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 .

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    $\begingroup$ Apply the inequality between geometric and arithmetic mean to both the product and its reciprocal, that can be written in analogous form; then use the sandwich theorem and $(1+r/n)^n\to e^r$. $\endgroup$ – Pietro Majer Jul 1 '11 at 17:25
  • $\begingroup$ Just from the fact that $e^x$ has slope $1$ at $(0,1)$, for any $\epsilon > 0$ if $a$ is sufficiently small then $e^{(1 -\epsilon)a_ir} < 1 + a_ir < e^{(1 + \epsilon)a_ir}$ whenever $0 < a_i r <a $. So multiply these together. $\endgroup$ – Michael Greenblatt Jul 1 '11 at 17:34
  • $\begingroup$ I think the two suggestions above don't quite meet the demands of the question (that you should just expand and look at coefficients of separate powers). $\endgroup$ – gowers Jul 1 '11 at 20:28
  • $\begingroup$ @gowers Well, showing $e^{x + y} = e^xe^y$ from the power series is not too bad, as are the inequalities used above. But I agree I did take a few liberties on his suggested path. $\endgroup$ – Michael Greenblatt Jul 1 '11 at 20:50

$\prod_{i=1}^n (1+a_ir)=1+\sum_{k=1}^n r^k\sum_{i_1 < \ldots < i_k}a_{i_1}\ldots a_{i_k}$.

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 $\delta^\Delta$.

  • $\begingroup$ I've been away from my computer all day, but the above is essentially the argument I had in mind ... $\endgroup$ – gowers Jul 1 '11 at 20:27
  • $\begingroup$ Thanks! I wrote your argument in detail by induction below, but this was exactly the idea I was looking for. $\endgroup$ – Phil Isett Jul 2 '11 at 2:57

Thanks, Anthony, for finding this solution. I was completely at a loss for how to handle all the indices. If you don't mind, I would like to write down one version of the argument that you've given in full detail.

Claim: Under the hypotheses of the question $1 = k! \sum_{i_1 < \ldots < i_k} a_{i_1} \cdots a_{i_k} + O(\max |a_i|) $ where the error is non-negative.

The claim is true without an error when $k = 1$, and follows from induction. If we write $1 = (\sum a_i)^{k+1} = (\sum a_i) ( \sum a_i )^k$ The induction hypothesis allows us to write this product as $(\sum a_i)\cdot(k! \sum_{i_1 < \ldots < i_k} a_{i_1} + O(\max |a_i|) ) = (\sum a_i)\cdot (k! \sum_{i_1 < \ldots < i_k} a_{i_1} ) + O(\max |a_i| ) $

If we now distribute out the product, we get the term we want $(k+1)! \sum_{i_1 < \ldots < i_k < i_{k+1} } a_{i_1} \cdots a_{i_{k+1} }$ from the products with no repeats and then an error coming from products with exactly one term repeated. Take whichever term is repeated and bound one copy of it in absolute value by $\max |a_i|$. Then the error is bounded by $\max |a_i| ( \sum |a_i| )^k = O(\max |a_i|)$.

Having this claim established and looking slightly more carefully at the dependence of the error on $k$ (the constant in the big O only grows like $C^k$), we also have prove the convergence that I was looking for (and we don't need non-negativity of the terms; just that $\sum |a_i|$ is bounded). In the non-negative case we can just observe the error is non-negative, so that the dominated convergence theorem applies (with respect to the finite measure $\frac{|r|^k}{k!}$), giving a small shortcut and a soft way to see the convergence without a rate.

All credit goes to Anthony Quas for the idea; I just thought the induction was a fairly clear way to get the details all down.


I record this answer because I think that Pietro Majer's comment can be made into a solution which meets the proposer's criterion of potentially being able to be used to define $e^r$ ( I had been thinking along the same lines, although I am not sure there would be an advantage over the usual $\lim_{n \to \infty} (1 + \frac{r}{n})^{n}$ definition). If $r > 0$, then the GM-AM inequality gives $\prod_{i=1}^{n}( 1+a_{i}r) \leq ( 1 + \frac{r}{n})^{n}$. If $r > 0$ and each $a_{i} < \frac{1}{r}$, we obtain $\prod_{i=1}^{n}( 1-a_{i}r) \leq ( 1 - \frac{r}{n})^{n}$. Taking reciprocals in the second case and using standard approximations gives $\prod_{i=1}^{n}( 1+ a_{i}r + \frac{a_{i}^{2}r^{2}}{1-a_{i}r}) \geq ( 1 +\frac{r}{n})^{n}$. Choose $\varepsilon > 0$, and suppose that $a_{i} < \frac{\varepsilon}{2}$ for each $i$, and that, furthermore, $1 - a_{i}^{2}r^{2} > \frac{1}{2}$ for each $i$. Then we obtain $\prod_{i=1}^{n}( 1+ a_{i}r + \frac{a_{i}^{2}r^{2}}{1-a_{i}r} ) \leq \left[ \prod_{i=1}^{n}(1 + a_{i}r) \right]. (1 + \frac{r^{2}\varepsilon}{n})^{n}$, so that $\lim_{max(a_{i}) \to 0} \prod_{i=1}^{n} (1 + a_{i}r) \geq \lim_{n \to \infty} (1 + \frac{r}{n})^{n}$. A similar argument can be devised for $r < 0$.


Knowing $\lim_{N\rightarrow\infty} \left(1+\frac{r}{N}\right)^N=e^r$ (the case where all $a_i$ are equal) we can use continuity and $\left(1+\frac{r}{N}\right)^{Na_i}\sim 1+a_ir$ if $N$ is very huge and $a_i\rightarrow 0$.


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