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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:

enter image description here

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 $\frac{5}{6} (1 + E(X))$

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    Let's flip your rectangle left-to-right and denote it $[1|2|3|\cdots]$. Then $[1|2|3|\cdots] = [1|1|1|\cdots] + [0|1|2|\cdots]$. The value of $[1|1|1|\cdots]$ is just $1$, while $[0|1|2|\cdots]$ is basically the original diagram shrunk to $\frac56$th of the area, so $[1|2|3|\cdots] = 1 + \frac56[1|2|3|\cdots]$. – 2017-01-27
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    @Rahul Splendid. That should be posted as an answer. – 2017-01-27
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    @Rahul, wow, mind blown. I can't believe you had time to read my question, understand it and come up with that on the spot. Thank you, that's a perfect example of the kind of stuff I was looking for. And yes, I agree with Graham; you should post it if you have time, I think (for what that's worth) that it's brilliant. – 2017-01-27
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    @Graham: If I posted an answer I would feel obligated to include a nice diagram instead of the above textual encoding, and I don't have time to do that. If anyone else wants to do so they have my full encouragement. – 2017-01-27

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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:

$$\begin{align}\mathsf E(X) ~&=~ \sum_{x=1}^\infty x\,\mathsf P(X=x)\\[1ex] &=~ \mathsf P(X=1)+\sum_{x=2}^\infty x\,\mathsf P(X=x)\\[1ex] &=~\sum_{x=1}^\infty \mathsf P(X=x) + \sum_{x=2}^\infty (x-1)\,\tfrac 56\,\mathsf P(X=x-1) \\[1ex] &=~1 + \tfrac 56\sum_{x=2}^\infty (x-1)\,\mathsf P(X=x-1) \\[1ex] &=~1 + \tfrac 56\sum_{y=1}^\infty y\,\mathsf P(X=y)\\[1ex] &=~ 1+\tfrac 56\mathsf E(X)\\[2ex]\therefore~\mathsf E(X) ~&=~6\end{align}$$

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    Thank you for coming back to write this answer and formalize Rahul's comment; this is exactly the kind of insight I was looking for. I'm very happy it can make sense using this visual model because I see it in my mind all the time and it really helps me a lot when thinking about probability. For some reason the P(success) * 1 + P(failure) * (1 + E(X)) wasn't immediately obvious, but now I can see it clearly. – 2017-01-27