Elementary methods for evaluating Jordan's sums and analogous Euler's type sums and for setting a sigma sum theorem

0.18 MB·English

by  Guy Bastien

1 F´evrier 2013
Elementary methods for evaluating Jordan’s sums
1 1 1 1 1 1
1+ +···+ and 1+ +···+
3 2n−1 n2a 3 2n−1 (2n−1)2a
nX≥1(cid:16) (cid:17) nX≥1(cid:16) (cid:17)
and analogous Euler’s type sums and for setting a σ-sum theorem
3
1
0
2 G. Bastien

n
a
J Institut Math´ematique de Jussieu et CNRS,

1 Universit´e Pierre et Marie Curie, 4 place Jussieu 75252 PARIS CEDEX 5
3
email adress: [email protected]

]
T
Abstract
N
h. Our aim is first to calculate the following sums :
t
a 1 1 1 1 1 1
m 1+ +···+ and 1+ +···+ (J)
3 2n−1 n2a 3 2n−1 (2n−1)2a
[ nX≥1(cid:16) (cid:17) nX≥1(cid:16) (cid:17)

1 when a is an integer ≥ 1, by mean of double sum methods or integral representation. The
v
values of sum in (J) are due P.F. Jordan, in Infinite sums of psi functions, Bulletin of
2
6 American Mathematical Society (79) 4, 1973 . We think that we give here the first simple
6
an elementary proof of (J) and of other formulas which are deduced in the litterature
7
. from Jordan ones. As a consequence of our calculations, we find an expression for the sums
1
1 1 1
0
1+ +···+ ·Moreover,inthelastsection,wefind somerelationsbetween
3 2 n (2n+1)2a
1 nX≥1(cid:16) (cid:17)
: 1 1 1
v the sums σ(s,t) := 1+ +···+ (with s and t integers, s ≥ 2), which
i 3t (2n−1)t ns
X
nX≥1(cid:16) (cid:17)
r 1 1 1
a givethesums 1+ +···+ · Finally,we prove a new additiverelationon
2 n (2n+1)2a+1
nX≥1(cid:16) (cid:17)
the σ(s,t) of same weight s+t which is a “σ-sum theorem” analogous to this one involving
the classical Euler’s sums ζ(s,t).
Key words and phrases : multiple zeta values, Jordan’s sums, Euler’s Sums, harmonic
and semi harmonic numbers, sum theorem.
Introduction and notations
We define harmonic numbers H and semi-harmonic numbers S by
n n
1 1
H = 0,H = H + (n ≥ 1), S = 0,S = S + (n ≥ 1) (1.1)
0 n n−1 0 n n−1
n 2n−1
and we put
S S
J(b) := n, J¯(b) := n · (1.2)
nb (2n−1)b
n≥1 n≥1
X X
1
These series converge when b > 1. and we denote them by “Jordan’s sums”
WewillgiveexpressionofJ(2a)andJ¯(2a)whenaisaninteger≥ 1.Inthepresentpaper,
weproposeanelementaryandautonomousproofoftheirvalue.In[5],R.Sitaramachandrarao
uses the values of J(2a) and J¯(2a) given by Jordan in [4] to evaluate some series analogous
∞
1 1 1 H
to the so called Euler’s sums 1+ +···+ , as, for example (−1)n−1 n· In
2 n nb n2a
nX≥1(cid:16) (cid:17) nX=1
this paper, we inverse the order of the proof, deducing for instance J(2a) from the previous
sum that we prove directly.
∞ ∞
1 1 1
We use λ(s) = = 1− ζ(s), where ζ(s) = for s > 1. We use
(2n−1)s 2s ns
n=1 (cid:18) (cid:19) n=1
X X
also the classical Euler star sum
b−1
H b 1
ζ∗(b,1) = n = 1+ ζ(b+1)− ζ(j)ζ(b+1−j) (1.3)
nb 2 2
nX≥1 (cid:16) (cid:17) Xj=2
when the integer b is greater than 1. This expression has a simpler form for even b ( say 2a)
a−1
ζ∗(2a,1) = (1+a)ζ(2a+1)− ζ(2j)ζ(2a+1−2j). (1.4)
j=1
X
We denote by ζ∗(b,1) the alternating corresponding sum, that is:
e ζ∗(b,1) = (−1)n−1Hn·
nb
n≥1
X
e
Besides the formulas giving J(2a) and J¯(2a), we obtain close formulas for the sums
H
n
when q in an integer ≥ 2. The result when q is odd is obtained by searching
(2n+1)q
n≥1
X
1 1 1
linear relations between the sums σ(s,t) := 1+ +···+ (with s and t
3t (2n−1)t ns
nX≥1(cid:16) (cid:17)
integers, s ≥ 2). We give also explicit values of some of these σ(s,t), which we call σ-Euler
sums in term of Riemann’s series or analogous ones and establish a sum theorem for all σ’
series of same weight. Finally, we sketch out a calculation of them when s+t is odd.
I. The results :
Principal formulas for Jordan’s and analogous sums and σ-Euler sums
A. Jordan’s sums
S 22a+1 −1 1 a−1
J(2a) = n = ζ(2a+1)− (22j+1 −1)ζ(2j +1)ζ(2a−2j) (a)
n2a 4 2
n≥1 j=1
X X
2
This formula becomes, by using λ’s series:
a−1
S
J(2a) = n = 22a−1λ(2a+1)− 22jλ(2j +1)ζ(2a−2j). (a′)
n2a
n≥1 j=1
X X
a−1
S 1 1
J¯(2a) = n = λ(2a)ln2+ λ(2a+1)− λ(2a−2j)ζ(2j +1).
(2n−1)2a 2 22j+1
n≥1 j=1
X X
(b)
These are, with slightly different notations, the formulas given by Jordan in [4].
B. Sums involving H ′s
n
We get two classes of closed formulas :
a−1
H
n
= −2λ(2a)ln2+2aλ(2a+1)−2 λ(2j)λ(2a+1−2j) . (c)
(2n+1)2a
n≥1 j=1
X X
a−2
H 1
n
= −2λ(2a−1)ln2+ a− λ(2a)− λ(2q +1)λ(2a−2q −1)
(2n+1)2a−1 2
nX≥1 (cid:16) (cid:17) Xq=1
(d)
When a = 2b it becomes:
b−1
H 1
n
= −2λ(4b−1)ln2+ 2b− λ(4b)−2 λ(2q +1)λ(4b−2q −1) ,
(2n+1)4b−1 2
nX≥1 (cid:16) (cid:17) Xq=1
(e)
and when a = 2b+1 :
H 1
n
=−2λ(4b+1)ln2+ 2b+ λ(4b+2)
(2n+1)4b+1 2
nX≥1 (cid:16) (cid:17) . (e′)
b−1
−λ2(2b+1)−2 λ(2q +1)λ(4b−2q +1)
q=1
X
C. Some close formulas for σ-Euler sums and sum theorem
Clearly, by definition, σ(2a,1) = J(2a). We prove the two relations :
a−1
σ(2,2a−1) = 2a(2a−1)λ(2a+1)−8 jλ(2a−2j)λ(2j +1) , (f)
j=1
X
22a−1(2a+1)
σ(2a−1,2) = −a22a−1λ(2a+1)+ λ(2)λ(2a−1)
3
a−2 . (g)
+ j22jλ(2j +1)ζ(2a−2j)
j=1
X
3
Theorem. [of the σ-sum] The sum of all the σ’s of same weight is calculable: for w ≥ 3,
we have:
w−2
σ(w−i,i) = (w−1)λ(w) . (h)
i=1
X
II. The first result: evaluation of sums involving the H ’s
2n
H
2n
The principal objective of this section is to evaluate the new sum Z(2a) := ·
n2a
n≥1
X
We begin by the sum ζ∗(2a,1).
H
II.1 Calculation of ζ∗(2a,1) := (−1)n−1 n
e n2a
n≥1
X
e Hn 1
By rational decomposition, we get = · So for a ≥ 1, by absolute
n q(q +n)
q≥1
convergence X
(−1)n−1 1 1 (−1)n−1
ζ∗(2a,1) = = · (3)
n2a−1 q(q +n) q n2a−1(q +n)
n≥1 q≥1 q≥1 n≥1
X X X X
e
We have the following decomposition in the variable n:
1 2a−2 (−1)j 1 1 1
= + − , (4)
n2a−1(q +n) nj+1q2a−j−1 q2a−1 n n+q
Xj=1 (cid:16) (cid:17)
which gives:
2a−2 (−1)j (−1)n−1 1 1
ζ∗(2a,1) = (−1)n−1 + −
nj+1q2a−j q2a n n+q
Xq≥1nX≥1 Xj=1 q≥X1,n≥1 (cid:16) (cid:17)
e 2a−2 (−1)n−1 (−1)n−1 1 1
= (−1)j + − (5)
nj+1q2a−j q2a n n+q
Xj=1 q≥X1,n≥1 q≥X1,n≥1 (cid:16) (cid:17)
2a−2
= (−1)jζ(j +1)ζ(2a−j)+u,
j=1
X
e
(−1)n−1 (−1)n−1 1 1
by defining, for s > 1, ζ(s) = and u := − · Remark
ns q2a n n+q
nX≥1 q≥X1,n≥1 (cid:16) (cid:17)
that if a = 1, the firstesum in the last line of (5) does not exist, as it was already the case
in (4). In the sequel, all sums indexed by a void set are identically 0. Now:
(−1)n−1 (−1)m m−1 (−1)q−1
u = = (6)
q2a−1n(n+q) m q2a−1(m−q)
n≥1q≥1 m≥1 q=1
XX X X
4
by defining n+q = m in the first summation. By new partial fractions decomposition, one
has:
2a−2
1 1 1 1 1
= + + , (7)
(m−q)q2a−1 qj+1m2a−j−1 m2a−1 q m−q
Xj=1 (cid:16) (cid:17)
which gives :
2a−2 (−1)m+q−1 (−1)m+q−1 1 1
u = + + · (8)
qj+1m2a−j m2a q m−q
Xj=1 mX>q≥1 mX>q≥1 (cid:16) (cid:17)
Let u′ be the first sum of (8) an u′′ the second one. The sum u′ splits in two parts : we
sum first from 1 to a−1, and from a to 2a−2. In the second part we put j = 2a−1−j′.
Then j +1 = 2a−j′ and 2a−j = j′ +1 : so this second part becomes (we replace index j′
a−1 (−1)m+q−1
by j) , which may be written, by exchange of mute indexes q and m,
mj+1q2a−j
j=1m>q≥1
X X
a−1 (−1)m+q−1
· So
qj+1m2a−j
j=1q>m≥1
X X
a−1 (−1)m+q−1
u′ = · (9)
qj+1m2a−j·
j=1q6=m
X X
Now we observe that
(−1)m+q
= ζ(j +1)ζ(2a−j)−ζ(2a+1).
qj+1m2a−j
q6=m
X
e e
So equality (9) becomes
a−1
u′ = − ζ(j +1)ζ(2a−j)+(a−1)ζ(2a+1). (10)
j=1
X
e e
In fact, we may also write :
a−1
u′ = − ζ(2j)ζ(2a+1−2j)+(a−1)ζ(2a+1). (10′)
j=1
X
e e
Now look at the sum u′′. We have :
(−1)m
u′′ = µ(m)
m2a
m≥1
X
where
m−1
1 1
µ(m) := (−1)q−1 + ·
q m−q
Xq=1 (cid:16) (cid:17)
5
Clearly, µ(m) vanishes if m is odd and
2m−1 (−1)q−1 1
µ(2m) = 2 = 2(H −H ) = 2(H −H )+ ·
2m−1 m−1 2m m
q m
q=1
X
As a result:
1 1
u′′ = 2(H −H )+
(2m)2a 2m m m
mX≥1 (cid:16) (cid:17) (11)
1 1 1
= Z(2a)− ζ∗(2a,1)+ ζ(2a+1),
22a−1 22a−1 22a
Remark that
1−(−1)m−1 1
ζ∗(2a,1)−ζ∗(2a,1) = H = Z(2a). (12)
m2a m 22a−1
m≥1
X
So collecting previous reesults we obtain :
2a−2 a−1
ζ∗(2a,1) = (−1)jζ(j +1)ζ(2a−j)− ζ(j +1)ζ(2a−j)+(a−1)ζ(2a+1)
j=1 j=1
X X
e e 1 e 1e
+(ζ∗(2a,1)−ζ∗(2a,1))− ζ∗(2a,1)+ ζ(2a+1).
22a−1 22a
This gives
e
2a−2 a−1
2ζ∗(2a,1) = (−1)jζ(j +1)ζ(2a−j)− ζ(j +1)ζ(2a−j)
(13)
j=1 j=1
X X
e +(1−21−2ae)ζ∗(2a,1)+(a−1+2−2ea)ζ(2a+e 1).
At this stage, using (1.4), we can write
2ζ∗(2a,1) = A+B, (14)
putting
2a+1 e
A = 2a− ζ(2a+1) = (2a+1)ζ(2a+1)−ζ(2a+1) (15)
22a
and (cid:16) (cid:17)
e
2a−2 a−1 a−1
1
B = (−1)jζ(j+1)ζ(2a−j)− ζ(j+1)ζ(2a−j)− 1− ζ(j+1)ζ(2a−j) (16)
22a−1
Xj=1 Xj=1 (cid:16) (cid:17)Xj=1
We will give a reeduction of B. We wreite firste(by consideration of parity) :
a−1 a−1
B = ζ(2j +1)ζ(2a−2j)− ζ(2j)ζ(2a+1−2j)
j=1 j=1
X X
a−e1 e a−1
1
− ζ(2j +1)ζ(2a−2j)− 1− ζ(2j +1)ζ(2a−2j)
22a−1
Xj=1 (cid:16) (cid:17)Xj=1 (17)
a−1 e e a−1
1
= ζ(2j +1)ζ(2a−2j)− ζ(2j)ζ(2a+1−2j)
22a−2j−1
j=1 j=1
X X
e a−1 e
1
− 1− ζ(2j +1)ζ(2a−2j).
22a−1
(cid:16) (cid:17)Xj=1
6
By replacing ζ(2j +1) by (1−2−2j)ζ(2j +1) in the first sum we obtain, by grouping with
the third one:
e
a−1 a−1
1
B = −1 ζ(2j +1)ζ(2a−2j)− ζ(2j)ζ(2a+1−2j)
22a−2j−1
Xj=1(cid:16) (cid:17) Xj=1
a−1 a−1 e
=− ζ(2j +1)ζ(2a−2j)− ζ(2j)ζ(2a+1−2j) (18)
j=1 j=1
X X
a−e1 e
=−2 ζ(2j)ζ(2a+1−2j),
j=1
X
e
the last equality resulting for the exchange j 7→ a−j. Using (14) and (15) we get finally:
a−1
1 1
ζ∗(2a,1) = a+ ζ(2a+1)− ζ(2a+1)− ζ(2j)ζ(2a+1−2j) . (19)
2 2
(cid:16) (cid:17) Xj=1
e e e
Examples
H 3 1 3 3 1 5
a = 1: (−1)n−1 n = ζ(3)− ζ(3) = × − ζ(3) = ζ(3);
n2 2 2 2 4 2 8
nX≥1 (cid:16) (cid:17)
a = 2: (−1)n−1Hn = 5ζe(5)− 1ζ(5)−ζ(2)ζ(3) = 59ζ(5)− 1ζ(2)ζ(3).
n4 2 2 32 2
n≥1
X
In fact formula (19) was oebtained in [2] bey Flajolet and Salvy, but they used there
residue’s theorem. And it was etablished by Sitaramachandrarao in [5] by using the result
for J(2a) given by Jordan in [4], but we will inverse the processes of Sitaramachandrarao,
because the proof in [4] is hard to understand.
II.2 The sum Z(2a)
Subtracting (19) from (1.4) and using (12) we get
2−2a+1Z(2a) =
a−1 a−1
1 1
(a+1)ζ(2a+1)− ζ(2j)ζ(2a+1−2j)− a+ ζ(2a+1)+ ζ(2a+1)+ ζ(2j)ζ(2a+1−2j)
2 2
Xj=1 (cid:16) (cid:17) Xj=1
e e
which gives
a−1
3 1 1
2−2a+1Z(2a) = a+ ζ(2a+1)− a+ ζ(2a+1)− ζ(2j)ζ(2a+1−2j).
2 2 22j−1
(cid:16) (cid:17) (cid:16) (cid:17) Xj=1
e
Replacing ζ(2a+1) by its expression in terms of ζ(2a+1), one obtains now :
e a−1
H 1
2n = (2a+1+22a+1)ζ(2a+1)− 22a−2jζ(2j)ζ(2a+1−2j) . (20)
n2a 4
n≥1 j=1
X X
7
This last expression is given in [5], Theorem 1, but using Bernoulli polynomials and integral
transformations. This relation (20) will be the cornerstone for our proof of first Jordan’s
formula.
Relations (20) et (1.4) give easily the formula:
a−1
H 2a+1
2n−1 = λ(2a+1)− ζ(2a+1−2j)λ(2j) , (20′)
(2n−1)2a 2
n≥1 j=1
X X
III. Proof of the first Jordan formula
In this section, we give a close formula for the sum J(2a) in terms of ζ and λ series,
first by using previous results and second by an integral representation of it.
III.1 First proof by sums of series
Clearly, we have
1
H = S + H ,
2n n n
2
which gives
H 1
2n = J(2a)+ ζ∗(2a,1). (21)
n2a 2
n≥1
X
By (20),(21), and (1.4) we obtain :
a−1
1
J(2a) = (2a+1+22a+1)ζ(2a+1)− 22a−2jζ(2j)ζ(2a+1−2j)
4
j=1
X
a−1
1 1
− ((a+1)ζ(2a+1))+ ζ(2j)ζ(2a+1−2j)
2 2
j=1
X
22a+1 −1 1 a−1
= ζ(2a+1)− (22a+1−2j −1)ζ(2j)ζ(2a+1−2j).
4 2
j=1
X
The transformation j 7→ a−j furnishes the final expression:
S 22a+1 −1 1 a−1
J(2a) = n = ζ(2a+1)− (22j+1 −1)ζ(2j +1)ζ(2a−2j) . (22)
n2a 4 2
n≥1 j=1
X X
This formula becomes, by using λ’s series:
a−1
S
J(2a) = n = 22a−1λ(2a+1)− 22jλ(2j +1)ζ(2a−2j) . (23)
n2a
n≥1 j=1
X X
Examples
S 7 S 31 7
n n
= ζ(3) ; = ζ(5)− ζ(3)ζ(2).
n2 4 n4 4 2
n≥1 n≥1
X X
III.2 Second method for calculating J(2a), by an integral representation.
We base this new method on the following claim:
8
π sin2nx
Claim. For n ≥ 0, we have the relation 2S = dx.
n
sinx
Z0
Proof. Denote by I the integral in the claim. For n ≥ 1, by the classical relation
n
sin2nx−sin2(n−1)x = sinxsin(2n−1)x, we have:
π 2
I −I = sin(2n−1)xdx = ,
n n−1
2n−1
Z0
and this prove our claim, since I = 0.
0
It follows from the claim the crucial integral representation for J(2a):
π ϕ (x)
a
2J(2a) = dx,
sinx
Z0
where
sin2px
ϕ (x) := · (24)
a p2a
p≥1
X
The function ϕ is expressible by Bernoulli polynomials, but we did not use them explicitly.
a
Infact,we usesuccessive partialintegrationsin(24),which correspond toinductionrelations
between this polynomials. Put
π ϕ (x)
a
α (a) = cos2nxdx.
n
sinx
Z0
ϕ (x)
a
We have 2J(2a) = α (a). For all a ≥ 1, the functions ψ : x 7→ are integrable on the
0 a
sinx
interval [0,π]. For a ≥ 2, this results from the inequality |sinpx| ≤ p|sinx|, which gives,
|ψ (x)| ≤ ζ(2a−1) < +∞.
a
And for a = 1, the integrability results from the classical expansion (for 0 ≤ x ≤ π):
1 sin2px
x(π −x) = = ϕ (x)
2 p2 2
p≥1
X
π
and from the fact that ψ (x) tends to when x tends to 0 or to π. Hence, by Riemann-
2
2
Lebesgue theorem, the sequence α (a) tends to 0 if n goes to infinity.
n
For n ≥ 1, we have :
π ϕ (x) π
a
α (a)−α (a) = (cos(2n−2)x−cos2nx)dx = 2 ϕ (x)sin(2n−1)x dx := 2v .
n−1 n a n
sinx
Z0 Z0
n ∞
So, by summation, α (a)−α (a) = 2 v . Then, α (a) = 2 v , and finally
0 n k 0 n
k=1 n=1
X X
∞
J(a) = v .
n
n=1
X
9
In the following, we drop the subscript a in ϕ .
a
π
Calculation of v = ϕ(x)sin(2n−1)x dx.
n
Z0
We proceed by successive integrations by part, involving the derivatives of ϕ :
∞ ∞
cos2kx sin2kx
ϕ(2j)(x) = (−1)j−122j−1 , ϕ(2j−1)(x) = (−1)j−122j−2
k2a−2j k2a−2j+1
k=1 k=1
X X
when 1 ≤ j ≤ a−1 for even derivatives and 1 ≤ j ≤ a for odd ones. We observe that if
j = a, the second relation makes sense only if x 6= 0 (mod π). In this case we have
∞
sin2kx
ϕ(2a−1)(x) = (−1)a−122a−2 ·
k
k=1
X
Using Fourier’s series or complex logarithm, this is (when 0 < x < π):
ϕ(2a−1)(x) = (−1)a−122a−3(π −2x).
In the following calculation, n appears only by mean of 2n−1, so we set 2n−1 = m.
Since ϕ(0) = ϕ(π) = 0, we obtain first :
ϕ(x)cosmx π 1 π 1 π
v = − + ϕ′(x)cosmxdx = ϕ′(x)cosmxdx.
n
m m m
h i0 Z0 Z0
1
When a = 1, this process stops (ϕ′(x) = (π −2x)) and we obtain :
2
1 π π 1 π π 1 π 2
v = −x cosmxdx = −x sinmx + sinmxdx = (24′)
n
m Z0 (cid:16)2 (cid:17) m2h(cid:16)2 (cid:17) i0 m2 Z0 m3
For a ≥ 2, we use two integrations by part :
1 π 1 π 1 π
v = ϕ′(x)sinmx − ϕ′′(x)sinmxdx = − ϕ′′(x)sinmxdx
n
m2 m2 m2
1 h i0π 1 Z0 π Z0
= ϕ′′(x)cosmx − ϕ′′′(x)cosmxdx
m3 m3
4h i10 π Z0
= − ζ(2a−2)− ϕ′′′(x)cosmxdx,
m3 m3
Z0
the last equality coming from the expression of ϕ′′ (remember that m is odd). When a = 2,
π
we stop. Eventually, we stop when the last integral is ϕ(2a−1)(x)cosmxdx and obtain
Z0
the relation:
a−1 22j (−1)a−1 π
v = − ζ(2a−2j)+ ϕ(2a−1)(x)cosmxdx.
n m2j+1 m2a−1
j=1 Z0
X
10