It begins with a limit

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An interesting thing happens if we examine the formula for compound exponential growth in the limit as n approaches infinity—that is, we divide the growth rate into infinitely small chunks and compound it an infinite number of times. Recall that the compound growth function is

$$A(t) = A_o \left(1 + \frac{r}{n}\right)^{nt}$$

For simplicity, we'll first let r = 1 and t = 1. We'll relax those constraints later. So we're interested in finding this limit:

If you aren't familiar with limit statements, the question it's asking is, "what does (1 + 1/n)ⁿ equal when n gets infinitely large?"

Finding e (one of many ways)

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Below is the rather lengthy process of finding e from the limit. I like to go through this with my advanced classes because it involves a wide variety of mathematical thinking, but doesn't require any tricks from calculus. Here we go:

First, we note that (1 + 1/n)ⁿ is a binomial that is being raised to the powers 1, 2, 3, ... So we'll treat the limit like a binomial expansion. Here are a few powers of the binomial (a + b):

$$ \begin{align} (a + b)^0 &= 1\\[3pt] (a + b)^1 &= a + b\\[3pt] (a + b)^2 &= a^2 + 2 ab + b^2 \\[3pt] (a + b)^3 &= a^3 + 3 a^2b + 3 ab^2 + b^3 \\[3pt] (a + b)^4 &= a^4 + 4 a^3b + 6 a^2b^2 + 4 ab^3 + b^4 \: \dots \end{align}$$

... and so on.

The coefficients

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Convince yourself that the above represents most of the expansions of the powers of (a + b). What it leaves out it the coefficients. For example, the coefficients (left to right) of the terms of the expansion (that means we multiply it completely through) of (a + b)⁴ are: 1, 4, 6, 4, 1. For our shorthand formula, we need a way to get the coefficients correct, too. Here's how.

If we write down the coefficients of the terms containing a and b in our binomial power expansions, they form Pascal's Triangle:

Pascal's triangle is formed by writing a 1, then 1   1 on the next line, then each

It's like magic, isn't it? but the trouble with Pascal's triangle is that it's necessary to know its (n-1)^th row to get the n^th row, and so on. We'd like to be able to calculate the coefficients of the n^th row directly, for very large n —without writing the whole triangle.

To do this, take a look at   $(a + b)^4,$ a term big enough to see some patterns:

Those coefficients are just the binomial coefficients, the number of ways of grouping k objects from a set of n, when order doesn't matter

In our final equation for the binomial expansions, k will count from 0 to n (the power of the binomial) as we move from left to right across the terms.

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Now that we have our coefficients, we can write a general expression for the n^th order binomial expansion (a + b)ⁿ . You should write a few terms for yourself and confirm that it works. Let's pause a moment here and remind ourselves, in this long-ish process, that we're trying to put our exponential expression (1+1/n)ⁿ into a form where we can properly evaluate the limit as   $n \rightarrow \infty$.

$$(a + b)^n = \sum_{k = 0}^n \frac{n!}{k! \, (n - k)!} a^{n - k} b^k$$

Terms of the binomial expansion

Now just a short detour: It will be convenient to express factorial expressions in a slightly different way. Because   $0 ≤ k ≤ n,$ and   $n! = 1\cdot 2\cdot 3\cdot \dots \cdot n,$ we have

$$n! = 1 \cdot 2 \cdot 3 \cdot \; \dots \; (n - k) \cdot (n - k + 1) \cdot \; \dots \; \cdot n \; \;$$

$$\color{#E90F89}{\text{ and }} \; \; (n - k)! = 1 \cdot 2 \cdot 3 \cdot \; \dots \; \cdot (n - k)$$

We can then write the factorial part of the binomial expansion as:

$$\frac{n!}{(n - k)!} = (n - k + 1) \cdot \; \dots \; \cdot n$$

This gives us a new way to express the binomial coefficients:

$$\binom{n}{k} = \frac{1}{k!} \frac{n!}{(n - k)!} = \frac{(n - k + 1) \cdot \; \dots \; \cdot (n - 2)(n - 1) n}{k!}$$

Now we're in a position to write down an expression for   $(a + b)^n$   for any n. Here it is:

$$(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \binom{n}{3} a^{n - 3} b^3 + \dots + \binom{n}{n} a^0 b^n$$

Now replace the binomial coefficients with our factorial expressions:

$$(a + b)^n = 1 a^n b^0 + n a^{n - 1} b + \frac{n(n - 1)}{2!} a^{n - 2} b^2 + \frac{n(n - 1)(n - 2)}{3!} a^{n - 3} b^3 + \dots + 1 a^0 b^n$$

Now plug in 1 for a and 1/n for b:

$$\left(1 + \frac{1}{n}\right)^n = 1^n \left( \frac{1}{n} \right)^0 + n 1^{n - 1} \left( \frac{1}{n} \right)^1 + \frac{n(n - 1)}{2!} 1^{n - 2} \left( \frac{1}{n} \right)^2 + \frac{n(n - 1)(n - 2)}{3!} 1^{n - 3} \left( \frac{1}{n} \right)^3 + \dots$$

... and simplify:

$$\left(1 + \frac{1}{n}\right)^n = \left( \frac{1}{n} \right)^0 + n\left( \frac{1}{n} \right)^1 + \frac{n(n - 1)}{2!}\left( \frac{1}{n} \right)^2 + \frac{n(n - 1)(n - 2)}{3!} \left( \frac{1}{n} \right)^3 + \dots$$

Now we're almost ready to evaluate this series as n approaches infinity. A little bit of creative factoring gives:

$$\left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{\left( 1 - \frac{1}{n} \right)}{2!} + \frac{\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{1}{2n} \right)}{3!} + \frac{\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{1}{2n} \right)\left( 1 - \frac{1}{3n} \right)}{4!} + \dots$$

Now it's easy to see that as   $n \rightarrow \infty,$   each term containing n tends toward zero, leaving:

$$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$$

which we can write as a tidy sum:

$$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = \sum_{n = 0}^{\infty} \; \frac{1}{n!}$$

Remember that 0! = 1. Notice that the terms of this sum get smaller and smaller. That is, the sum converges. Values of the sum are shown in the table, and you can see that it converges to the number 2.718 ..., which we call e, the base of all continuously-growing exponential functions. The magenta numbers in the growing sum show the digits that are "locked in" and won't change further as more terms are added to the sum.

n	n!	1/n!	sum of terms
0	1	1.000000000000000	1.0000000000
1	1	1.000000000000000	2.0000000000
2	2	0.500000000000000	2.5000000000
3	6	0.166666666666667	2.6666666667
4	24	0.041666666666667	2.7083333333
5	120	0.008333333333333	2.7166666667
6	720	0.001388888888889	2.7180555556
7	5040	0.00019841268413	2.7182539683
8	40320	0.000024801587302	2.7182787698
9	362880	0.000002755731922	2.7182815256
10	3628800	0.000000025052108	2.7182818011
11	39916800	0.0000000275573192	2.7182818262
12	479001600	0.000000002087676	2.7182818283
13	6227020800	0.000000000160590	2.7182818284
14	87178291200	0.000000000011471	2.7182818285
15	1307674368000	0.000000000000765	2.7182818285

We began this long process by asking about the limit of the function   $f(t) = A_o (1 + 1/n)^{nt}$   as n becomes infinitely large. We found that   $f(t) = (1 + 1/n)^n$   approaches the transcendental number e as n gets large.

If we had left the A_o, r and t in as we went through the same procedure, we'd have arrived at the expression for continuous growth below.