# 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 a fraction, 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[1] (sequence A003417 in OEIS):

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

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

$e = [ 1 ; \textbf{0.5} , 12 , 5 , 28 , 9 , 44 , 13 , 60 , 17 , \ldots , \textbf{4(4n-1)} , \textbf{4n+1} , \ldots]. \,$

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:

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

## 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:

$e = \sum_{k=0}^\infty \frac{1}{k!}$,[2] and
$e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k}{k!}$

Other series include the following:

$e = \left [ \sum_{k=0}^\infty \frac{1-2k}{(2k)!} \right ]^{-1}$ [3]
$e = \frac{1}{2} \sum_{k=0}^\infty \frac{k+1}{k!}$
$e = 2 \sum_{k=0}^\infty \frac{k+1}{(2k+1)!}$
$e = \sum_{k=0}^\infty \frac{3-4k^2}{(2k+1)!}$
$e = \sum_{k=0}^\infty \frac{(3k)^2+1}{(3k)!}$
$e = \left [ \sum_{k=0}^\infty \frac{4k+3}{2^{2k+1}\,(2k+1)!} \right ]^2$
$e = \sum_{k=1}^\infty \frac{k^n}{B_n(k!)}$ where $B_n$ is the $n^{th}$ Bell number. Some few examples: (for n = 2..7)
$e = \sum_{k=1}^\infty \frac{k^2}{2(k!)}$
$e = \sum_{k=1}^\infty \frac{k^3}{5(k!)}$
$e = \sum_{k=1}^\infty \frac{k^4}{15(k!)}$
$e = \sum_{k=1}^\infty \frac{k^5}{52(k!)}$
$e = \sum_{k=1}^\infty \frac{k^6}{203(k!)}$
$e = \sum_{k=1}^\infty \frac{k^7}{877(k!)}$

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

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

## 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 [4][5]

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

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,[6][7]

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

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]

$e= \lim_{n \to \infty}(p_n \#)^{1/p_n}$

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:

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

## In trigonometry

Trigonometrically, e can be written as the sum of two hyperbolic functions:

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

## Notes

1. Sandifer, Ed (Feb 2006). "How Euler Did It: Who proved e is Irrational?" (PDF). MAA Online. Retrieved 2010-06-18.
2. Brown, Stan (2006-08-27). "It’s the Law Too — the Laws of Logarithms". Oak Road Systems. Retrieved 2008-08-14.
3. 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.
4. J. Sondow, A faster product for pi and a new integral for ln pi/2, Amer. Math. Monthly 112 (2005) 729–734.
5. 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.
6. 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.
7. Khattri, Sanjay. "From Lobatto Quadrature to the Euler constant e" (PDF).
8. S. M. Ruiz 1997