# $\lim_{m \to \infty}{\prod_{i=0}^{tm-1}(1+\frac{r(i)}{m})} = e^{\int_{0}^{t} r(s)ds}$?

How can I deduce the following equation ?:

$$\lim_{m \to \infty}\prod_{i = 0}^{{\large tm - 1}} \left[\,1 + \frac{r\left(\,i\,\right)}{m}\,\right] = \exp\left(\,\int_{0}^{t}r\left(\,s\,\right)\,\mathrm{d}s\,\right)$$

where $t > 0$ and $r\left(\,i\,\right)$ a real valued function.

I can figure out the series expansion of $\mathrm{e}$, and it makes somehow sense as you can think of $r$ as the average "rate of interest" for example, but what exactly is the math behind it ?.

• It's generally a bad idea to put only mathjax in the title, because your post won't show up in any searches and it's not very descriptive of the question. See how to choose a good title :) – Dando18 Jun 21 '17 at 14:04
• Will keep this in mind going forward – Hekri Jun 21 '17 at 14:32

There is a suspected typo in the OP. The argument of $r$ should be $i/m$, not $i$ alone.
$$\log\left(\prod_{i=0}^{\lfloor tm-1\rfloor} \left(1+\frac{r(i/m)}{m}\right)\right)=\sum_{i=0}^{\lfloor tm-1\rfloor} \log\left(1+\frac{r(i/m)}{m}\right)$$
Now use the fact that $\log\left(1+\frac{r(i/m)}{m}\right)=\frac{r(i/m)}{m}+O\left(\frac{r(i/m)}{m}\right)^2$ as $m\to \infty$.