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](https://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/69224#69224) .



A limiting product formula for the exponential function
$\prod\_{i=1}^n (1+a\_ir)=1+\sum\_{k=1}^n r^k\sum\_{i\_1 < \ldots < i\_k}a\_{i\_1}\ldots a\_{i\_k}$.


Notice that $1^k=\left(\sum\_{i=1}^n a\_i\right)^k =k!\sum\_{i\_1 < \ldots < i\_k}a\_{i\_1}\ldots a\_{i\_k}+\text{other terms}$, where the other terms are (positive) terms with a repeated $a\_i$. It follows that the $k$th term in the original sum is at most $1/k!$ so this gives the upper bound. 


To show the original claim it suffices to bound the terms with repeated $a\_i$'s in terms of $\delta=\max a\_i$. More specifically given $\epsilon>0$ and $k$ it suffices to show that for any finite sequence of positive numbers summing to 1 whose maximum is less than $\delta$, one has
$$\left(\sum a\_i\right)^k -k!\sum\_{a\_{i\_1} < \ldots < a\_{i\_k}}a\_{i\_1}\ldots a\_{i\_k} < \epsilon$$. 


This summation consists of terms of a number of "types" e.g. (2,1,1,1,1,1) represents terms in which one $a\_i$ occurs twice and 5 other $a\_i$'s occur once each; $(5,3,2,1,1)$ represents terms of the form $a\_{i\_1}^5a\_{i\_2}^3a\_{i\_3}^2a\_{i\_4}a\_{i\_5}$ where there is no longer a requirement that the $i\_j$ are increasing; instead the $i\_j$ should be increasing within the terms that have the same power.


For a fixed $k$, the number of types is finite. So that it suffices to show that for each type, the contribution goes to 0 uniformly as $\delta\to 0$. Clearly for the type $(p\_1,p\_2,\ldots,p\_r)$ you can bound the term $a\_{i\_1}^{p\_1}\ldots a\_{i\_r}^{p\_r}$ by $\delta^{\sum p\_i-r}a\_{i\_1}\ldots a\_{i\_r}$. Let $\Delta=\sum p\_i-r$ (this is at least 1 for all non-trivial types). The summation is then bounded above by $\delta^\Delta\sum\_{a\_{i\_j}\text{ distinct}}a\_{i\_1}\ldots a\_{i\_r}$ which is at most $\delta^\Delta$.


