k)=\sum_{k=1}^nkP(X=k)+nP(X\geq n+1)\quad (*)$$
So now your problem is to prove that 
$$\lim_{n\to\infty}nP(X\geq n+1)=0$$
But you are given that the expectation $E(X)$ exists, which by definition means that the series
$$\sum_{k=1}^{\infty}kP(X=k)$$
is convergent, which implies that $nP(X=n)$ tends to zero as $n\to\infty$. Since
$$nP(X\geq n+1)\leq nP(X\geq n)$$
we deduce that $nP(X\geq n+1)$ tends to zero as well, and letting $n\to\infty$ in (*) gives the desired result.