Proof of Formula for Continuous Compounding

Copyright 1995 by Campbell R. Harvey and Stephen Gray. All rights reserved. No part of this lecture may be reproduced without the permission of the author.

Latest Revision: August 1996.

When there are n compounding periods per year, we saw that the effective annual interest rate is equal to (1+R/n)n - 1

We wish to show that if interest compounds continuously, then the effective annual interest rate is equal to eR - 1

We can prove this, if we can show that as there are more and more compounding periods per year, then the effective annual interest rate moves closer and closer to eR - 1

In the language of Calculus, this is the same as showing that:

This is equivalent to:

In the special case where R is zero, both sides of this equation are equal, so it is true. We will examine what happens when R is nonzero.

If we take the natural logarithm of the left hand side, we get:

You may recall, from Calculus, that the definition of a derivative is:

The expression on the right is called a difference quotient. The trick in our proof is to make our equation look like a difference quotient. In order to do this, we need two more tricks.

The first is to multiply and divide the right hand side of the equation by R. With some basic algebra, the equation becomes:

The second trick is to realize that ln(1) is equal to zero, so that if we subtract it from the top of the fraction, it will not change the value.

Our equation now looks like:

Note that if we let h = (R/n) and x = 1 then the bracketed expression on the right is just a difference quotient. Our equation becomes:

As n gets large, then h = R/n will go to zero.

This last equality holds, because (1/x) is the derivative of ln(x) and we set x = 1.

We are almost done. We have shown that as n gets large, the natural logarithm of (1+R/n)n goes to R. Since eln(x) = x, we can conclude that: