Is there a way to visualise $ \sum_{k \geq 0} \mathbb{P}(X>k) = \mathbb{E} [X] $? Let X be a random variable taking values in N
$ \sum_{k \geq 0} \mathbb{P} (X>k) \dots (0)$
$= \sum_{k \geq 0} \sum_{i \geq k+1} \mathbb{P}(X=i) \dots (1)$ 
$= \sum_{i \geq 1} \sum_{k=0}^{i-1}\mathbb{P}(X=i) \dots (2) $
$= \sum_{i \geq1} i \mathbb{P}(X=i) \dots (3) $
$= \mathbb{E} [X]  $
I am having a problem with the double sum, I don't see how to think of coming up with this, or how do we go from (0) to (2) , the way I reason by default is like :
$\mathbb{E} [X] = \sum_{k \in N} k \mathbb{P}(X=k) = 1 P(1) + 2 P(2) + ... $
then I block
 A: I'm not sure if this is "visual" in the sense that you want, but I think the following is a nice illustration. As you note, we have $\mathbb{E}[X]=\sum_{n=1}^{\infty}n\mathbb{P}(X=n)$. Think of this as summing all the terms in the following picture:
$$
\begin{matrix}
\mathbb{P}(X=1)&&\\
\mathbb{P}(X=2)&\mathbb{P}(X=2)&\\
\mathbb{P}(X=3)&\mathbb{P}(X=3)&\mathbb{P}(X=3)\\
\vdots&&&\ddots
\end{matrix}
$$
Now, if you sum these terms row by row, you get $\sum_{n=1}^{\infty}n\mathbb{P}(X=n)$. If you instead sum column by column, you get $\sum_{n=1}^{\infty}\mathbb{P}(X\ge n)$.
You can picture any exchange of double sums in a similar manner. The mathematical justification for this and much more general types of exchanges is given by Fubini's and/or Tonelli's theorem.
A: 0->1
What is P(X>K)?  For example P(X>3)=P(4)+P(5)+P(6)........... 
or P(X>k)=$\sum_{r=k+1}^{\infty} P(r) $.Now can you viisualize how we moved from 0->1

1->2
How many times do we count P(3), for P(X>0),P(X>1) and P(X>2).(3 times)
How many times do we count P(k),for P(X>0),P(X>1),P(X>2)..........P(X>(k-2)) and P(X>(K-1)) ,hence k times.
Now can you visualize 1->2
A: $$\sum_{k\ge 0}\mathbb{P}(X>k)$$
$$=\sum_{k>0}\mathbb{P}(X\ge k)$$
$$=\mathbb{P}(X\ge 1)+\mathbb{P}(X\ge 2)+\mathbb{P}(X\ge 3)+\dots$$
$$=\mathbb{P}(X=1)+2\mathbb{P}(X\ge 2)+\mathbb{P}(X\ge 3)+\dots$$
$$=\mathbb{P}(X=1)+2\mathbb{P}(X=2)+3\mathbb{P}(X\ge 3)+\dots$$
$$\vdots$$
$$=\mathbb{P}(X=1)+2\mathbb{P}(X=2)+3\mathbb{P}(X=3)+\dots+k\mathbb{P}(x=k)+\dots$$
Edit:
$$\mathbb{P}(X\ge 1)=\mathbb{P}(X=1)+\mathbb{P}(X\ge 2)$$
