Applications of Series — Question 9

*Question 9 A principal $P$ earns annual rate $r$ compounded $n$ times per year, so $A_n=P(1+r/n)^n$. Use the series for $\log(1+u)$ to derive the continuous-compounding limit as $n\to\infty$.

*Question 9 – Solution

For sufficiently large $n$, $1+r/n>0$, and $$\log\left(\frac{A_n}{P}\right)=n\log\left(1+\frac rn\right).$$

$$\begin{align*} n\log\left(1+\frac rn\right)&=n\left(\frac rn-\frac{r^2}{2n^2}+\frac{r^3}{3n^3}+O(n^{-4})\right)\\&=r-\frac{r^2}{2n}+\frac{r^3}{3n^2}+O(n^{-3})\longrightarrow r. \end{align*}$$

Continuity of the exponential function gives $$\frac{A_n}{P}\longrightarrow e^r,\qquad\boxed{A_n\longrightarrow Pe^r}.$$ The correction $-r^2/(2n)$ also explains the approach rate. For positive $r$, discrete compounding approaches the continuous value from below.