By the way, does anyone know how to prove in an elementary way (i.e. expanding) that $\prod_1^n (1+a_i r)$ tends to $e^r=\sum \frac{r^k}{k!}$ as you let $\max|a_i|\to 0$ with $0\leq a_i \leq 1$ and $\sum a_i = 1$? An easy solution goes by writing the product with the exponential function so that you get the exponential of $\sum \log(1+a_i r) = \sum \int_0^1 \frac{a_i r}{(1+s a_i r)} ds$.

You can then integrate by parts (i.e. Taylor expand) to obtain $\sum a_ i r − \sum \int_0^1 (1−s)\frac{(a_i r)2}{(1+s a_i r)2}ds$. Now, $\sum a_i r = r$ is the main term. After you take $\max|a_i|$ to be less than $.5/|r|$, the error term is bounded in absolute value by $C \sum |a_i r|^2 \leq \max|a_i|\cdot \sum |a_i| |r|^2 \leq C |r|^2 \max |a_i|$.

I was hoping to find an elementary proof of this convergence by expanding the product $\prod_1^n (1+a_i r)$ and gathering terms with a common power of $r$. In particular, it would be nice to prove the convergence of this limit without the exponential function, since then the limit could be considered a definition of $e^r$. The case when all of the $a_i$ are equal is done in Rudin's "Principles of Mathematical Analysis".

The motivation for this problem comes from compound interest, which I described in a different thread here: Generalizing a problem to make it easier .