Introduction on Bernoulli's numbers
(Click
here for a Postscript version of this page.)
1 Introduction
Bernoulli's numbers play an important and quite mysterious role in
mathematics and in various places like analysis, number theory and
differential topology. They first appeared in Ars Conjectandi,
page 97, a famous (and posthumous) treatise published in 1713, by Jakob
Bernoulli (16541705) when he studied the sums of powers of consecutive
integers
s_{p}(n)=

n1
å
k=1

k^{p},


(1)

where p and n are two given positive integers.
Bernoulli's numbers also appear in the computation of the
numbers
and in the expansion of many usual functions as tan(x), tanh(x), 1/sin(x),
¼
Perhaps one of the most important result is EulerMaclaurin
summation formula, where Bernoulli's numbers are contained and which
allows to accelerate the computation of slow converging series (see the
essay on Euler's constant at
[9]). They also appear in
numbers theory (Fermat's theorem) and in many other domains and have
caused the creation of a huge literature (see the 2700 and more entries
enumerated in [6]).
According to Louis Saalschültz
[17], the term
Bernoulli's numbers was used for the first time by Abraham De
Moivre (16671754) and also by Leonhard Euler (17071783) in 1755.
2 Bernoulli's approach
During the first years of the Calculus period and in its first integral
computations of the function x® x^{p},
Pierre de Fermat (16011665) in 1636 had to evaluate the sums
s_{p}(n) defined by (1). You can see
this by replacing the area under the curve x®
x^{p} by it's rectangular approximations and naturally comes the
need to compute s_{p}(n).
Also in 1631, Johann Faulhaber (15801635) developed explicit formulas
for these sums up to p=17 (read the excellent
[12] for the beginnings of
integration and [18] for some
excerpts of Bernoulli's work). Thus, it was already known to Jakob
Bernoulli that






1
2

n^{2}

1
2

n=

n(n1)
2






1
3

n^{3}

1
2

n^{2}+

1
6

n=

n(n1)(2n
1)
6






1
4

n^{4}

1
2

n^{3}+

1
4

n^{2}=

n^{2}(n1)^{2}
4






1
5

n^{5}

1
2

n^{4}+

1
3

n^{3}

1
30

n=

n(n1)(2n
1)(3n^{2}
3n 1)
30






1
6

n^{6}

1
2

n^{5}+

5
12

n^{4}

1
12

n^{2}=

n^{2}(2n^{2}2n
1)(n1)^{
2}
12







Jakob Bernoulli then, empirically, noticed that the polynomials
s_{p}(n) have the form
s_{p}(n)=

1
p+1

n^{p+1}

1
2

n^{p}+

p
12

n^{p1}+0×n^{p
2}+...


In this expression, the numbers
(1,1/2,1/12,0,...) are appearing and do not
depend on p. More generally, the sums s_{p}(n) can be written in
the form



p
å
k=0


B_{k}
k!


p!
(p+1k)!

n^{p+1k}





B_{0}
0!


n^{p+1}
p+1

+

B_{1}
1!

n^{p}+

B_{2}
2!

pn^{p1}+

B_{3}
3!

p(p1)n^{p
2}+...+

B_{p}
1!

n



where the B_{k}are numbers which are independent of p and called
Bernoulli's numbers.
We find by identification
B_{0}=1,B_{1}=

1
2

,B_{2}=

1
6

,B_{3}=0,...


To illustrate the usefulness of his formula, Bernoulli computed the
astonishing value of s_{10}(1000) with little effort (in less
than ''half a quarter of an hour'' he says ...
[18])
91409924241424243424241924242500


(you can check it !). To achieve this he needed to find B_{0}up to
B_{10}.
There is good evidence that the famous Japanese mathematician, Seki
Takakazu (16421708) also discovered Bernoulli's numbers at the same
time. The famous Indian mathematician Srinivasa Ramanujan (18871920)
independently studied and rediscovered those numbers in 1904. He wrote
one of his first article on this subject in 1911
[15].
3 A more modern definition
An equivalent definition of the Bernoulli's numbers is obtained from the
series expansion of the function
z/(e^{z}1):
G(z)=

z
e^{z}1

=

¥
å
k=0

B_{k}

z^{k}
k!

 z <
2p


(2)

In other words, the generating function of the Bernoulli's numbers
B_{k} is z/(e^{z}1). The
first terms of the expansion of this function are
G(z)=

æ
è

1+

z
2!

+

z^{2}
3!

+

z^{3}
4!

+...

ö
ø

1

=1

z
2

+

z^{2}
12



z^{4}
720

+

z^{6}
30240



z^{8}
1209600

+...


which permit to obtain the first value of the Bernoulli's numbers:


B_{1}=

1
2

, B_{2}=

1
6

, B_{3}=0,




B_{5}=0,
B_{6}=

1
42

, B_{7}=0,
B_{8}=

1
30

.



Further, we observe that
G(z)+

z
2

=

z
2


æ
è

2
e^{z}1

+1

ö
ø

=

z
2


e^{z/2}+e^{z/2}
e^{z/2}e^{
z/2}

=

z
2

coth

z
2

,


where coth is the hyperbolic tangent, hence G(z)+z/2 is an even function
and consequently every Bernoulli's numbers of the form B_{2k+1}(k
> 0) is null.
3.1 Bernoulli's polynomials
With a little modification it's possible to define Bernoulli's
polynomials B_{k}(x) by
G(z,x)=

ze^{zx}
e^{z}1

=

¥
å
k=0

B_{k}(x)

z^{k}
k!



and G(z,0)=G(z) hence the value of the function B_{k}(x) at x=0 is
B_{k}and because

¶G
¶x

(z,x)=zG(z,x)=

¥
å
k=0


dB_{k}
dx

(x)

z^{k}
k!



it follows the important relation

dB_{k}
dx

(x)=kB_{k1}(x).


Then, it's easy to deduce that B_{k}(x) are polynomials of degree
k, and the first one are














x^{4}2x^{3}+x
^{2}

1
30





x^{5}

5
2

x^{4}+

5
3

x^{3}

1
6

x






Thanks to Bernoulli's polynomials, it's possible to rewrite the
expression of the sums s_{p}(n) as
s_{p}(n)=

n1
å
k=0

k^{p}=

1
p+1

(
B_{p+1}(n)B_{p+1}).


There are many relations with these polynomials, for
example








 B_{2k}
k=1,2,... and 0 < x < 1,




(12
^{1k})B_{k}
k=0,1,...






Consult [1] for other
formulas.
4 Properties
4.1 Recurrence relation
In the expression
s_{p}(n)=

p
å
k=0


B_{k}
k!


p!
(p+1k)!

n^{p+1k}


we let n=1, giving
0=

p
å
k=0


B_{k}
k!


1
(p+1k)!



or equivalently
B_{p}=

1
p+1


p1
å
k=0

\binomp+1kB_{k}.


(3)

The recurrence relation (3) allows an easy
generation of Bernoulli's numbers and shows that the numbers
B_{p} are all rational numbers. It's convenient to rewrite this
relation as the symbolic equation
and expand the binomial (B+1)^{p+1}where each power
B^{k}must be replaced by B_{k}.
Example 1 With p=4 we have
5B_{4}+10B_{3}+10B_{2}+5B_{1}+B
_{0}=0


thus
5B_{4}+

æ
è

0+10.

1
6

5.

1
2

+1

ö
ø

=0
therefore
B_{4}=

1
30

.


4.2 Bernoulli's numbers and the zeta
function
In 1735, the solution of the Basel problem, expressed by Jakob
Bernoulli some years before, was one of Euler's most sensational
discovery. The problem was to find the limit of
he found it to be p^{2}/6. He also
discovered the values of the sums
for k up to 13 ([8] and for more
details [7]).
It's an extraordinary result that z(2k) can be
expressed with Bernoulli's numbers ; the values of these sums are given
by
z(2k)=

¥
å
n=1


1
n^{2k}

=

4^{k}B_{2k}
p^{2k}
2(2k)!

k > 0,


(4)

(a proof based on two different expansions of z cot(z) is given in
[4] p. 383). No similar expression is
known for the odd values of the Zeta function.
On the extension of this function to negative values, we also
have
z(12k)=


B_{2k}
2k

k > 0,


which may also be used to compute B_{2k}(see
[5]).
4.3 Asymptotic expansion of
Bernoulli's numbers
From the previous relation (4) with the Zeta
function, it's clear that
 B_{2k} =

2(2k)!
(2p)^{2k}

z(2k)


(5)

and because, when k becomes large, thanks to Stirling's formula
we have
 B_{2k} ~ 4

æ
è

k
pe

ö
ø

2k


Ö

pk

.


In [16], the following
results describes how the numerator N_{2k} of B_{2k}
grows with k:
log N_{2k} = 2klog(k)+O(k).


4.4 Bounds
It may be useful to estimate bounds for B_{2k}, to achieve this
we use the following relation between the function
z(s) and the alternating series
z_{a}(s)






¥
å
n=1


(1)^{n
1}
n^{s}




and since
1

1
2^{s}

< z_{a}(s) < 1,


we have the bounds for z(s)

12^{s}
12^{1s}

< z(s) <

1
12^{1s}

.


If we use this relation with (5), the bounds for
B_{2k} are therefore

2(2k)!(14^{
k})
(2p)^{2k}(1
2.4^{k})

<  B_{2k} <

2(2k)!
(2p)^{2k}(1
2.4^{k})

.


5 Clausenvon Staudt's theorem
The following famous and important theorem was published in 1840 by Karl
von Staudt (17981867) and it allows to compute easily the fractional
part of Bernoulli's numbers (thus it also permits to compute the
denominator of those numbers). This theorem was discovered the same year,
independently, by Thomas Clausen (18011885).
Theorem 2 The value B_{2k}, added to the sum of the
inverse of prime numbers p such that (p1)
divides 2k, is an integer. In other words,
B_{2k}
º

å
(p1)
 2k


1
p

mod 1


Proof. A complete proof is given in
[11], p. 91.
When k > 1, we observe that the primes p= 2,3 are such as
(p1) divides 2k. Let's illustrate this
theorem with a few examples. For k=1, it becomes



å
(p1)
2


1
p

=

1
2

+

1
3

=

5
6

mod 1






for k=5



å
(p1)
10


1
p

=

1
2

+

1
3

+

1
11

=

61
66

mod 1






and for k=8



å
(p1)
16


1
p

=

1
2

+

1
3

+

1
5

+

1
17

=

47
510

mod 1






Corollary 3 (due to Rado 1934) For every prime numbers k of
the form 3n+1
Proof. It's an easy consequence of the Staudt's theorem since
p1 divides 2k=2(3n+1) only if
p1 is one of 1,2,3n+1,6n+2, that is p is one
of 2,3,3n+2,6n+3. But 6n+3 is divisible by 3 and 3n+2 is divisible by 2
because 3n+1 is prime so the only primes p candidates are 2 and 3.
Example 4 The first primes of the form 3n+1 are
7,13,19,31,37,43,61,67,... hence we have
B_{14} º B_{26}
º B_{38}
º B_{62}
º B_{74}
º B_{86}
º B_{122}
º B_{134}
º

1
6

mod 1.


Clausenvon Staudt's theorem also permits to compute exactly a
Bernoulli's number as soon as a sufficiently good approximation of it is
known.
6 Expansion of usual functions
In a previous section we gave the definition

z
e^{z}1

=

¥
å
k=0

B_{k}

z^{k}
k!

z <
2p


and the consequence

z
2

coth

z
2

=

¥
å
k=0

B_{2k}

z^{2k}
(2k)!

.


It obviously leads to the two following expansions



¥
å
k=0

4^{k}B_{2k}

z^{2k}
(2k)!

=1+

z^{2}
3



z^{4}
45

+

2z^{6}
945

...
 z < p





¥
å
k=0

(4)^{k}B_{2k}

z^{2k}
(2k)!

=1

z^{2}
3



z^{4}
45



2z^{6}
945

...
 z < p



where cot(z)=cos(z)/sin(z)=icoth(iz) is the cotangent function.
Now it's possible to find the expansion for tanh(z) and tan(z), if we
observe that
2coth(2z)coth(z)=2

cosh(2z)
sinh(2z)



cosh(z)
sinh(z)

=

cosh^{2}(z)+sinh^{2}(z)
sinh(z)cosh(z)



cosh(z)
sinh(z)

=tanh(z)


so that



¥
å
k=1

4^{k}(4^{k}1)B
_{2k}

z^{2k1}
(2k)!

=z

z^{3}
3

+

2z^{5}
15



17z^{7}
315

+... z <

p
2






¥
å
k=1

(4)^{k}(1
4^{k})B_{2k}

z^{2k1}
(2k)!

=z+

z^{3}
3

+

2z^{5}
15

+

17z^{7}
315

+... z <

p
2

.



Bernoulli's numbers also occur in the expansions of other classical
functions

z
sin(z)

,

z
sinh(z)

,log

æ
è

sin(z)
z

ö
ø

,log(cos(z)),log

æ
è

tan(z)
z

ö
ø

,...


6.1 Series
Setting z=1 or z=1 in the expansion
(2) of
z/(e^{z}1) leads to the fast
converging series

1
e1

=

¥
å
k=0


B_{k}
k!

=

1
2

+

¥
å
k=1


B_{2k}
(2k)!



and

B_{2k}
(2k)!

~

2
(4p^{2})^{k}

»

2
39.48^{k}

.


In one of his famous notebook, Ramanujan stated without proof the
following result:
Theorem 5 Let ( a,b) two positive real numbers such as
ab=p^{2}, let n > 1 an integer,
then
a^{n}

¥
å
k=1


k^{2n1}
e^{2ak}1

(b)
^{n}

¥
å
k=1


k^{2n1}
e^{2bk}1

=(
a^{n}(b)
^{n})

B_{2n}
4n

.


Proof. See [3].
Corollary 6 Let n ³ 1,
then

¥
å
k=1


k^{4n+1}
e^{2pk}1

=

B_{4n+2}
8n+4

.


Proof. Just apply the theorem with
a=b=p and replace n by 2n+1.
It's interesting to compare this result with the classical integral
representation valid for n ³ 1

ó
õ

¥
0


x^{2n1}
e^{2px}1

dx=(1)^{n1}

B_{2n}
4n



which implies that for n ³0

ó
õ

¥
0


x^{4n+1}
e^{2px}1

dx=

B_{4n+2}
8n+4

.


7 EulerMaclaurin formula
Let f(x) be a function of class C^{2p+2} on an interval [a,b] and
let h=(ba)/m a subdivision of this interval
into m equal parts then we have the important result:
Theorem 7 There exist 0 < J <
1 and



1
h


ó
õ

b
a

f(x)dx+

1
2

(f(a)+f(b)) +

p
å
k=1


h^{2k1}
(2k)!

B_{2k}(f^{(2k1)}(b)
f^{(2k1)}(a))




+

h^{2p+2}
(2p+2)!

B_{2p+2}

m1
å
k=0

f^{(2p+2)}(a+kh+Jh).



Proof. A proof is given in
[10].
This formula was first studied by Euler in 1732 and independently by
Colin Maclaurin (16981746) in 1742
[13]. Euler used it to
compute sums of slow converging series and Maclaurin used it as a
numerical quadrature formula.
With the same conditions, setting n=m+1,a=1,b=n,h=1 the theorem becomes:
Theorem 8

n
å
k=1

f(k)=

ó
õ

n
1

f(x)dx+

1
2

( f(1)+f(n))+

p
å
k=1


B_{2k}
(2k)!

( f^{(2k1)}(n)
f^{(2k1)}(1))+R
_{n}(f,p),


where R_{n}(f,p) is the remainder bounded by
R_{n}(f,p) £

2
(2p)^{2p}


ó
õ

n
1

 f^{(2p+1)}(x)dx


7.1 Applications

f(x)=x^{2}, the remainder is null since f^{(p)}(x)=0
for p > 2:



ó
õ

n
1

x^{2}dx+

1
2

( 1+n^{2}) +

B_{2}
2

(2n2)+0





n^{3}1
3

+

1+n^{2}
2

+

n1
6

=

n(n+1)(2n+1)
6

.




f(x)=1/x,
f^{(2k1)}(x)=(2k
1)!/x^{2k}, EulerMaclaurin formula
yields for a given p:

n
å
k=1


1
k

log(n)=

1
2

+

1
2n

+

p
å
k=1


B_{2k}
2k


æ
è

1

1
n^{2k}

ö
ø

+R_{n}(f,p)


when n® ¥, the
left hand side of the equality tends to g
(Euler's constant) and the equality gives
g =

1
2

+

p
å
k=1


B_{2k}
2k

+R_{¥}(f,p),


finally

n
å
k=1


1
k

log(n)=g+

1
2n



p
å
k=1


B_{2k}
2k


1
n^{2k}

+(
R_{n}(f,p)R_{
¥}(f,p)) .


(check that
R_{n}(f,p)R_{
¥}(f,p) =O(1/n^{2p+2}) )

f(x)=log(x), with the same method (left as exercise) EulerMaclaurin
formula becomes

n
å
k=1

log(k)=nlog(n)n+

log(n)
2

+log(

Ö

2p

)+

p
å
k=1


B_{2k}
2k(2k1)


1
n^{2k1}

+O

æ
è

1
n^{2p+1}

ö
ø



so, for example, with p=3
log(n!)=nlog(n)n+log(

Ö

2pn

)+

1
12n



1
360n^{3}

+

1
1260n^{5}

+O

æ
è

1
n^{7}

ö
ø



and taking the exponential
n!=n^{n}e^{n}

Ö

2pn

exp

æ
è

1
12n



1
360n^{3}

+

1
1260n^{5}

+O

æ
è

1
n^{7}

ö
ø

ö
ø

.


This is the asymptotic Stirling formula. Using the series
expansion of the exponential function near the origin, it's more
convenient to write it as
n!=n^{n}e^{n}

Ö

2pn


æ
è

1+

1
12n

+

1
288n^{2}



139
51840n^{3}



571
2488320n^{4}

+

163879
209018880n^{5}

+O

æ
è

1
n^{6}

ö
ø

ö
ø



8 Bernoulli's numbers and Fermat's last
theorem
The famous Fermat's last theorem states that the
equation
never has nonzero integer solutions for n > 2. Since Fermat expressed
this result around 1630, the pursuit of a proof occupied generations of
mathematicians.
A big step was made in 1850 by Ernst Kummer (18101893) when he proved
Fermat's theorem for n=p, whenever p is, what is called today, a
regular prime. Kummer gave the beautiful regularity
criterion:
p is
a regular
prime if
and only
if p does
not
divide the
numerator
of B_{2},B_{4},...,B
_{p3}.


He showed that all primes before 37 where regular, hence Fermat's theorem
was proved for those primes. 37 is the first non regular prime because it
divides the numerator of
B_{32}=

7709321041217
510

=

208360028141×37
510

.


The next irregular primes (less than 300) are
59,67,101,103,131,149,157,233,257,263,271,283,293,...


For example, 157 divides the numerators of B_{62}and
B_{110}.
Thanks to arithmetical properties of Bernoulli's numbers, Johann Ludwig
Jensen (18591925) proved in 1915 that the number of irregular
primes is infinite. Even if it's probable that the number of regular
primes is infinite, a proof remains unknown
[16].
9 The first Bernoulli's numbers
9.1 First numbers
Here is the list of the first Bernoulli's numbers. Except for
B_{1} numbers of the form B_{2k+1} are
null.



























































26315271553053477373/1919190







261082718496449122051/13530






More numbers are given in
[1] and in
[17].
9.2 Some computations
Bernoulli himself computed the numbers that now bear his name up to
B_{10}. Later, Euler computed these numbers up to B_{30},
then Martin Ohm extended the calculation up to B_{62} in 1840
[14]. A few years later, in 1877, Adams
made the impressive computation of all Bernoulli's numbers up to
B_{124} (or B_{62}^{*}
according to his convention)
[2]. For instance, the
numerator of B_{124} has 110 digits and the denominator is the
number 30.
In 1996, Simon Plouffe and Greg J. Fee computed B_{200000} a huge
number of about 800000 digits, the computation took about 2 hours on a
work station. In 2002, the same authors improved the record to
B_{600000} which has 2727474 digits by a 12 hours computation on
a personal computer. The method is based on the formula
(5) which allow a direct computation of the
required number without the need to compute the previous ones.
References

[1]

M. Abramowitz and I. Stegun, Handbook of Mathematical
Functions, Dover, New York, (1964)

[2]

J.C. Adams, On the calculation of Bernoulli's numbers up to
B_{62} by means of Staudt's theorem, Rep. Brit.
Ass., (1877)

[3]

B.C. Berndt, Ramanujan's Notebooks, Part II, SpringerVerlag,
New York, (1989)

[4]

J.M. Borwein and P.B. Borwein, Pi and the AGM  A study in Analytic
Number Theory and Computational Complexity, A WileyInterscience
Publication, New York, (1987)

[5]

P. Borwein, An efficient algorithm for the Riemann Zeta
function, (1995)

[6]

K. Dilcher, A Bibliography of Bernoulli Numbers, World Wide
Web site at the address:
http://www.mscs.dal.ca/~dilcher/bernoulli.html

[7]

W. Dunham, Euler The Master of Us All, The Mathematical
Association of America, (1999)

[8]

L. Euler, Introduction à l'analyse infinitésimale (french
traduction by Labey), Barrois, ainé, Librairie, (original 1748,
traduction 1796), vol. 1

[9]

X. Gourdon and P. Sebah, Numbers, Constants and
Computation, World Wide Web site at the address:
http://numbers.computation.free.fr/Constants/constants.html,
(1999)

[10]

R.L. Graham, D.E. Knuth and O. Patashnik, Concrete
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[11]

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