e (Base of Natural Logarithm)

Suppose your bank account grows at 100% interest per year, but instead of waiting until year-end, the bank compounds it twice a year at 50% each time. You end up with more money than the once-a-year version. Compound it four times, twelve times, daily, every millisecond — and the final balance keeps creeping upward. Does it grow without bound? It turns out it does not: the process converges to a specific real number, denoted $e$, that sits at the heart of every phenomenon in which change is proportional to the current amount.

Compound growth and the limiting sequence

Start with $1$ and apply an interest rate of $\frac{1}{n}$ a total of $n$ times over one year. The balance after one year is:

$$a_n \;\coloneqq\; \left(1 + \frac{1}{n}\right)^n.$$

A few computed values make the convergence concrete:

$n$	$a_n$
$1$	$2.000\,000$
$2$	$2.250\,000$
$10$	$2.593\,742$
$100$	$2.704\,814$
$10^4$	$2.718\,146$
$10^6$	$2.718\,280$

The sequence is climbing toward something near $2.718$. To define $e$ rigorously you need to know the limit exists — that $(a_n)$ is convergent.

The sequence converges

You will show $(a_n)$ is monotonically increasing and bounded above, which forces it to converge by the completeness of $\mathbb{R}$ established in Real Numbers.

Monotone increasing

Apply the AM–GM inequality to the $n+1$ positive numbers consisting of $n$ copies of $\left(1 + \tfrac{1}{n}\right)$ and one copy of $1$:

$$\underbrace{\frac{n \cdot \left(1 + \frac{1}{n}\right) + 1}{n+1}}_{\text{arithmetic mean}} \;\geq\; \underbrace{\left[\left(1+\frac{1}{n}\right)^n \cdot 1\right]^{\frac{1}{n+1}}}_{\text{geometric mean}}.$$

The left side simplifies to $\dfrac{n + 2}{n+1} = 1 + \dfrac{1}{n+1}$, so raising both sides to the $(n+1)$-th power gives:

$$\left(1 + \frac{1}{n+1}\right)^{n+1} \;\geq\; \left(1 + \frac{1}{n}\right)^n,$$

i.e.\ $a_{n+1} \geq a_n$. The sequence is monotonically increasing.

Bounded above by $3$

Expand $a_n$ using the binomial theorem:

$$\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^{n} \frac{1}{k!} \cdot \underbrace{\frac{n(n-1)\cdots(n-k+1)}{n^k}}_{\leq\, 1}. \tag{1}$$

Because each factor $\frac{n - j}{n} \leq 1$, the product in $(1)$ is at most $1$ for every $k$, giving:

$$a_n \;\leq\; \sum_{k=0}^{n} \frac{1}{k!} \;\leq\; 1 + 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}} \;<\; 3,$$

where the last bound uses $k! \geq 2^{k-1}$ for $k \geq 1$, so each term $\frac{1}{k!}$ is at most $\frac{1}{2^{k-1}}$, and the resulting geometric series sums to $2$.

An increasing sequence that stays below $3$ must converge. This gives the right to write the next definition.

Definition of $e$

Euler’s number $e$ is the real number

$$e \;\coloneqq\; \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n. \tag{2}$$

The series representation

Look again at inequality $(1)$. As $n \to \infty$, the ratio $\frac{n(n-1)\cdots(n-k+1)}{n^k}$ tends to $1$ for every fixed $k$ (it is a product of $k$ terms each approaching $1$). One can show that the bound $a_n \leq \sum_{k=0}^{n} \frac{1}{k!}$ together with the lower bound that follows from keeping only finitely many terms both squeeze toward the same value. The result is:

$$e \;=\; \sum_{k=0}^{\infty} \frac{1}{k!} \;=\; \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \tag{3}$$

where $0! \coloneqq 1$ by convention.

The denominators $k!$ grow faster than any fixed exponential, so the series converges extremely rapidly. Summing the first eight terms already gives:

$$1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} + \frac{1}{5040} \;\approx\; 2.71827,$$

accurate to five decimal places. The series form $(3)$ is often the most convenient representation for theoretical work.

Numerical value and irrationality

To twelve decimal places:

$$e \approx 2.718\,281\,828\,459\ldots$$

$e$ is irrational. The proof uses the series $(3)$. Suppose for contradiction that $e = \frac{p}{q}$ for positive integers $p, q$. Multiply both sides of $(3)$ by $q!$:

$$q!\cdot e \;=\; \underbrace{\sum_{k=0}^{q} \frac{q!}{k!}}_{\text{integer}} + \underbrace{\sum_{k=q+1}^{\infty} \frac{q!}{k!}}_{\text{tail}}.$$

Since $e = p/q$, the left side $q! \cdot e = (q-1)!\cdot p$ is an integer, and the first sum on the right is also an integer (each $\frac{q!}{k!}$ is a product of consecutive integers for $k \leq q$). Therefore the tail must also be an integer. But:

$$\text{tail} = \frac{1}{q+1} + \frac{1}{(q+1)(q+2)} + \cdots \;<\; \frac{1}{q+1} \cdot \frac{1}{1 - \frac{1}{q+1}} = \frac{1}{q} \;\leq\; 1,$$

and the tail is clearly positive. A positive quantity strictly less than $1$ cannot be an integer — contradiction. Therefore $e \notin \mathbb{Q}$.

In fact $e$ is transcendental (not a root of any polynomial with integer coefficients), but establishing this requires tools beyond the current prerequisites.

Why $e$ is the natural base

You might wonder what makes $e$ special over, say, $2$ or $10$. The answer lies in calculus: among all bases $b > 0$, the function $x \mapsto b^x$ has the simplest derivative — with no extra multiplicative constant — precisely when $b = e$. Likewise, the logarithm with base $e$ (the natural logarithm) has derivative $\frac{1}{x}$ with no extra factor. Any other base introduces a correction proportional to a logarithm of that base.

You will explore these properties in detail in Exponential Functions and Logarithms.

Summary

• Euler’s number $e$ is defined by the limit $\displaystyle e \coloneqq \lim_{n\to\infty}\!\left(1 + \tfrac{1}{n}\right)^n$, which converges because the sequence is monotone increasing and bounded above by $3$.
• Equivalently, $\displaystyle e = \sum_{k=0}^{\infty} \frac{1}{k!}$, a rapidly converging series whose partial sums provide arbitrarily accurate approximations.
• $e \approx 2.71828$ to five decimal places.
• $e$ is irrational: assuming $e = p/q$ leads to a contradiction because the tail of the series is a positive number less than $1$.
• $e$ is the uniquely natural base for exponential and logarithmic functions — a fact made precise when those functions are defined analytically.