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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
The number e can be represented as an infinite simple continued fraction (sequence A003417 in the OEIS ):
e
=
[
2
;
1
,
2
,
1
,
1
,
4
,
1
,
1
,
6
,
1
,
1
,
8
,
1
,
…
,
1
,
2n
,
1
,
…
]
{\displaystyle e=[2;1,{\textbf {2}},1,1,{\textbf {4}},1,1,{\textbf {6}},1,1,{\textbf {8}},1,\ldots ,1,{\textbf {2n}},1,\ldots ]\,}
Here are some infinite generalized continued fraction expansions of e . The second of these can be generated from the first by a simple equivalence transformation . The third one – with ... 6, 10, 14, ... in it – converges very quickly.
e
=
2
+
1
1
+
1
2
+
2
3
+
3
4
+
4
⋱
e
=
2
+
2
2
+
3
3
+
4
4
+
5
5
+
6
⋱
{\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {2}{3+{\cfrac {3}{4+{\cfrac {4}{\ddots }}}}}}}}}}\qquad e=2+{\cfrac {2}{2+{\cfrac {3}{3+{\cfrac {4}{4+{\cfrac {5}{5+{\cfrac {6}{\ddots \,}}}}}}}}}}}
e
=
1
+
2
1
+
1
6
+
1
10
+
1
14
+
1
⋱
{\displaystyle e=1+{\cfrac {2}{1+{\cfrac {1}{6+{\cfrac {1}{10+{\cfrac {1}{14+{\cfrac {1}{\ddots \,}}}}}}}}}}}
e
2
m
/
n
=
1
+
2
m
(
n
−
m
)
+
m
2
3
n
+
m
2
5
n
+
m
2
7
n
+
m
2
⋱
{\displaystyle e^{2m/n}=1+{\cfrac {2m}{(n-m)+{\cfrac {m^{2}}{3n+{\cfrac {m^{2}}{5n+{\cfrac {m^{2}}{7n+{\cfrac {m^{2}}{\ddots \,}}}}}}}}}}}
Setting m =x and n =2 yields
e
x
=
1
+
2
x
(
2
−
x
)
+
x
2
6
+
x
2
10
+
x
2
14
+
x
2
⋱
{\displaystyle e^{x}=1+{\cfrac {2x}{(2-x)+{\cfrac {x^{2}}{6+{\cfrac {x^{2}}{10+{\cfrac {x^{2}}{14+{\cfrac {x^{2}}{\ddots \,}}}}}}}}}}}
As an infinite series
The number e is also equal to the sum of the following infinite series :
e
=
[
∑
k
=
0
∞
(
−
1
)
k
k
!
]
−
1
{\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}\right]^{-1}}
e
=
[
∑
k
=
0
∞
1
−
2
k
(
2
k
)
!
]
−
1
{\displaystyle e=\left[\sum _{k=0}^{\infty }{\frac {1-2k}{(2k)!}}\right]^{-1}}
[ 1]
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
)
!
{\displaystyle e=\sum _{k=0}^{\infty }{\frac {(3k)^{2}+1}{(3k)!}}}
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
=
−
12
π
2
[
∑
k
=
1
∞
1
k
2
cos
(
9
k
π
+
k
2
π
2
−
9
)
]
−
1
/
3
{\displaystyle e=-{\frac {12}{\pi ^{2}}}\left[\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\ \cos \left({\frac {9}{k\pi +{\sqrt {k^{2}\pi ^{2}-9}}}}\right)\right]^{-1/3}}
e
=
∑
k
=
1
∞
k
2
2
(
k
!
)
{\displaystyle e=\sum _{k=1}^{\infty }{\frac {k^{2}}{2(k!)}}}
e
=
∑
k
=
1
∞
k
2
(
k
−
1
)
!
{\displaystyle e=\sum _{k=1}^{\infty }{\frac {k}{2(k-1)!}}}
e
=
∑
k
=
1
∞
k
3
5
(
k
!
)
{\displaystyle e=\sum _{k=1}^{\infty }{\frac {k^{3}}{5(k!)}}}
e
=
∑
k
=
1
∞
k
4
15
(
k
!
)
{\displaystyle e=\sum _{k=1}^{\infty }{\frac {k^{4}}{15(k!)}}}
e
=
∑
k
=
1
∞
k
n
B
n
(
k
!
)
{\displaystyle e=\sum _{k=1}^{\infty }{\frac {k^{n}}{B_{n}(k!)}}}
where
B
n
{\displaystyle B_{n}}
is the
n
t
h
{\displaystyle n^{th}}
Bell number .
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 [ 2]
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 }}.}
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}}
and
e
=
lim
n
→
∞
n
n
!
n
{\displaystyle e=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}}
(both by Stirling's formula ).
The symmetric limit,
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]}
[ 3]
may be obtained by manipulation of the basic limit definition of e . Another limit is
e
=
lim
n
→
∞
(
p
n
#
)
1
/
p
n
{\displaystyle e=\lim _{n\to \infty }(p_{n}\#)^{1/p_{n}}}
[ 4]
where
p
n
{\displaystyle p_{n}}
is the n th prime and
p
n
#
{\displaystyle p_{n}\#}
is the primorial of the n th prime.
Finally, the famous sentence:
e
=
lim
n
→
∞
(
1
+
1
n
)
n
{\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}}
Notes
^ 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; pages 34-39.
^ J. Sondow, A faster product for pi and a new integral for ln pi/2, Amer. Math. Monthly 112 (2005) 729-734.
^ H. J. Brothers and J. A. Knox, New closed-form approximations to the Logarithmic Constant e. The Mathematical Intelligencer , Vol. 20, No. 4, 1998; pages 25-29.
^ S. M. Ruiz 1997