Assuming X is a discrete random variable over the natural numbers using the fact that $n=\sum^n_{k=1}1$ show $E[X] = \sum^\infty_{n=1}P[X\ge n]$
$E[X] = \sum_{n=1}^\infty P[X\ge \sum^n_{k=1}1]$ I'm not sure how to move out the summation in the inequality over to the left. Is this the right approach?
Note that
$$ E[X] = \sum_{n=1}^{\infty} n P[X = n] = \sum_{n=1}^{\infty} \left( \sum_{k=1}^n P[X = n] \right) = \sum_{k=1}^{\infty} \left( \sum_{n=k}^{\infty} P[X = n] \right) = \sum_{k=1}^{\infty} P[X \geq k] $$
where the change of order in the summation is justified by the fact that everything is non-negative.
In terms of characteristic functions: $$X = \sum_{n\geq 1} n 1_{X=n} = \sum_{n\geq 1} \sum_{k=1}^n 1_{X=n} = \sum_{k\geq 1}\sum_{n=k}^\infty 1_{X=n}= \sum_{k\geq 1} 1_{X\geq k}.$$ Now take expectation.
