# Euler’s Number as the Limit of a Sequence
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https://algebrica.org/euler-number-limit-sequence/
Fetched from algebrica.org post 15726; source modified 2026-03-20T18:15:05.
Concept
The structure of the entry is shown in the conceptual map, where each branch represents a core component and the sub-nodes highlight the specific notions discussed.
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The following concepts, [Binomial Theorem](../binomial-theorem.md), [Geometric Series](../geometric-series.md), [Sequences](../sequences.md), are required as prerequisites for this entry.
## 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 = \\left(1 + \\frac{1}{n}\\right)^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 := \\lim\_{n \\to \\infty} \\left(1 + \\frac{1}{n}\\right)^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](../integers.md). Transcendence is a stronger property: it means that \\(e\\) is not the root of any non-zero [polynomial equation](../polynomial-equations.md) 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.
\\[
\\begin{align} n = 1:& \\quad a\_1 = \\left(1 + \\frac{1}{1}\\right)^1 = 2 \\[6pt] n = 10:& \\quad a\_{10} = \\left(1 + \\frac{1}{10}\\right)^{10} \\approx 2.59374 \\[6pt] n = 100:& \\quad a\_{100} = \\left(1 + \\frac{1}{100}\\right)^{100} \\approx 2.70481 \\[6pt] n = 1000:& \\quad a\_{1000} = \\left(1 + \\frac{1}{1000}\\right)^{1000} \\approx 2.71692 \\end{align}
\\]
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](../binomial-theorem.md). Applied to the expression \\(\\left(1 + \\frac{1}{n}\\right)^n\\), the expansion gives the following:
\\[
a\_n = \\sum\_{k=0}^{n} \\binom{n}{k} \\frac{1}{n^k} = \\sum\_{k=0}^{n} \\frac{1}{k!} \\cdot \\frac{n(n-1)\\cdots(n-k+1)}{n^k}
\\]
Each factor of the form:
\\[
\\frac{n(n-1)\\cdots(n-k+1)}{n^k}
\\]
can be written as a product of \\(k\\) terms of the type \\(\\left(1 - \\frac{j}{n}\\right)\\), for \\(j = 0, 1, \\ldots, k-1\\). The expansion therefore takes the form:
\\[
a\_n = \\sum\_{k=0}^{n} \\frac{1}{k!} \\prod\_{j=0}^{k-1} \\left(1 - \\frac{j}{n}\\right)
\\]
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 \\(\\left(1 - \\frac{j}{n}\\right)\\) is replaced by \\(\\left(1 - \\frac{j}{n+1}\\right)\\), 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 \\(\\left(1 - \\frac{j}{n}\\right)\\) is at most \\(1\\), we obtain the following estimate:
\\[
a\_n = \\sum\_{k=0}^{n} \\frac{1}{k!} \\prod\_{j=0}^{k-1} \\left(1 - \\frac{j}{n}\\right) < \\sum\_{k=0}^{n} \\frac{1}{k!}
\\]
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 \\cdots k\\) is at least \\(2\\). This gives:
\\[
\\sum\_{k=0}^{n} \\frac{1}{k!} \\leq 1 + \\sum\_{k=1}^{n} \\frac{1}{2^{k-1}} = 1 + \\sum\_{k=0}^{n-1} \\frac{1}{2^k}
\\]
The sum on the right is a partial sum of a [geometric series](../geometric-series.md) with ratio \\(\\frac{1}{2}\\). Its value is:
\\[
\\sum\_{k=0}^{n-1} \\frac{1}{2^k} = 2\\left(1 - \\frac{1}{2^n}\\right) < 2
\\]
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:
\\[
\\sum\_{k=0}^{n} \\frac{1}{k!}
\\]
that appears as an upper bound for \\(a\_n\\) is itself a partial sum of the series:
\\[
\\sum\_{k=0}^{\\infty} \\frac{1}{k!} = 1 + 1 + \\frac{1}{2!} + \\frac{1}{3!} + \\cdots
\\]
This series converges, and its sum is exactly \\(e\\). In fact, the following identity shows that the two definitions are equivalent:
\\[
e = \\lim\_{n \\to \\infty} \\left(1 + \\frac{1}{n}\\right)^n = \\sum\_{k=0}^{\\infty} \\frac{1}{k!}
\\]
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\\)](https://spot.colorado.edu/~baggett/chap2.pdf)
- **University of Connecticut, K. Conrad**. [Irrationality of \\(\\pi\\) and \\(e\\)](https://kconrad.math.uconn.edu/blurbs/analysis/irrational.pdf)
