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How a sequence reveals $e$

Euler’s number, denoted by $e$, is one of the most important constants in mathematics. There are several equivalent ways to introduce it: through infinite series, through the natural exponential function, or through the limit of a sequence. This page focuses on the latter approach.

We consider the sequence $a_{n}$ with $n \in \mathbb{N}$ defined by the following expression: \(a_{n} = (( 1 + \frac{1}{n} ))^{n}\) As shown in the sections below, this sequence is strictly increasing and bounded above. By the monotone convergence theorem, it therefore converges to a finite limit. That limit is taken as the definition of Euler’s number, and we write: \(e := \underset{n \rightarrow \infty}{lim} (( 1 + \frac{1}{n} ))^{n}\) The symbol $:=$ indicates that this is a definition: the number $e$ is introduced as the value to which the sequence converges. Its decimal expansion begins as $e \approx 2.71828$, and $e$ can be shown to be both irrational and transcendental. Irrationality means that $e$ cannot be expressed as a ratio of two integers. Transcendence is a stronger property: it means that $e$ is not the root of any non-zero polynomial equation with rational coefficients.

The graph below illustrates how the terms of the sequence behave as $n$ grows. The values increase rapidly for small $n$, then rise more slowly, approaching $e$ from below without ever reaching it.

Each term $a_{n}$ is strictly less than $e$, and the gap closes as $n$ grows, though the rate of convergence is slow enough that even large values of $n$ yield only a rough approximation of the limit.

The following table of values illustrates how the sequence behaves for increasing indices. \(n = 1 : & a_{1} = (( 1 + \frac{1}{1} ))^{1} = 2 \\ n = 10 : & a_{10} = (( 1 + \frac{1}{10} ))^{10} \approx 2.59374 \\ n = 100 : & a_{100} = (( 1 + \frac{1}{100} ))^{100} \approx 2.70481 \\ n = 1000 : & a_{1000} = (( 1 + \frac{1}{1000} ))^{1000} \approx 2.71692\) The terms increase steadily and approach $e \approx 2.71828$ from below, with each successive value capturing more decimal places of the limit. The convergence is monotone but slow: even at $n = 1000$, the approximation agrees with $e$ only to the second decimal place.

Demonstrating the monotonicity of the sequence

To prove that the sequence $a_{n}$ is strictly increasing, we expand each term using the Binomial Theorem. Applied to the expression $(( 1 + \frac{1}{n} ))^{n}$, the expansion gives the following:

Each factor of the form:

can be written as a product of $k$ terms of the type $( 1 - \frac{j}{n} )$, for $j = 0 , 1 , \ldots , k - 1$. The expansion therefore takes the form:

Now consider the analogous expression for $a_{n + 1}$, obtained by replacing $n$ with $n + 1$ throughout. Two things happen: the upper limit of the sum increases by one, adding a new positive term, and each existing factor $( 1 - \frac{j}{n} )$ is replaced by $( 1 - \frac{j}{n + 1} )$, which is strictly larger since the subtracted quantity decreases.

Every term in the sum for $a_{n + 1}$ is therefore strictly greater than the corresponding term in the sum for $a_{n}$, and the sum itself contains one additional positive term. It follows that $a_{n + 1} > a_{n}$ for all $n \in \mathbb{N}$, so the sequence is strictly increasing.

Demonstrating the boundedness of the sequence

It remains to show that the sequence is bounded. Since $a_{1} = 2$ and the sequence is strictly increasing, we have $a_{n} > 2$ for all $n \geq 1$. It therefore suffices to establish an upper bound. We claim that $a_{n} < 3$ for all $n \in \mathbb{N}$. Starting from the expansion derived in the previous section, and observing that each factor $( 1 - \frac{j}{n} )$ is at most $1$, we obtain the following estimate:

To bound this sum from above, we use the inequality $k ! \geq 2^{k - 1}$, which holds for all $k \geq 1$ and follows from the fact that each of the $k - 1$ factors in $2 \cdot 3 \hdots k$ is at least $2$. This gives:

The sum on the right is a partial sum of a geometric series with ratio $\frac{1}{2}$. Its value is:

Combining these estimates, we conclude that $a_{n} < 1 + 2 = 3$ for all $n \in \mathbb{N}$. Together with the lower bound $a_{n} > 2$, this confirms that the sequence is bounded.

Connection with the series definition of $e$

The proof of boundedness reveals something more than a mere upper estimate. The quantity:

that appears as an upper bound for $a_{n}$ is itself a partial sum of the series:

This series converges, and its sum is exactly $e$. In fact, the following identity shows that the two definitions are equivalent:

The series representation, which arises from the Taylor expansion of the exponential function, converges considerably faster than the sequence $a_{n}$ and provides a more efficient route to computing decimal approximations of $e$.

Selected references

• University of Colorado, L. Baggett. Definition of the Number $e$
• University of Connecticut, K. Conrad. Irrationality of $\pi$ and $e$