Imagine a generous bank that pays 100% interest a year. You deposit 1 dollar in January and, in December, you have 2 dollars. Now the bank offers something else: instead of 100% at the end of the year, it pays 50% every six months, and the interest from the first half also earns interest in the second. Is it worth it? And if it paid every month, every day, every second? Could you become infinitely rich?

That is the question Jakob Bernoulli asked himself around 1683 and published in the Acta Eruditorum in 1690. It looks like bookkeeping, but the answer is one of the most important numbers in mathematics.

Doing the sum

If the year is divided into $n$ equal periods, each period pays $\frac{100\%}{n}$, that is, the balance is multiplied by $1+\frac1n$. After $n$ periods, 1 dollar becomes

$$a_n=\left(1+\frac1n\right)^n.$$

With two periods, $a_2=1.5^2=2.25$: better than 2. With more periods, better still:

Interest paid $n$ Balance at year end
once 1 2.000000
every half-year 2 2.250000
every quarter 4 2.441406
monthly 12 2.613035
daily 365 2.714567
hourly 8,760 2.718127

The balance grows, but less and less. Going from monthly to daily interest earns only about 10 cents more; from daily to hourly, less than half a cent. The balance seems to approach a ceiling near $2.718$.

Why there is a ceiling

Bernoulli did not have the word limit in today’s sense, but he had Newton’s binomial. Expanding $a_n$:

$$a_n=1+n\cdot\frac1n+\frac{n(n-1)}{2}\cdot\frac1{n^2}+\frac{n(n-1)(n-2)}{2\cdot3}\cdot\frac1{n^3}+\cdots$$

Tidying up each term,

$$\begin{aligned}a_n={}&1+1+\frac1{2!}\left(1-\frac1n\right)\\&+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots\end{aligned}$$

Two things happen as $n$ increases. The brackets $\left(1-\frac kn\right)$ get closer to 1, so each term grows; and new terms appear, all positive. So $a_n$ only increases. On the other hand, each bracket is less than 1, so

$$a_n<1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}+\cdots$$

And that sum is easy to control: since $3!=6>4$, $4!=24>8$ and, in general, $k!\geq 2^{k-1}$,

$$a_n<1+\left(1+\frac12+\frac14+\frac18+\cdots\right)=1+2=3.$$

Bernoulli concluded exactly this: the balance always rises, but stays between 2 and 3. However finely the bank divides the year, 1 dollar never becomes 3. A sequence that only grows and never passes a ceiling must approach a value:

$$\boxed{e=\lim_{n\to\infty}\left(1+\frac1n\right)^n=2.718281828\ldots}$$

Interest at every instant

The calculation above also shows the series that gives $e$ directly:

$$e=1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}+\cdots=2+0.5+0.1\overline{6}+0.041\overline{6}+\cdots$$

With only eight terms we already have $2.71825$: the series converges much faster than $(1+1/n)^n$.

For any rate $r$ a year, the same reasoning gives $\left(1+\frac rn\right)^n\to e^r$. This is the formula for continuously compounded interest: a capital $C$, invested for $t$ years, becomes $Ce^{rt}$. At 100% a year, continuous interest earns $e-1\approx 71.8\%$ more than the deposit, against 100% for the yearly version. Paying interest at every instant is good, but it is not magic.

Exercise: at 12% a year, how much does 1,000 dollars earn in one year, paid once, monthly and continuously?

Once: $1{,}000\cdot1.12=1{,}120$ dollars. Monthly, the monthly rate is $1\%$: $1{,}000\cdot1.01^{12}\approx1{,}126.83$ dollars. Continuously: $1{,}000\cdot e^{0.12}\approx1{,}127.50$ dollars. The difference between monthly and continuous is less than 70 cents.

Exercise: how long does it take a capital to double with continuous interest of 7% a year?

We want $e^{0.07t}=2$, that is, $0.07t=\ln 2\approx0.693$. Hence $t\approx9.9$ years. This is the origin of economists’ “rule of 70”: the doubling time is roughly 70 divided by the rate in percent.

Sources: Jakob Bernoulli, “Quaestiones nonnullae de usuris”, Acta Eruditorum (1690) · Eli Maor, e: The Story of a Number (Princeton, 1994) · biography of Jacob Bernoulli, MacTutor.

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