We continue the discussion that was started in the previous section: Simple versus Compound Interest.
In a nutshell:
You invest some money (the principal ) in a bank which pays interest.
After a specified time (the compounding period ), you add in the interest that has been earned on the principal.
After another compounding period, you add in the interest earned on the principal plus accumulated interest.
Repeat.
Compound interest, by definition, is interest calculated on the principal amount together with accumulated interest.
Interest can be added in at different fixed intervals: annually, monthly, weekly, daily, and so on.
The following are equivalent:
Adding in interest more and more often produces a slightly better accumulation. (Most banks add in interest daily.)
What happens as you increase the number of times that interest is added in each year?
Your earnings will get bigger. How much bigger? Will your earnings increase without bound?
Equivalently, what happens as you add in interest over shorter and shorter periods of time?
In other words, what happens as you let the compounding period approach zero?
You'll end up with more money. How much more? Will you get rich?
The amount of money produced under these situations does NOT increase without bound! (Wishful thinking, but doesn't happen.)
Instead, the irrational number ‘$\,\text{e}\,$’ emerges to describe the limiting amount earned.
The resulting formula is called the Continuous Compounding Formula, and is the subject of this section.
As $\,n\,$ gets large, $\displaystyle\,\frac{r}{n}\,$ gets small.
That is, as $\displaystyle\,n\rightarrow\infty\,$, we have $\displaystyle\,\frac{r}{n}\rightarrow 0\,$.
As $\displaystyle\,\frac{r}{n}\,$ approaches zero, $\displaystyle\,1 + \frac{r}{n}\,$ approaches $\,1\,$.
That is, as $\,\displaystyle\frac rn\rightarrow 0\,$, we have $\,\displaystyle 1 + \frac rn \rightarrow 1\,$.
As $\,n\,$ gets big, $\,nt\,$ gets big.
That is, as $\,n\rightarrow\infty\,$, we have $\,nt\rightarrow\infty\,$.
So, as we add interest in more and more frequently (let $\,n\rightarrow\infty\,$), here's what happens in the Compound Interest Formula: $$A=P{ \bigl( \underbrace{1+\frac{r}{n}}_{\text{base approaches $1$}} \bigr) }^{ \overbrace{nt}^{\text{exponent approaches $\infty$}} }$$
We're looking at a ‘$\,1^\infty\,$’ form—which is just a shorthand for saying that the base approaches $\,1\,$,
and the exponent approaches $\,\infty\,$.
But, here's the important question:
There are lots of things that make the ‘$\,1^\infty\,$’ form both interesting and difficult to analyze. All of the following are true:
Depending on precisely how the base approaches $\,1\,$, and how the exponent approaches infinity,
the form $\,1^\infty\,$ can approach different numbers!
These three different $\,1^\infty\,$ forms approach (respectively) $\,\text{e}\,$, $\,\infty\,$, and $\,1\,$:
For us, the investigation will involve noting the similarity of our situation to the definition of the irrational number $\,\text{e}\,$.
Compare these:
$$\lim_{n\rightarrow\infty} P(1 + \frac rn)^{nt} = \ ? \quad \quad \text{ and } \quad \quad \lim_{n\rightarrow\infty} (1+\frac 1n)^n := \text{e}$$
There are some pesky differences:
So, you want $\displaystyle\,\frac rn\,$ to ‘look like’ $\displaystyle\,\frac 1n\,$? Make it happen!
Define a new variable $\,m\,$ by this relationship:
$$\text{Let}\ \frac 1m := \frac rn\,.\quad \quad \quad \quad (*)$$
Recall that ‘ $:=$ ’ means ‘equal, by definition’.
(By the way, this is the tricky part of the derivation.)
Crossmultiplication in equation (*) gives $\,n = mr\,$.
Now, rewrite the Compound Interest Formula in terms of $\,m\,$ instead of $\,n\,$, and do a bit of algebra: $$ P(1 + \frac rn)^{nt} \quad = \quad P(1 + \frac 1m)^{mrt} \quad = \quad P\left[(1 + \frac 1m)^m\right]^{rt} $$
Taking reciprocals in equation (*), we have $\displaystyle\,m = \frac nr\,$.
So, as $\,n\,$ goes to
infinity, $\,m\,$ also goes to infinity.
And (here's the punch line!) as $\,m\,$ goes to infinity,
$\displaystyle P\left[(1 + \frac 1m)^m\right]^{rt} \rightarrow P{\text{e}}^{rt}\,$ !!
Suppose that \$1000 is invested at $3\%$ annual interest.
What is the accumulation after ten years if compounded monthly, daily, and continuously?
Compounded monthly:
$\displaystyle A = P(1 + \frac rn)^{nt} = 1000(1 + \frac {0.03}{12})^{12\cdot 10} = \$1,349.35$
Compounded daily:
$ \displaystyle A = 1000(1 + \frac {0.03}{365})^{365\cdot 10} = \$1,349.84$
Compounded continuously:
$\displaystyle A = P{\text{e}}^{rt} = 1000{\text{e}}^{0.03\cdot10} = \$1,349.86$
Not much difference! You won't get rich if your bank decides to
compound continuously!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
