List of representations of e

The mathematical constant e can be represented in a variety of ways as a real number. Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. Using calculus, e may also be represented as an infinite series, infinite product, or other sort of limit of a sequence.

As a continued fraction

Euler proved that the number e is represented as the infinite simple continued fraction (sequence A003417 in the OEIS):

$e=[2;1,2,1,1,4,1,1,6,1,1,8,1,\ldots ,1,2n,1,\ldots ].$

Its convergence can be tripled by allowing just one fractional number:

$e=[1;1/2,12,5,28,9,44,13,60,17,\ldots ,4(4n-1),4n+1,\ldots ].$

Here are some infinite generalized continued fraction expansions of e. The second is generated from the first by a simple equivalence transformation.

$e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {2}{3+{\cfrac {3}{4+{\cfrac {4}{5+\ddots }}}}}}}}}}=2+{\cfrac {2}{2+{\cfrac {3}{3+{\cfrac {4}{4+{\cfrac {5}{5+{\cfrac {6}{6+\ddots \,}}}}}}}}}}$

$e=2+{\cfrac {1}{1+{\cfrac {2}{5+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+\ddots \,}}}}}}}}}}=1+{\cfrac {2}{1+{\cfrac {1}{6+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+\ddots \,}}}}}}}}}}$

This last, equivalent to [1; 0.5, 12, 5, 28, 9, ...], is a special case of a general formula for the exponential function:

$e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}$

Conjectures

There are also continued fraction conjectures for e. For example, a computer program developed at the Israel Institute of Technology has came up with:

$e=3+{\cfrac {-1}{4+{\cfrac {-2}{5+{\cfrac {-3}{6+{\cfrac {-4}{7+\ddots \,}}}}}}}}$

in 2019. This conjecture has now been proved, according to

As an infinite series

The number e can be expressed as the sum of the following infinite series:

$e^{x}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}$ for any real number x.

In the special case where x = 1 or −1, we have:

$e=\sum _{k=0}^{\infty }{\frac {1}{k!}}$, and

$e^{-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}.$

Other series include the following:

$e=\left[\sum _{k=0}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{-1}$

$e={\frac {1}{2}}\sum _{k=0}^{\infty }{\frac {k+1}{k!}}$

$e=2\sum _{k=0}^{\infty }{\frac {k+1}{(2k+1)!}}$

$e=\sum _{k=0}^{\infty }{\frac {3-4k^{2}}{(2k+1)!}}$

$e=\sum _{k=0}^{\infty }{\frac {(3k)^{2}+1}{(3k)!}}=\sum _{k=0}^{\infty }{\frac {(3k+1)^{2}+1}{(3k+1)!}}=\sum _{k=0}^{\infty }{\frac {(3k+2)^{2}+1}{(3k+2)!}}$

$e=\left[\sum _{k=0}^{\infty }{\frac {4k+3}{2^{2k+1}\,(2k+1)!}}\right]^{2}$

$e=\sum _{k=0}^{\infty }{\frac {k^{n}}{B_{n}(k!)}}$ where $B_{n}$ is the nth Bell number.

Consideration of how to put upper bounds on e leads to this descending series:

$e=3-\sum _{k=2}^{\infty }{\frac {1}{k!(k-1)k}}=3-{\frac {1}{4}}-{\frac {1}{36}}-{\frac {1}{288}}-{\frac {1}{2400}}-{\frac {1}{21600}}-{\frac {1}{211680}}-{\frac {1}{2257920}}-\cdots$

which gives at least one correct (or rounded up) digit per term. That is, if 1 ≤ n, then

$e<3-\sum _{k=2}^{n}{\frac {1}{k!(k-1)k}}<e+0.6\cdot 10^{1-n}\,.$

More generally, if x is not in {2, 3, 4, 5, ...}, then

$e^{x}={\frac {2+x}{2-x}}+\sum _{k=2}^{\infty }{\frac {-x^{k+1}}{k!(k-x)(k+1-x)}}\,.$

As an infinite product

The number e is also given by several infinite product forms including Pippenger's product

$e=2\left({\frac {2}{1}}\right)^{1/2}\left({\frac {2}{3}}\;{\frac {4}{3}}\right)^{1/4}\left({\frac {4}{5}}\;{\frac {6}{5}}\;{\frac {6}{7}}\;{\frac {8}{7}}\right)^{1/8}\cdots$

and Guillera's product

$e=\left({\frac {2}{1}}\right)^{1/1}\left({\frac {2^{2}}{1\cdot 3}}\right)^{1/2}\left({\frac {2^{3}\cdot 4}{1\cdot 3^{3}}}\right)^{1/3}\left({\frac {2^{4}\cdot 4^{4}}{1\cdot 3^{6}\cdot 5}}\right)^{1/4}\cdots ,$

where the nth factor is the nth root of the product

$\prod _{k=0}^{n}(k+1)^{(-1)^{k+1}{n \choose k}},$

as well as the infinite product

$e={\frac {2\cdot 2^{(\ln(2)-1)^{2}}\cdots }{2^{\ln(2)-1}\cdot 2^{(\ln(2)-1)^{3}}\cdots }}.$

More generally, if 1 < B < e² (which includes B = 2, 3, 4, 5, 6, or 7), then

$e={\frac {B\cdot B^{(\ln(B)-1)^{2}}\cdots }{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^{3}}\cdots }}.$

As the limit of a sequence

The number e is equal to the limit of several infinite sequences:

$e=\lim _{n\to \infty }n\cdot \left({\frac {\sqrt {2\pi n}}{n!}}\right)^{1/n}$ and

$e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}$ (both by Stirling's formula).

The symmetric limit,

$e=\lim _{n\to \infty }\left[{\frac {(n+1)^{n+1}}{n^{n}}}-{\frac {n^{n}}{(n-1)^{n-1}}}\right]$

may be obtained by manipulation of the basic limit definition of e.

The next two definitions are direct corollaries of the prime number theorem

$e=\lim _{n\to \infty }(p_{n}\#)^{1/p_{n}}$

where $p_{n}$ is the nth prime and $p_{n}\#$ is the primorial of the nth prime.

$e=\lim _{n\to \infty }n^{\pi (n)/n}$

where $\pi (n)$ is the prime counting function.

Also:

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

In the special case that $x=1$, the result is the famous statement:

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

The ratio of the factorial $n!$, that counts all permutations of an orderet set S with cardinality $n$, and the derangement function $!n$, which counts the amount of permutations where no element appears in its original position, tends to $e$ as $n$ grows.

$e=\lim _{n\to \infty }{\frac {n!}{!n}}.$

In trigonometry

Trigonometrically, e can be written in terms of the sum of two hyperbolic functions,

$e^{x}=\sinh(x)+\cosh(x),$

at x = 1.