probability - Find the Mean for Non-Negative Integer-Valued Random Variable

Let $X$ be a non-negative integer-valued random variable with finite mean.
Show that
$$E(X)=\sum^\infty_{n=0}P(X>n)$$

This is the hint from my lecturer.

"Start with the definition $E(X)=\sum^\infty_{x=1}xP(X=x)$. Rewrite the series as double sum."

For my opinion. I think the double sum have the form of $\sum\sum f(x)$, but how to get this form? And how to continue?

Answer

\begin{array}
& & 0P(X=0) & + & 1P(X=1) & + & 2 P(X=2) & + & 3P(X=3) & + & \cdots \\[18pt]
= & & & P(X=1) & + & P(X=2) & + & P(X=3) & + & \cdots \\
& & & & + & P(X=2) & + & P(X=3) & + & \cdots \\
& & & & & & + & P(X=3) & + & \cdots\\
& & & & & & & & + & \cdots
\end{array}

The sum in the first row is $P(X>0)$; that in the second row is $P(X>1)$; that in the third row is $P(X>2)$, and so on.