Random multiplication and Euler's number I was toying with random number generators and wrote a program to begin with a random number between 0 and a, and multiply term n by a random number between 0 and a to attain term n+1 (using a uniform distribution to determine the random numbers).  Through experiment up to 500,000 terms, it appears that when a is below e, as n approaches infinity, term n will almost surely approach 0, and when a is above e, term n will almost surely approach infinity.  Is this correct?  If so, I would be interested in seeing a proof.
 A: I think that you're right.
Let $X_i \sim U(0, a)$, and let $M_n = \prod_{i=1}^n X_i$. We can write
$$M_n = a^n \prod_{i=1}^n \frac{X_i}{a}$$
which implies that
\begin{align*}
 -\log(M_n) &= -n \log(a) + \sum_{i=1}^{n} \left[-\log \left(\frac{X_i}{a} \right)  \right] \\
&=n \left(-\log(a) + \frac{\sum_{i=1}^n \left[-\log \left(\frac{X_i}{a}  \right)  \right]}{n}   \right)
\end{align*}
or that
$$\frac{-\log(M_n)}{n} = -\log(a) + \frac{\sum_{i=1}^n \left[-\log \left(\frac{X_i}{a}  \right)  \right]}{n}.$$
We note that $\frac{X_i}{a} \sim U(0, 1)$, whence $-\log(X_i/a) \sim \operatorname{Exp}(1)$ (see here for a proof, for instance). Thus, the strong law of large numbers tells us that the fraction on the far right converges to $1$ almost surely, and hence, 
$$\frac{-\log(M_n)}{n} \to 1 - \log(a)$$
with probability 1. From here, we can break this into cases based on what $\log(a)$ is. If $\log(a) > 1$ ${(\Leftrightarrow a > e)}$, then the expression on the right is negative, which implies that $\log(M_n)$ diverges to $\infty$ almost surely, and therefore $M_n \to \infty$. Conversely, if $\log(a) < 1$, then $\log(M_n)$ must diverge to $-\infty$, and therefore $M_n \to 0$ almost surely.
