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volume 7, issue 1, article 35, 2006.

Received 04 January, 2006;

accepted 18 January, 2006.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

AROUND APÉRY’S CONSTANT

WALTHER JANOUS

Ursulinengymnasium Fürstenweg 86 A-6020 Innsbruck Austria.

EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 020-06

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Around Apéry’s Constant Walther Janous

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J. Ineq. Pure and Appl. Math. 7(1) Art. 35, 2006

http://jipam.vu.edu.au

Abstract

In this note we deal with some aspects of Apéry’s constantζ(3).

2000 Mathematics Subject Classification:49J40, 90C33, 47H10.

Key words: Apéry’s constant, Harmonic numbers, Infinite sums, Integrals.

This research is partially supported by grant 04-39-3265-2/03 (11.XII.2003) of the Federal Ministry of Education and Sciences, Bosnia and Herzegowina.

Dedicated to Professor Gerd Baron on the occasion of his 65th birthday.

1. Introduction

Since Apéry’s miraculous proof (1979) that the valueζ(3)of Rieman’sζ-function is irrational, Apéry’s constant ζ(3) has been the focus of attention for many mathematicians. (An extensive list of results and references are found in Sec- tion 1.6 of the highly recommended encyclopedic book [2].)

It is the purpose of this note to extend some of these results. Thereby we will also obtain a new infinite sum rapidly converging toζ(3).

At the end of this note we raise two questions for further investigation.

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2. Two Multisums

Recently in [1] the proof of (2.1)

∞

X

i=1

∞

X

j=1

1

ij(i+j) = 2ζ(3),

whereζ(3) = 1.202056903...,was posed as a problem. Although this result was published earlier (see [2, p. 43]) it is worthwhile reconsidering in the following more general way.

Theorem 2.1. Forr ≥1the multisum

Sr =

∞

X

k1=1

· · ·

∞

X

kr=1

1

k₁· · ·k_r(k₁+· · ·+k_r)

attains the valuer!ζ(r+ 1).

Proof. Firstly we rewrite the multisum as an integral as follows

S_r =

∞

X

k1=1

· · ·

∞

X

kr=1

1 k₁· · ·k_r

Z 1

0

x^k¹^+···+k^r⁻¹dx

that is (upon interchanging of summation and integration), S_r =

Z 1

0

1 x

∞

X

k1=1

x^k¹ k₁ · · ·

∞

X

kr=1

x^k^r k_r dx.

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Due to

∞

X

j=1

x^j

j =−ln(1−x), we get

S_r = (−1)^r Z 1

0

ln(1−x)^r x dx.

Substitutingx= 1−tyields

S_r = (−1)^r Z 1

0

ln(t)^r 1−tdt.

This and the known result ([2, p. 47]) Z 1

0

ln(t)^r

1−tdt= (−1)^rr!ζ(r+ 1) readily yield the claim.

Subsequently we will deal with a ‘relative’ ofS_r, namely the multisum T_r =

∞

X

k1=1

· · ·

∞

X

kr=1

(−1)^k¹^+···+k^r k₁· · ·k_r(k₁+· · ·+k_r).

As it will turn out, matters are here more involved. Indeed, we will prove now

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Theorem 2.2. Forr ≥1,

(2.2) T_r = (−1)^r r!ζ(r+ 1)− r

r+ 1(ln 2)^r+1

−

∞

X

k=1 r

X

m=1

r(r−1)...(r−m+ 1)

2^kk^m+1 (ln 2)^r−m

!

holds.

Proof. Proceeding as in the previous proof we get

T_r = (−1)^r Z 1

0

ln(1 +x)^r

x dx.

Substitution ofx=e^−t−1yields T_r= (−1)^r

Z ln(1/2)

0

(−t)^r

e^−t−1(−e^−t)dt, that is,

T_r = Z 0

ln(1/2)

t^r 1−e^tdt.

Upon expanding _1−e¹ t as a geometric series we arrive at T_r =

∞

X

k=0

Z 0

−ln 2

t^re^ktdt.

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Integration by parts leads to the identity (we suppress integration constants) Z

t^re^ktdt=e^kt t^r k +

r

X

m=1

(−1)^mr(r−1)...(r−m+ 1) k^m+1 t^r−m

! ,

wherek >0.

Therefore a straightforward simplification yields

T_r= (−1)^r (ln 2)^r+1 r+ 1 +

∞

X

k=1

r!

k^r+1

−

∞

X

k=1

1 2^k

(ln 2)^r

k +

r

X

m=1

r(r−1)...(r−m+ 1)

k^m+1 (ln 2)^r−m

!!

.

Since

∞

X

k=1

1

k2^k = ln 2, we finally get the claimed identity (2.2).

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3. A New Formula for Apéry’s Constant

Theorem2.2enables us to obtain a new way to expressζ(3)by a fast converging series.

Indeed, lettingr = 2we get

T₂ = 2 ζ(3)−(ln 2)³

3 −

∞

X

k=1

1 2^k

ln 2 k² + 1

k³ !

.

Furthermore [2, p. 43], reports

T2 = 1 4ζ(3).

Therefore the following holds.

Theorem 3.1.

(3.1) ζ(3) = 8

7

(ln 2)³

3 +

∞

X

k=1

1 2^k

ln 2 k² + 1

k³ !

.

This formula should be compared with the following one (see [5]) ζ(3) = 2

3(ln 2)³+ 4

∞

X

k=1

(−1)^k+1 k³2^k ^2k_k.

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4. Further Observations

• FromS₂ +T₂ = ⁹₄ζ(3)we infer

2 X

i,j≥1 i+jeven

1

ij(i+j) = 9 4ζ(3) that is (we puti+j = 2k),

∞

X

k=1 2k−1

X

j=1

1

2(2k−j)jk = 9 8ζ(3),

i.e. ∞

X

k=1

1 2k

2k−1

X

j=1

1 2k

1

j + 1 2k−j

= 9 8ζ(3).

This can be summarized as

∞

X

k=1

1

k²H_2k−1 = 9 4ζ(3), where

H_n=

n

X

j=1

1 j denotes then-th harmonic number.

In a similar wayS₂−T₂ = ⁷₄ζ(3)implies the formula

∞

X

k=1

1

(2k+ 1)²H_2k = 7 16ζ(3).

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From these two formulae we get easily Theorem 4.1.

(4.1)

∞

X

k=1

1

(2k−1)²H2k−1 = 21 16ζ(3) and

(4.2)

∞

X

k=1

1

(2k)²H_2k= 11 16ζ(3).

Adding these two identities yields (4.3)

∞

X

k=1

1

k²H_k= 2ζ(3), a result already known to L. Euler.

• [4, p. 499, item 2.6.9.14] reads T₂ =

Z 1

0

ln(1 +x)²

x dx= 2

∞

X

k=1

(−1)^kψ(k)

k² − π²γ 6 ,

where ψ(z) = (ln Γ(z))⁰ and γ denote the digamma function and the Euler- Mascheroni constant, resp.

Therefore there holds the curious identity

∞

X

k=1

(−1)^kψ(k) k² = 1

8ζ(3) + π²γ 12 .

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Because ofψ(k) =−γ+Hk−1 it reads in equivalent form (4.4)

∞

X

k=1

(−1)^k

k² H_k = 5 8ζ(3)

• Recently [3] posed the problem of proving the identity

∞

X

n=0

1

(2n+ 1)³ = 7 Z π/4

0

ln(cosx) ln(sinx) cosxsinx dx.

We show that it implies a remarkable result for two doublesums.

Indeed, we firstly note

∞

X

n=0

1

(2n+ 1)³ =

∞

X

k=1

1 k³ −

∞

X

k=1

1 (2k)³, that is

∞

X

n=0

1

(2n+ 1)³ = 7 8ζ(3).

Therefore, the identity under consideration in fact means ζ(3) = 8

Z π/4

0

ln(cosx)

cosx · ln(sinx) sinx dx.

Lettingf(x) = ^ln(cos_cos_x^x) and settingz =π/2−xwe obtain Z π/4

0

f(x)fπ 2 −x

dx= Z π/2

π/4

fπ 2 −z

f(z)dz

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whence

ζ(3) = 4 Z π/2

0

ln(cosx)

cosx · ln(sinx) sinx dx.

Next, we substitutesinx=√ w.

Fromcosxdx= ₂^√¹_wdwandcosx=√

1−wwe getdx= ₂^√_w^√¹_1−wdw.

This in turn yields ζ(3) = 4

Z 1

0

ln(√

1−w)

√1−w ·ln(√

√ w)

w · 1

2√ w√

1−wdw, that is

ζ(3) = 1 2

Z 1

0

ln(1−w) 1−w · lnw

w dw.

Upon rewriting this as

ζ(3) = 1 2

Z 1

0

ln(1−w)

w · lnw 1−wdw

and developing the two factors of the integrand we get ζ(3) = 1

2 Z 1

0

−

∞

X

i=1

wⁱ⁻¹ i

!

−

∞

X

j=1

(1−w)^j−1 j

! dw,

that is

ζ(3) = 1 2

∞

X

i=1

∞

X

j=1

1 ij

Z 1

0

wⁱ⁻¹(1−w)^j−1dw.

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Keeping in mind that Z 1

0

wⁱ⁻¹(1−w)^j−1dw= (i−1)!(j −1)!

(i+j−1)! ,

we arrive at the formula

(4.5) ζ(3) = 1

2

∞

X

i=1

∞

X

j=1

1 ij² ^i+j−1_j .

Equation (4.5) andζ(3) = ¹₂S₂give the two noteworthy identities (4.6)

∞

X

i=1

∞

X

j=1

1

ij² ^i+j−1_j =

∞

X

i=1

∞

X

j=1

1 ij(i+j)

and (4.7)

Z 1

0

ln(1−z) z · lnz

1−zdz = Z 1

0

ln(1−z)

z ln(1−z)dz.

• Finally, Theorem 4.1 enables us to prove the following finite analogon of the initial formula (2.1) of the present note, namely

(4.8)

∞

X

i=1 i

X

j=1

1

ij(i+j) = 5 4ζ(3).

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Indeed, (4.2) and (4.3) imply

∞

X

k=1

1 k²

1

k+ 1 +· · ·+ 1 2k

= 3 4ζ(3).

However,

1 k²

1

k+ 1 +· · ·+ 1 2k

= 1 k

k

X

j=1

1 k(k+j) and

1 k

k

X

j=1

1

k(k+j) = 1 k

k

X

j=1

1 j

1 k − 1

k+j

= 1 k²Hk−

k

X

j=1

1 kj(k+j)

readily lead to

2ζ(3)−

∞

X

k=1 k

X

j=1

1

kj(k+j) = 3 4ζ(3) as claimed.

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5. Two Questions for Further Research

• The results of Theorem 4.1 may be regarded as special cases of the more general sums

S_a,b=

∞

X

k=1

1

(ak−b)²H_ak−b, where0≤b < aare entire numbers.

Problem 1. DetermineSa,bfora≥3in terms of ’familiar’ expressions.

• Let, in analogy toS_randT_r,U_r,sdenote the multisum

Ur,s =

∞

X

k1=1

· · ·

∞

X

kr=1

(−1)^k¹^+···+k^s k₁· · ·k_r(k₁+· · ·+k_r),

wherer≥1and0≤s≤r.

Problem 2. DetermineU_r,sin the spirit of Theorem2.2.

In other words evaluate the integrals

Ir,s= Z 1

0

ln(1−x)^sln(1 +x)^r−s

x dx

forr≥1and0≤s ≤r.

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References

[1] M. BENCZE, Problem 2984, Crux Math., 29 (2004), 431.

[2] S.R. FINCH, Mathematical Constants, Cambridge Univ. Press, Cambridge 2003.

[3] N. LORD, Problem 89.D, part (b), Math. Gaz., 89 (2005), 115.

[4] A.P. PRUDNIKOV et al., Integrals and Series (Elementary Functions) (in Russian), Nauka, Moscow 1981.

[5] E.W. WEISSTEIN, Apéry’s Constant, MathWorld–A Wolfram Web Re- source. http://mathworld.wolfram.com/AperysConstant.

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