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Here is a problem that sounds like it belongs in a bank and ends up inside every differential equation ever written.

You deposit 1 unit into an account that pays 100% interest per year. After one year you have 2. Simple enough.

Now the bank says: we’ll give you 50% every six months instead. After six months you have 1.5. After another six months, 50% of 1.5 is 0.75, so you end up with 2.25. Better.

Quarterly: 25% four times. You end up with $(1 + 1/4)^4 \approx 2.4414$. Better still.

Monthly: $(1 + 1/12)^{12} \approx 2.6130$.

Daily: $(1 + 1/365)^{365} \approx 2.7146$.

Every second: $(1 + 1/31536000)^{31536000} \approx 2.71828$.

The numbers are converging. The question Jacob Bernoulli asked in 1683 is: what do they converge to? What happens when you compound continuously - infinitely many compounding periods, each infinitesimally small?

The answer is a specific irrational number. Euler called it $e$. It is approximately $2.71828182845904523536\ldots$

The Limit That Defines It

The compounding sequence gives us a precise definition:

$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n.$$

This is not a definition by decree. It is the answer to Bernoulli’s compound interest question, and it will turn out to be the same number that satisfies several other remarkable properties we’ll meet below. But first: does this limit even exist? The expression $(1 + 1/n)^n$ is the product of $n$ terms each slightly greater than 1 - so it could conceivably grow without bound. We need to prove it converges.

Proof That the Limit Exists

We show two things: the sequence $a_n = (1 + 1/n)^n$ is monotone increasing and bounded above. A monotone bounded sequence must converge - this is one of the fundamental completeness properties of the real numbers.

Step 1: The sequence is increasing

Expand $a_n$ using the binomial theorem:

$$\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \cdot \frac{1}{n^k}.$$

The $k$-th term is:

$$\frac{n!}{k!(n-k)!n^k} = \frac{1}{k!} \cdot \frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k} = \frac{1}{k!}\cdot\left(1\right)\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right).$$

When you pass from $a_n$ to $a_{n+1}$, two things happen: you get one more term (the $k = n+1$ term, which is positive), and every existing term gets larger because factors of the form $(1 - j/n)$ become $(1 - j/(n+1))$, which are closer to 1. So every term increases and a new positive term is added. Therefore $a_{n+1} > a_n$.

Step 2: The sequence is bounded above by 3

From the binomial expansion, the $k$-th term is at most $\frac{1}{k!}$, since each factor $(1 - j/n) < 1$. So:

$$a_n \leq \sum_{k=0}^{n} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}.$$

Now $k! \geq 2^{k-1}$ for $k \geq 1$ (easy by induction), so $\frac{1}{k!} \leq \frac{1}{2^{k-1}}$. Therefore:

$$a_n \leq 1 + \sum_{k=1}^{n} \frac{1}{2^{k-1}} = 1 + \frac{1 - (1/2)^n}{1 - 1/2} < 1 + 2 = 3.$$

Conclusion

The sequence $(1 + 1/n)^n$ is monotone increasing and bounded above by 3. By the Monotone Convergence Theorem, it converges. The number it converges to is what we call $e$.

We also learn: $2 < e < 3$. (The lower bound $e > 2$ is immediate since $a_2 = (1+1/2)^2 = 2.25 > 2$.)

Discomfort check. The binomial expansion gave us $a_n \leq \sum_{k=0}^n \frac{1}{k!}$. As $n \to \infty$, the right side converges too - this is the series representation $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots$. In fact the sequence $(1+1/n)^n$ and the partial sums of this series both converge to the same limit. This series representation converges quickly - four terms already give $1 + 1 + 0.5 + 0.1\overline{6} = 2.6\overline{6}$, six terms give $2.71805\ldots$

Three Definitions, One Number

$e$ has a habit of appearing from multiple directions, seemingly unrelated. Each of the following is a legitimate definition, and they all produce the same number.

Definition 1 (Compound interest limit): $$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n.$$

This generalizes directly to any $x$. Replace 1 with $x$ in the interest rate - now the bank pays $x \cdot 100%$ per year, compounded continuously. The same argument gives:

$$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n.$$

The derivation is one substitution: let $m = n/x$, so as $n \to \infty$, $m \to \infty$ and $n = mx$:

$$\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^{mx} = \left[\lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m\right]^x = e^x.$$

This is the form that appears when deriving the Poisson distribution - the term $(1 - \lambda/n)^n$ is exactly $e^{-\lambda}$ in the limit, because $(1 + x/n)^n \to e^x$ with $x = -\lambda$.

Definition 2 (Series): $$e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots$$

Definition 3 (Logarithmic integral): $e$ is the unique positive real number satisfying: $$\int_1^e \frac{dt}{t} = 1.$$

Definition 3 is perhaps the most geometric. The function $f(t) = 1/t$ has a hyperbolic graph. The area under this curve from $t = 1$ to some endpoint $x$ defines the natural logarithm: $\ln(x) = \int_1^x \frac{dt}{t}$. Then $e$ is defined as the unique $x$ where this area equals exactly 1. This definition requires no prior knowledge of exponentiation - you define the logarithm first as an integral, then define $e$ as a specific area, and only then introduce $e^x$ as the inverse of $\ln(x)$.

All three definitions give the same number. The equivalence of Definition 1 and Definition 2 follows from the binomial expansion we did above (taking $n \to \infty$, the partial sums converge to the same limit). The equivalence with Definition 3 requires showing that $\int_1^{xy} \frac{dt}{t} = \int_1^x \frac{dt}{t} + \int_1^y \frac{dt}{t}$, which follows by splitting the integral at $x$ and substituting $t = xu$ in the second piece.

The Self-Derivative Property

Here is why $e$ becomes essential in calculus rather than just a curiosity.

What is $\frac{d}{dx}(a^x)$ for an arbitrary base $a > 0$? Using the definition of the derivative:

$$\frac{d}{dx}(a^x) = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}.$$

The factor $a^x$ comes outside cleanly. What remains is $L(a) = \lim_{h \to 0} \frac{a^h - 1}{h}$, a constant depending only on the base $a$, not on $x$. So every exponential function is proportional to its own derivative:

$$\frac{d}{dx}(a^x) = L(a) \cdot a^x.$$

Now the question: is there a base where $L(a) = 1$, so that the function is literally equal to its own derivative?

The intuition. $L(a)$ measures how steeply $a^h$ climbs away from $1$ at $h = 0$. When $a = 1$, there is no climb at all: $1^h = 1$ always, so $L(1) = 0$. As $a$ grows, the climb gets steeper. Somewhere between $2$ and $3$ - the steepness crosses $1$. That crossing point is $e$.

To see why it must cross exactly at $e$, use Definition 3. We defined $e$ as the unique positive number with $\int_1^e \frac{dt}{t} = 1$, and set $\ln(x) = \int_1^x \frac{dt}{t}$.

Proof that $L(e) = 1$.

Substitute $u = e^h - 1$ in the limit, so $e^h = 1 + u$ and $h = \ln(1 + u)$. As $h \to 0$, $u \to 0$:

$$L(e) = \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{u \to 0} \frac{u}{\ln(1 + u)}.$$

It suffices to show $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$. Using the integral definition $\ln(1+u) = \int_1^{1+u} \frac{dt}{t}$, and noting that on $[1, 1+u]$ (for $u > 0$) the integrand satisfies $\frac{1}{1+u} \leq \frac{1}{t} \leq 1$:

$$\frac{u}{1+u} \leq \ln(1+u) \leq u.$$

Dividing by $u > 0$:

$$\frac{1}{1+u} \leq \frac{\ln(1+u)}{u} \leq 1.$$

As $u \to 0^+$ both bounds approach $1$. By the Squeeze Theorem, $\frac{\ln(1+u)}{u} \to 1$. A symmetric argument handles $u \to 0^-$. Therefore $L(e) = 1$. $\blacksquare$

Why the squeeze works geometrically. The quantity $\ln(1+u) = \int_1^{1+u} \frac{dt}{t}$ is the area under the curve $y = 1/t$ over a tiny interval of width $u$. Near $t = 1$, the curve $y = 1/t$ is nearly flat - its value is close to $1$ everywhere on $[1, 1+u]$. So the area is sandwiched between two thin rectangles: the inner one (height $\frac{1}{1+u}$, width $u$) and the outer one (height $1$, width $u$). As $u \to 0$, both rectangles have area approaching $u \cdot 1 = u$, and the curve area is trapped between them. The ratio $\ln(1+u)/u$ cannot escape $1$.

$e$ is the unique such base. For any $a > 0$, write $a = e^{\ln a}$, so $a^h = e^{h \ln a}$. Substituting $k = h \ln a$:

$$L(a) = \lim_{h \to 0} \frac{e^{h \ln a} - 1}{h} = \lim_{k \to 0} \frac{e^k - 1}{k / \ln a} = \ln a \cdot \underbrace{\lim_{k \to 0} \frac{e^k - 1}{k}}_{= L(e) = 1} = \ln a.$$

So $L(a) = \ln a$ for every base $a > 0$. Setting $L(a) = 1$ requires $\ln a = 1$, which has the unique solution $a = e$.

The natural logarithm $\ln a$ is not a separate construction that happens to appear here. It literally is $L(a)$ - the proportionality constant between an exponential and its own derivative. This is the real reason $\ln$ is called “natural”: it emerges directly from the question of which base makes differentiation clean.

A Brief History

The number was not discovered in one place.

John Napier (1614) invented logarithms for computational convenience - to turn multiplication into addition - without knowing $e$ explicitly. His logarithms used a base close to but not exactly $e$.

Jacob Bernoulli (1683) stumbled onto the compound interest limit while studying continuous growth, computed that it was between 2 and 3, and left it there without naming it or realizing its significance.

Leibniz (1690) encountered the number in connection with the curve $y = 1/x$ and the integral $\int_1^x \frac{dt}{t}$.

Leonhard Euler (1731) systematized all of this. He named the constant $e$ in a letter - almost certainly for “exponential” rather than for himself, despite the coincidence - proved it was the base of natural logarithms, established the series representation, and computed it to 23 decimal places. He also proved the identity $e^{i\pi} + 1 = 0$, which ties $e$ to the geometry of the complex plane.

Charles Hermite (1873) proved $e$ is transcendental: not only irrational, but not the root of any polynomial with rational coefficients. This is a much stronger statement than irrationality. The proof is involved but the implication is striking - $e$ cannot be described by any finite algebraic expression.

Why It Appears Everywhere

Once you know $\frac{d}{dx}(e^x) = e^x$, the cascade begins.

Differential equations. The equation $y' = y$ has solution $y = Ce^x$. More generally, $y' = ky$ (exponential growth or decay) has solution $y = Ce^{kx}$. This is the model for population growth, radioactive decay, charging capacitors, cooling, compound interest. Any system where the rate of change is proportional to the current value produces $e$.

Complex exponentials. Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ connects $e$ to circles and oscillations. This is not a definition - it is a theorem, following from the Taylor series of $e^x$, $\cos x$, and $\sin x$. It means that $e^x$ for complex $x$ encodes rotation.

Probability. The normal distribution $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ is built from $e$. So is the Poisson distribution $e^{-\lambda}\lambda^k / k!$. The central limit theorem - the reason bell curves appear everywhere - is saturated with $e$.

Information theory. The entropy $-\sum p_i \ln p_i$ uses the natural logarithm, base $e$. Measuring information in nats (natural units, using $\ln$) is mathematically cleaner than measuring in bits (using $\log_2$).

All of these appearances trace back to the same root: $e$ is the fixed point of differentiation, and differentiation is the language in which the universe’s laws are written.

Summary

Definition Formula
Compound interest limit $e = \lim_{n \to \infty}(1 + 1/n)^n$
Series $e = \sum_{k=0}^\infty 1/k!$
Logarithmic integral $\int_1^e \frac{dt}{t} = 1$
Self-derivative $\frac{d}{dx}e^x = e^x$

$e \approx 2.71828$. It is irrational. It is transcendental. It is the natural base for logarithms and exponentials. It is the solution to the continuous compounding problem Bernoulli left unsolved. It is the only number that makes the derivative of an exponential equal to that exponential.

Once you have limits, $e$ is inevitable. Once you have $e$, calculus becomes fluent.

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