A product for 1/e? This question is related to this question but I see that one part is really not a statistics question.
That $\lim_{n \to \infty} (1 - 1/n)^n = 1/e $ is clear. 
What is not clear to me is under what circumstances

$$\lim_{n \to \infty} \prod_{i=1}^n (1 - F_i) = 1/e. $$

Let me give some examples of candidates for $F_i$: 


*

*Subdivide the standard normal curve into (say) 100 subintervals from, say,  Z = -5 to Z = 5. With each subinterval associate a number $F_i$ equal to the area under the curve in that subinterval.  

*[removed]

*For $\frac{1}{2}\int_{-\pi/2}^{\pi/2}\cos x ~dx$ subdivide the interval $[ -\pi/2 < x < \pi/2]$ and find $F_i $ in a similar way. 
So either I am making a consistent mistake in calculating (not out of the question) or there is something general that I am missing. 
Can someone suggest why this is true, if it is, and how it generalizes? 
Thank you.
 A: I am not sure it helps in a statistical context, but I think the precise condition is that 
$\sum_{n=1}^{\infty} \frac{1}{n}\left( \sum_{i=1}^{\infty} F_{i}^{n} \right) = 1.$ You get this by taking the log, and looking at the Taylor series for $\log(1-x).$
In response to comment: The thing to note is that $\log(1 - x ) = -\sum_{k=1}^{\infty} \frac{x^{k}}{k}$ when $|x| < 1.$ Assuming each $F_{i}$ has absolute value less that $1,$ you require $-1 = \lim_{i \to \infty} \sum_{i=1}^{n} - \log(1-F_{i}) = 1.$ In other words, you want $\sum_{i=1}^{\infty} - \log(1-F_{i}) = 1$, assuming the left hand series converges. Hence you want $\sum_{i=1}^{\infty} \sum_{n=1}^{\infty} \frac{F_{i}^{n}}{n} =1$ assuming all relevant series converge. If $0 \leq F_{i} <1$ for each $i,$ and all partial sums $\sum_{i=1}^{M} \sum_{n=1}^{N} \frac{F_{i}^{n}}{n}$ are less then or equal to $1,$ then it is permissible to change the order of summation and in that case you require precisely that the original double sum converges to $1$ (in which case the order of summation does not matter).
A: There's nothing special about $1/e$, just use a product for $e^x$ function and plug $x=-1$ into the expression. A relevant topic that answers this particular question is
Exponential Function as an Infinite Product
