A useful fact: as \(n \rightarrow \infty\), \[(1+r/n)^n \rightarrow e^r\]
- We say that an annual rate of \(r\) compounded continuously produces an effective annual rate of \(e^r-1\).
- For example, 12% compounded continuously implies an effective annual rate of \(e^{0.12}-1 = 0.12750\).
- Given an effective annual rate of \(r_a\), the corresponding continuously compounded rate is \(\log (1+r_a)\).