I have to show that if $X \ge 0$ then $$\sum_{n=1}^\infty P(X \ge n) \le E[X] \le 1 + \sum_{n=1}^\infty P(X \ge n).$$
I know that if the random variable $X$ takes only values in $\mathbb N$, we have:
$$E[X] = \sum_{n=1}^\infty n P(X=n) = \sum_{n=1}^\infty \sum_{k=1}^j P(X=k) = \sum_{k=1}^\infty \sum_{n=k}^\infty P(X = n) = \sum_{n=1}^\infty P(X \ge n)$$
How can I use this idea to prove my inequality
$$\begin{align}E[X]&=\int XdP\\&=\sum_{n\geq0}\int_{n\leq X<n+1}XdP\\&\geq\sum_{n>0}\int_{n\leq X<n+1}ndP\\&=\sum_{n>0}nP(n\leq X<n+1)\\&=\sum_{n>0}P(X\geq n)\end{align}$$
The other side is similar.
$$\begin{align}E[X]&=\int XdP\\&=\sum_{n\geq0}\int_{n\leq X<n+1}XdP\\&\leq\sum_{n\geq 0}\int_{n\leq X<n+1}(n+1)dP\\&=\sum_{n>0}nP(n\leq X<n+1)+\sum_{n\geq0}P(n\leq X<n+1)\\&=1+\sum_{n>0}P(X\geq n)\end{align}$$