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I was doing some software engineering and wanted to have a thread do something in the background to basically just waste CPU time for a certain test. While I could have done something really boring like for(i < 10000000) j = 2 * i , I ended up having the program start with $1$, and then for a million steps choose a random real number r in the interval $[0,R]$ (uniformly distributed) and multiply the result by r at each step. When $R = 2$, it converged to $0$. When $R = 3$, it exploded to infinity. So of course, the question anyone with a modicum of curiosity would ask: for what $R$ do we have the transition. And then, I tried the first number between $2$ and $3$ that we would all think of, Euler's number $e$, and sure enough, this conjecture was right. Would love to see a proof of this.

Now when I should be working, I'm instead wondering about the behavior of this script. Ironically, rather than wasting my CPUs time, I'm wasting my own time. But it's a beautiful phenomenon. I don't regret it. :)

EDIT: I saw that you solved it yourself. Congrats! I'm posting this anyway because I was most of the way through typing it when your answer hit. :)

Infinite products are hard, in general; infinite sums are better, because we have lots of tools at our disposal for handling them. Fortunately, we can always turn a product into a sum via a logarithm.

Let $X_i sim operatornameUniform(0, r)$, and let $Y_n = prod_i=1^n X_i$. Note that $log(Y_n) = sum_i=1^n log(X_i)$. The eventual emergence of $e$ as important is already somewhat clear, even though we haven't really done anything yet.

The more useful formulation here is that $fraclog(Y_n)n = frac 1 n sum log(X_i)$, because we know from the Strong Law of Large Numbers that the right side converges almost surely to $mathbb E[log(X_i)]$. We have
$$mathbb E log(X_i) = int_0^r log(x) cdot frac 1 r , textrm d x = frac 1 r [x log(x) - x] bigg|_0^r = log(r) - 1.$$

If $r < e$, then $log(Y_n) / n to c < 0$, which implies that $log(Y_n) to -infty$, hence $Y_n to 0$. Similarly, if $r > e$, then $log(Y_n) / n to c > 0$, whence $Y_n to infty$. The fun case is: what happens when $r = e$?

I found the answer! One starts with the uniform distribution on $ [0,R] $. The natural logarithm pushes this distribution forward to a distribution on $ (-infty, ln(R) ] $ with density function given by $ p(y) = e^y / R, y in (-infty, ln(R)] $. The expected value of this distribution is $ int_-infty^ln(R)cfracy e^yR dy = ln(R) - 1 $. Solving for zero gives the answer to the riddle! Love it!