I need to prove that for a random variable $X$ with $$ X \geqslant 0 $$ it's true that $$ \sum_{n=1}^{\infty} P(X \geqslant n) \leqslant E[X] \leqslant 1 + \sum_{n=1}^{\infty} P(X \geqslant n) $$
I proved that
$$
E[X] = \int_{0}^{\infty} P(X \geqslant t) dt
$$
so I have the first part of the inequality proved by comparing the integral with the sum. I'm not sure how to go about proving the other part, i.e.
$$
E[X] \leqslant 1 + \sum_{n=1}^{\infty} P(X \geqslant n)
$$
I was trying to move some things around and reason why $\int_{0}^{\infty} P(X \geqslant t) dt - \sum_{n=1}^{\infty} P(X \geqslant n) \leqslant 1$ but I didn't get far.
I was thinking of (somehow) using the fact that the integral of the pdf of $X$ is bounded by 1 but no luck. Any suggestions for how to get started on this?
Split the integral up and estimate the probability $P(X\ge t)$ for $t\in[n,n+1]$ from above by $P(X\ge n)$.
\begin{align*} EX &= \int_0^1P(X\ge t)\,dt + \int_1^\infty P(X\ge t)\,dt\\ &\le 1 + \sum_{n=1}^\infty\int_{n}^{n+1}P(X\ge t)\,dt \\ &\le 1 + \sum_{n=1}^\infty\int_n^{n+1}P(X\ge n)\,dt \\ &= 1 + \sum_{n=1}^\infty P(X\ge n) \end{align*}