List of representations of e

Euler proved that the number e is represented as the infinite simple continued fraction[1] (sequence A003417 in the OEIS):

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}}$,[3] and

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

Other series include the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

and Guillera's product [5][6]

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

where the nth factor is the nth root of the product

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

as well as the infinite product

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

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

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

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

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

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

The symmetric limit,[7]

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

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

The next two definitions are direct corollaries of the prime number theorem[8]

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

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

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

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

Also:

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

In the special case that ${\displaystyle x=1}$, the result is the famous statement:

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

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

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

at x = 1.