Show that if $X\ge 0$ , $E(X)\le \sum_{n=0}^{\infty}P(X>n)$ If $X$ is a random variable and also let $X\ge 0$. 
I want to show $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.
 A: You don't specify anything about $X$. Is it the general case/what is its support? 
If it is for discrete r.v.s, can you say something about the relationship between 
$\sum_{n = 0}^{\infty} n P(X = n)$  and $\sum_{n = 0}^{\infty} P(X > n)$?
Consider:
\begin{eqnarray}
&0 P(0)& +\, 1& P(1)& +\, 2 &P(2)& +\, 3 &P(3)& +& ...&\\
& &\\
&= &  [ &P(1)& +&P(2)& +&P(3)& +& ...&]\\
& &  [   & & +&P(2)& +&P(3)& + &...&]\\
& &  [   & &  &    & +&P(3)& + &...&]\\
& &  [& & & & & & &...&]
\end{eqnarray}
Can you see a way to do it now? (Though I believe this approach establishes a stronger result than you have)
To consider it more generally than the discrete case, see here, and then adapt the above idea. 
That is, can you see how establish a relationship between $∑^∞_{n=0}P(X>n)$ and a similar-looking integral that would the give the required inequality?
A: Define the sets $A_n=\{x\in \mathbb{R}:x>n\}$, for $n=0,1,2\dots$. 
For any fixed $\omega$, let $n_0$ be the smallest integer such that $X(\omega)\leq n_0$. Since $X(\omega)\geq 0$, we have
$$
  X(\omega)\leq n_0 = \sum_{n=0}^{n_0} I_{A_n} (X(\omega)) = \sum_{n=0}^\infty I_{A_n} (X(\omega)) \, ,
$$
yielding
$$
  \mathrm{E}[X]\leq \sum_{n=0}^\infty \mathrm{E}[I_{A_n} (X)]=\sum_{n=0}^\infty P(X>n) \, .
$$
A: You have
$$EX=\int_0^{\infty}xdF(x)$$
Notice that $dF(x)=-d(1-F(x))$ and that $P(X>t)=1-F(t)$ and use integration by parts. 
Now show that for monotone decreasing positive function 
$$\sum_{n=0}^\infty f(n)\ge\int_0^{\infty} f(t) dt$$
Combine these two results and you get your desired result. Hint for the second, recall Riemman sums.
