Concretize this geometric interpretation of the expected number of trials for a certain result. I came across André Nicolas' answer to this question: Determining Convergence of Power Series
I have this mental visual representation of what expected value (in this case the expected number of rolls of a die before getting a 6, for ex) looks like, and it's something like this:

The 1 rectangle is $\frac{1}{6}$ of the total area (the probability of getting a six in one throw), the 2 rectangle is $\frac{1}{6}$ of $\frac{5}{6}$ of the total area, etc...without bounds, meaning that the subdivisions keep going forever.
If we assume that the total area (of the sample space) is 1, then the expected number of throws before getting a six is going to be the sum of [each subrectangle's area times that subrectangle's value].
What I'm trying to do is reconcile this geometric/visual model with the formula $$ E(X) = P(success) \cdot 1 + P(failure) \cdot (1 + E(X))$$
The formula does make sense, don't get me wrong, but only superficially (for me); it's not crystal clear. At some point in the algebra I go "hmm...why does this make sense again?"
I guess what I'm asking is whether there's an obvious way to see for example that the sum of each rectangle times its area (minus rectangle 1), is $\frac{5}{6} (1 + E(X))$
 A: Okay.  You have 
$$\begin{array}{|r|l:l|} \mathsf P(X=x) & \tfrac 16 & \tfrac 5{6^2} & \tfrac {5^2}{6^3} & \tfrac {5^3}{6^4} & \cdots\\ \hline x & 1 & 2 & 3 &4 & \cdots \end{array}$$
Which is clearly
$$\begin{array}{|r|l:l|} \mathsf P(X=x) & \tfrac 1{6} & \tfrac 5{6}\cdot\tfrac 1{6} & \tfrac 5{6}\cdot\tfrac 5{6^2} & \tfrac 5{6}\cdot\tfrac {5^2}{6^3} & \cdots\\ \hline x & 1+0 & 1+1 & 1+2 & 1+3 & \cdots \end{array}$$
Or as Rahul put it, succinctly:

denote it $[1|2|3|⋯]$. Then $[1|2|3|⋯]=[1|1|1|⋯]+[0|1|2|⋯]$. The value of $[1|1|1|⋯]$ is just $1$, while $[0|1|2|⋯]$ is basically the original diagram shrunk to $\frac 56$-th of the area, so $[1|2|3|⋯]=1+\tfrac 56 [1|2|3|⋯]$
  .

So you should be able to visualise the following:
Since $\mathsf P(X=x)= \tfrac 56\mathsf P(X=x-1)$, for $x>1$ then:
$$\begin{align}\mathsf E(X) ~&=~ \sum_{x=1}^\infty x\,\mathsf P(X=x)\\[1ex] &=~ \mathsf P(X=1)+\sum_{x=2}^\infty x\,\mathsf P(X=x)\\[1ex] &=~\sum_{x=1}^\infty \mathsf P(X=x) + \sum_{x=2}^\infty (x-1)\,\tfrac 56\,\mathsf P(X=x-1) \\[1ex] &=~1 + \tfrac 56\sum_{x=2}^\infty (x-1)\,\mathsf P(X=x-1) \\[1ex] &=~1 + \tfrac 56\sum_{y=1}^\infty y\,\mathsf P(X=y)\\[1ex] &=~ 1+\tfrac 56\mathsf E(X)\\[2ex]\therefore~\mathsf E(X) ~&=~6\end{align}$$
