Equivalent rate with compounding interest to a continuously compounding rate for a given period

Question

I'm reading a book called An Introduction to Quantitative Finance and in it the author notes the following:

---

Result Suppose the continuously compounded rate for period T is r. Then the equivalent rate r_m with compounding frequency m is

$$ r_{m} = m(e^{(r/m)}-1). $$

Proof The result follows immediately by noting that

$$ e^{rT}= (1 + r_{m}/m)^{mT}. $$

---

This is confusing to me because I thought that $e^{rT}$ is the limit of $(1 + r_{m}/m)^{mT}$ as m approaches infinity, not equivalent to it.

Any help would be appreciated!

Answer 1

Here you have a unit of time (month, year, for example). Call the unit of time the base period.

• The continuously compounded rate $r$ yields, over a period of length $T$ units of time, $e^{rT}$ per unit of money.
• The base period rate $r_1$ yields $(1+r_1)^T$ in the same period of length $T$
• The base period rate that compounds every $1/m$-th of the base period yields, over a period of length $T$, $(1+\tfrac{r_m}{m})^{Tm}$ units of money.

If these rates are to be equivalent (whether the investent is continuous, every unit of time or every $1/m$-th of a unit of time) then $$e^{rT}=(1+r_1)^T=(1+\tfrac{r_m}{m})^{mT}$$

The sequence $r_m$ also satisfies $r_m\xrightarrow{m\rightarrow\infty}r$. Indeed, $$r_m=m(e^{r/m}-1)=r\frac{e^{\tfrac{r}{m}}-1}{\tfrac{r}{m}}\xrightarrow{m\rightarrow\infty}r$$ since $\lim_{h\rightarrow0}\frac{e^h-1}{h}$ is the value of the derivative at $t=0$ of the function $f(t)=e^t$.

The convexity of the exponential function implies in fact that $r_m$ is monotone decreasing.