List of representations of e Last updated July 25, 2025 As a continued fraction Euler proved that the number e is represented as the infinite simple continued fraction [ 1] (sequence A003417 in the OEIS ) :
e = [ 2 ; 1 , 2 , 1 , 1 , 4 , 1 , 1 , 6 , 1 , 1 , 8 , 1 , … , 1 , 2 n , 1 , … ] = 2 + 1 1 + 1 2 + 1 1 + 1 1 + 1 4 + 1 1 + 1 1 + 1 6 + 1 1 + 1 1 + 1 8 + ⋱ {\displaystyle {\begin{aligned}e&=[2;1,2,1,1,4,1,1,6,1,1,8,1,\ldots ,1,2n,1,\ldots ]\\[8pt]&=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{6+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{8+{{} \atop \ddots }}}}}}}}}}}}}}}}}}}}}}}\end{aligned}}} Here are some infinite generalized continued fraction expansions of e . The second is generated from the first by a simple equivalence transformation .
e = 2 + 1 1 + 1 2 + 2 3 + 3 4 + 4 5 + ⋱ = 2 + 2 2 + 3 3 + 4 4 + 5 5 + 6 6 + ⋱ {\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {2}{3+{\cfrac {3}{4+{\cfrac {4}{5+{{} \atop \ddots }}}}}}}}}}}=2+{\cfrac {2}{2+{\cfrac {3}{3+{\cfrac {4}{4+{\cfrac {5}{5+{\cfrac {6}{6+{{} \atop \ddots }\,}}}}}}}}}}} e = 2 + 1 1 + 2 5 + 1 10 + 1 14 + 1 18 + ⋱ = 1 + 2 1 + 1 6 + 1 10 + 1 14 + 1 18 + ⋱ {\displaystyle e=2+{\cfrac {1}{1+{\cfrac {2}{5+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+{{} \atop \ddots }\,}}}}}}}}}}=1+{\cfrac {2}{1+{\cfrac {1}{6+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{18+{{} \atop \ddots }\,}}}}}}}}}}} This last non-simple continued fraction (sequence A110185 in the OEIS ) , equivalent to e = [ 1 ; 0.5 , 12 , 5 , 28 , 9 , . . . ] {\displaystyle e=[1;0.5,12,5,28,9,...]} , has a quicker convergence rate compared to Euler's continued fraction formula [ clarification needed ] and is a special case of a general formula for the exponential function :
e x / y = 1 + 2 x 2 y − x + x 2 6 y + x 2 10 y + x 2 14 y + x 2 18 y + ⋱ {\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+{{} \atop \ddots }}}}}}}}}}}} As an infinite series The number e can be expressed as the sum of the following infinite series :
e x = ∑ k = 0 ∞ x k k ! {\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:
e = ∑ k = 0 ∞ 1 k ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}} , [ 2] ande − 1 = ∑ k = 0 ∞ ( − 1 ) k k ! . {\displaystyle e^{-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}.} Other series include the following:
e = [ ∑ k = 0 ∞ 1 − 2 k ( 2 k ) ! ] − 1 {\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{-1}} [ 3] e = 1 2 ∑ k = 0 ∞ k + 1 k ! {\displaystyle e={\frac {1}{2}}\sum _{k=0}^{\infty }{\frac {k+1}{k!}}} e = 2 ∑ k = 0 ∞ k + 1 ( 2 k + 1 ) ! {\displaystyle e=2\sum _{k=0}^{\infty }{\frac {k+1}{(2k+1)!}}} e = ∑ k = 0 ∞ 3 − 4 k 2 ( 2 k + 1 ) ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {3-4k^{2}}{(2k+1)!}}} e = ∑ k = 0 ∞ ( 3 k ) 2 + 1 ( 3 k ) ! = ∑ k = 0 ∞ ( 3 k + 1 ) 2 + 1 ( 3 k + 1 ) ! = ∑ k = 0 ∞ ( 3 k + 2 ) 2 + 1 ( 3 k + 2 ) ! {\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)!}}} e = [ ∑ k = 0 ∞ 4 k + 3 2 2 k + 1 ( 2 k + 1 ) ! ] 2 {\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {4k+3}{2^{2k+1}\,(2k+1)!}}\right]^{2}} e = ∑ k = 0 ∞ k n B n ( k ! ) {\displaystyle e=\sum _{k=0}^{\infty }{\frac {k^{n}}{B_{n}(k!)}}} where B n {\displaystyle B_{n}} is the n th Bell number .e = ∑ k = 0 ∞ 2 k + 3 ( k + 2 ) ! {\displaystyle e=\sum _{k=0}^{\infty }{\frac {2k+3}{(k+2)!}}} [ 4] Consideration of how to put upper bounds on e leads to this descending series:
e = 3 − ∑ k = 2 ∞ 1 k ! ( k − 1 ) k = 3 − 1 4 − 1 36 − 1 288 − 1 2400 − 1 21600 − 1 211680 − 1 2257920 − ⋯ {\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
e < 3 − ∑ k = 2 n 1 k ! ( k − 1 ) k < e + 0.6 ⋅ 10 1 − n . {\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
e x = 2 + x 2 − x + ∑ k = 2 ∞ − x k + 1 k ! ( k − x ) ( k + 1 − x ) . {\displaystyle 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 ( 2 1 ) 1 / 2 ( 2 3 4 3 ) 1 / 4 ( 4 5 6 5 6 7 8 7 ) 1 / 8 ⋯ {\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 [ 6] [ 7]
e = ( 2 1 ) 1 / 1 ( 2 2 1 ⋅ 3 ) 1 / 2 ( 2 3 ⋅ 4 1 ⋅ 3 3 ) 1 / 3 ( 2 4 ⋅ 4 4 1 ⋅ 3 6 ⋅ 5 ) 1 / 4 ⋯ , {\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 n th factor is the n th root of the product
∏ k = 0 n ( k + 1 ) ( − 1 ) k + 1 ( n k ) , {\displaystyle \prod _{k=0}^{n}(k+1)^{(-1)^{k+1}{n \choose k}},} as well as the infinite product
e = 2 ⋅ 2 ( ln ( 2 ) − 1 ) 2 ⋯ 2 ln ( 2 ) − 1 ⋅ 2 ( ln ( 2 ) − 1 ) 3 ⋯ . {\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 < e 2 (which includes B = 2 , 3 , 4 , 5 , 6 , or 7 ), then
e = B ⋅ B ( ln ( B ) − 1 ) 2 ⋯ B ln ( B ) − 1 ⋅ B ( ln ( B ) − 1 ) 3 ⋯ . {\displaystyle e={\frac {B\cdot B^{(\ln(B)-1)^{2}}\cdots }{B^{\ln(B)-1}\cdot B^{(\ln(B)-1)^{3}}\cdots }}.} Also
e = lim n → ∞ ∏ k = 0 n ( n k ) 2 / ( ( n + α ) ( n + β ) ) ∀ α , β ∈ R {\displaystyle e=\lim \limits _{n\rightarrow \infty }\prod _{k=0}^{n}{n \choose k}^{2/{((n+\alpha )(n+\beta ))}}\ \forall \alpha ,\beta \in {\mathbb {R}}} As the limit of a sequence The number e is equal to the limit of several infinite sequences :
e = lim n → ∞ n ⋅ ( 2 π n n ! ) 1 / n {\displaystyle e=\lim _{n\to \infty }n\cdot \left({\frac {\sqrt {2\pi n}}{n!}}\right)^{1/n}} ande = lim n → ∞ n n ! n {\displaystyle e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}} (both by Stirling's formula ).The symmetric limit, [ 8]
e = lim n → ∞ [ ( n + 1 ) n + 1 n n − n n ( n − 1 ) n − 1 ] {\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 [ 9]
e = lim n → ∞ ( p n # ) 1 / p n e = lim n → ∞ n π ( n ) / n = lim n → ∞ n n / p n {\displaystyle {\begin{aligned}e&=\lim _{n\to \infty }(p_{n}\#)^{1/p_{n}}\\e&=\lim _{n\to \infty }n^{\pi (n)/n}\\&=\lim _{n\to \infty }n^{n/p_{n}}\end{aligned}}} where p n {\displaystyle p_{n}} is the n th prime , p n # {\displaystyle p_{n}\#} is the primorial of the n th prime, and π ( n ) {\displaystyle \pi (n)} is the prime-counting function .
Also:
e x = lim n → ∞ ( 1 + x n ) n . {\displaystyle 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 → ∞ ( 1 + 1 n ) n . {\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.} The ratio of the factorial n ! , that counts all permutations of an ordered set S with cardinality n , and the subfactorial (a.k.a. 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 → ∞ n ! ! n . {\displaystyle e=\lim _{n\to \infty }{\frac {n!}{!n}}.} As a limiting probability If we consider an event which has a probability of 1 n {\displaystyle {\frac {1}{n}}} of occurring in any one trial, then the probability of the event not occurring in n trials tends to 1/e . That is, lim n → ∞ ( 1 − 1 n ) n = 1 e {\displaystyle \lim _{n\to \infty }\left(1-{\frac {1}{n}}\right)^{n}={\frac {1}{e}}}
As a ratio of ratios A unique representation of e can be found within the structure of Pascal's triangle , as discovered by Harlan Brothers . Pascal's triangle is composed of binomial coefficients , which are traditionally summed to derive polynomial expansions. However, Brothers identified a product-based relationship between these coefficients that links to e . Specifically, the ratio of the products of binomial coefficients in adjacent rows of Pascal's triangle tends to e as the row number n increases:
P b ( n ) = ∑ k = 0 n log b ( n k ) A = P b ( n − 1 ) , B = P b ( n ) , C = P b ( n + 1 ) x = ( A − B ) + ( C − B ) ∼ 1 {\displaystyle {\begin{aligned}P_{b}(n)&=\sum _{k=0}^{n}\log _{b}{\binom {n}{k}}\\A&=P_{b}(n-1),B=P_{b}(n),C=P_{b}(n+1)\\x&=(A-B)+(C-B)\sim 1\\\end{aligned}}} For b = e , exp x ∼ e {\displaystyle b=e,\exp x\sim e} .
The details of this relationship and its proof are outlined in the discussion on the properties of the rows of Pascal's triangle . [ 11] [ 12]
In trigonometry Trigonometrically, e can be written in terms of the sum of two hyperbolic functions ,
e x = sinh ( x ) + cosh ( x ) , {\displaystyle e^{x}=\sinh(x)+\cosh(x),} at x = 1 .
Notes ↑ Sandifer, Ed (Feb 2006). "How Euler Did It: Who proved e is Irrational?" (PDF) . MAA Online. Retrieved 2017-04-23 . ↑ Brown, Stan (2006-08-27). "It's the Law Too — the Laws of Logarithms" . Oak Road Systems. Archived from the original on 2008-08-13. Retrieved 2008-08-14 . ↑ Formulas 2–7: H. J. Brothers, Improving the convergence of Newton's series approximation for e , The College Mathematics Journal , Vol. 35, No. 1, (2004), pp. 34–39. ↑ Formula 8: A. G. Llorente, A Novel Simple Representation Series for Euler's Number e , preprint, 2023. ↑ "e" , Wolfram MathWorld : ex. 17, 18, and 19, archived from the original on 2023-03-15 . ↑ J. Sondow, A faster product for pi and a new integral for ln pi/2 , Amer. Math. Monthly 112 (2005) 729–734. ↑ J. Guillera and J. Sondow, Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent , Ramanujan Journal 16 (2008), 247–270. ↑ H. J. Brothers and J. A. Knox, New closed-form approximations to the Logarithmic Constant e , The Mathematical Intelligencer , Vol. 20, No. 4, (1998), pp. 25–29. ↑ Ruiz, Sebastian Martin (1997). "81.27 A result on prime numbers" . The Mathematical Gazette . 81 (491). Cambridge University Press: 269. doi :10.2307/3619207 . ↑ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole Cengage Learning. p. 742. ↑ Brothers, Harlan (2012). "Pascal's Triangle: The Hidden Stor-e". The Mathematical Gazette . 96 : 145– 148. doi :10.1017/S0025557200004204 . ↑ Brothers, Harlan (2012). "Math Bite: Finding e in Pascal's Triangle". Mathematics Magazine . 85 (1): 51. doi :10.4169/math.mag.85.1.51 . This page is based on this
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