There has been some discussion recently about the theorem
$$\lim_{n \rightarrow \infty} \prod_{i = 1}^n \left( 1 + \frac{x_i}{n} \right) = \exp \left( \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i = 1}^n x_i \right)$$
and the grounds of it. At the beginning of this discussion, it was stipulated that the set ${x_j}$ be bounded ... and I think this condition surely must ensure the holding of the theorem. But the question is this: what set ${x_j}$ might be devised that would foil this theorem? - a set such that it does not hold.
I think I might be getting to the heart of what's been bugging me about this. I saw the question, & I devised a rather laboured combinatorial argument to show that the binomial expansion of the LHS morphs into the Taylor expansion for exponential on the RHS, even when the $x$ is an infinite set, one element per bracket on the LHS instead of a single value in every bracket, and the arithmetic mean of the set is substituted for the single value of $x$ on the RHS; and then I saw the proofs involving taking the logarithm of both sides, & expressing the logarithm as a Taylor series on the LHS; and I thought "how silly of me using that clumsy combinatorial method!": but! ... there was something about these logarithmic proofs that I just kept noticing in the periphery of my vision - and it is is that none of them when the LHS is developed as the taylor series of a logarithm actually uses at any point the fact that it is particularly the logarithm - the series could have prettymuch any coefficients, and the reasoning would be exactly the same!
Anyway, now, to my mind, the solution is this: that what this theorem is ultimately essentially saying is that the exponential function is, essentially, the function that has the property that $\operatorname{f}(x+y)\equiv\operatorname{f}(x)\operatorname{f}(y)$, the property that $\operatorname{f}(0)=1$, and the property that $\operatorname{f^\prime}(0)=1$. Of course we know that the exponential function has these properties: what I am saying is that the theorem that is the subject of this post (and of the other post mentioned) is saying simply that ... just simply that.
You've got the sum $$\sum_{j=1}^n{x_j\over n}$$ and if this sum is finite in the limit as $n\rightarrow\infty$ you can forget about the higher degree terms, and it doesn't matter whether it's $\operatorname{log}(1+x)$ or $x/(1-x)$ or whatever it is as long as it goes through the origin, because by reason of each term in the sum having an $n$ that's going to infinity in the denominator its only the first derivative that counts anymore; and when it's exponentiated, by reason of the second & third properties of the exponential cited above, each term becomes $$\left(1+{x_j\over n}\right) ,$$ and by reason of the first property all these brackets become multiplied together.
Whence you arrive at the theorem that is the subject of this post ... & of that other.