proof of values of the Riemann zeta function in terms of Bernoulli numbers

proof of values of the Riemann zeta function in terms of Bernoulli numbers

This article proves part of the

theorem

given in the article.

Theorem 1

.

For any

positive integer

n

ζ

⁢

(

2

⁢

n

)

=

(

2

⁢

π

)

2

⁢

n

⁢

|

B

2

⁢

n

|

2

⁢

(

2

⁢

n

)

!

where

B

2

⁢

n

is the

2

⁢

n

th

Bernoulli number

.

Proof.

The method is as follows. Using

Fourier series

together with

induction

on

n

, we derive a

formula

for the

Bernoulli periodic function

B

2

⁢

n

⁢

(

x

)

involving an infinite sum. On setting

x

to

0

, this sum reduces to a constant times the appropriate

zeta function

, and the result follows.

We first compute the Fourier series for

B

2

⁢

(

x

)

.

B

2

⁢

(

x

)

is periodic with period

1

, so

c

n

=

∫

0

1

B

2

⁢

(

x

)

⁢

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

⁢

𝑑

x

=

∫

0

1

x

2

⁢

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

⁢

𝑑

x

-

∫

0

1

x

⁢

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

⁢

𝑑

x

+

1

6

⁢

∫

0

1

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

⁢

𝑑

x

We have

∫

0

1

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

⁢

𝑑

x

=

0

∫

0

1

x

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

d

x

=

-

1

2

⁢

π

⁢

n

x

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

|

0

1

+

1

2

⁢

π

⁢

i

⁢

n

∫

0

1

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

d

x

=

i

2

⁢

π

⁢

n

∫

0

1

x

2

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

d

x

=

-

1

2

⁢

π

⁢

i

⁢

n

x

2

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

|

0

1

+

2

2

⁢

π

⁢

i

⁢

n

∫

0

1

x

e

-

2

⁢

π

⁢

i

⁢

n

⁢

x

d

x

=

1

2

⁢

π

2

⁢

n

2

+

i

2

⁢

π

⁢

n

so that

c

n

=

1

2

⁢

π

2

⁢

n

2

But then

b

n

=

c

n

-

c

-

n

=

0

for all

n

,

a

0

=

0

, and for

n

>

0

,

a

n

=

c

n

+

c

-

n

=

1

π

2

⁢

n

2

(where

a

n

are the

coefficients

of

cos

and

b

n

the coefficients of

sin

in the Fourier series). Thus

B

2

⁢

(

x

)

=

∑

k

=

1

∞

1

π

2

⁢

k

2

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

=

1

π

2

⁢

∑

k

=

1

∞

1

k

2

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

Using this case as an

inductive hypothesis

, assume that for some

n

≥

2

B

2

⁢

(

n

-

1

)

⁢

(

x

)

=

(

-

1

)

n

⁢

2

⋅

(

2

⁢

(

n

-

1

)

)

!

(

2

⁢

π

)

2

⁢

(

n

-

1

)

⁢

∑

k

=

1

∞

1

k

2

⁢

(

n

-

1

)

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

Then on

(

0

,

1

)

B

2

⁢

n

′′

⁢

(

x

)

=

(

2

⁢

n

)

⁢

(

2

⁢

n

-

1

)

⁢

B

2

⁢

(

n

-

1

)

⁢

(

x

)

=

(

-

1

)

n

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

(

n

-

1

)

⁢

∑

k

=

1

∞

1

k

2

⁢

(

n

-

1

)

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

and thus

B

2

⁢

n

⁢

(

x

)

=

(

-

1

)

n

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

(

n

-

1

)

⁢

∫

∫

∑

k

=

1

∞

1

k

2

⁢

(

n

-

1

)

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

⁢

d

⁢

x

⁢

d

⁢

x

Since

n

≥

2

, the sum

converges absolutely

, so we can move the sum outside the

integrals

, and we get

B

2

⁢

n

⁢

(

x

)

=

(

-

1

)

n

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

(

n

-

1

)

⁢

∑

k

=

1

∞

1

k

2

⁢

(

n

-

1

)

⁢

∫

∫

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

⁢

𝑑

x

⁢

𝑑

x

=

(

-

1

)

n

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

(

n

-

1

)

⁢

∑

k

=

1

∞

1

k

2

⁢

(

n

-

1

)

⁢

-

1

4

⁢

π

2

⁢

k

2

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

=

(

-

1

)

n

+

1

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

n

⁢

∑

k

=

1

∞

1

k

2

⁢

n

⁢

cos

⁡

(

2

⁢

π

⁢

k

⁢

x

)

Thus we have established this formula for all

n

≥

1

. Setting

x

=

0

, then, we get

B

2

⁢

n

=

(

-

1

)

n

+

1

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

n

⁢

∑

k

=

1

∞

1

k

2

⁢

n

=

(

-

1

)

n

+

1

⁢

2

⋅

(

2

⁢

n

)

!

(

2

⁢

π

)

2

⁢

n

⁢

ζ

⁢

(

2

⁢

n

)

or, trivially rewriting,

ζ

⁢

(

2

⁢

n

)

=

(

-

1

)

n

+

1

⁢

(

2

⁢

π

)

2

⁢

n

⁢

B

2

⁢

n

2

⁢

(

2

⁢

n

)

!

But clearly

ζ

⁢

(

2

⁢

n

)

>

0

for

n

≥

1

, so it must be that the

B

2

⁢

n

alternate in sign, and thus

ζ

⁢

(

2

⁢

n

)

=

(

2

⁢

π

)

2

⁢

n

⁢

|

B

2

⁢

n

|

2

⁢

(

2

⁢

n

)

!

Note that as a effect of this proof, we see that the even-index Bernoulli numbers alternate in sign!

Title

proof of values of the

Riemann zeta function

in terms of Bernoulli numbers

Canonical name

ProofOfValuesOfTheRiemannZetaFunctionInTermsOfBernoulliNumbers

Date of creation

2013-03-22 17:46:37

Last modified on

2013-03-22 17:46:37

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Proof

Classification

msc 11M99

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