One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means
Available proofs of the result of the type "at least one of the odd zeta values ζ(5), ζ(7),…, ζ(s) is irrational" make use of the saddle-point method or of linear independence criteria, or both. These two remarkable techniques are, however, counted as highly non-elementary, therefore leavi...
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nasplib_isofts_kiev_ua-123456789-2094362025-11-22T01:00:47Z One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means Zudilin W. Available proofs of the result of the type "at least one of the odd zeta values ζ(5), ζ(7),…, ζ(s) is irrational" make use of the saddle-point method or of linear independence criteria, or both. These two remarkable techniques are, however, counted as highly non-elementary, therefore leaving the partial irrationality result inaccessible to the general mathematics audience in all its glory. Here we modify the original construction of linear forms in odd zeta values to produce, for the first time, an elementary proof of such a result — a proof whose technical ingredients are limited to the prime number theorem and Stirling's approximation formula for the factorial. I thank Stéphane Fischler, Tanguy Rivoal, Johannes Sprang, and the anonymous referees for their feedback on the manuscript. 2018 Article One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means / W. Zudilin // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 10 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 11J72; 11M06; 33C20 arXiv: 1801.09895 https://nasplib.isofts.kiev.ua/handle/123456789/209436 https://doi.org/10.3842/SIGMA.2018.028 en Symmetry, Integrability and Geometry: Methods and Applications application/pdf Інститут математики НАН України |
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Available proofs of the result of the type "at least one of the odd zeta values ζ(5), ζ(7),…, ζ(s) is irrational" make use of the saddle-point method or of linear independence criteria, or both. These two remarkable techniques are, however, counted as highly non-elementary, therefore leaving the partial irrationality result inaccessible to the general mathematics audience in all its glory. Here we modify the original construction of linear forms in odd zeta values to produce, for the first time, an elementary proof of such a result — a proof whose technical ingredients are limited to the prime number theorem and Stirling's approximation formula for the factorial. |
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Zudilin W. One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means Symmetry, Integrability and Geometry: Methods and Applications |
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Zudilin W. |
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Zudilin W. |
| title |
One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means |
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One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means |
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One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means |
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One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means |
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One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means |
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one of the odd zeta values from ζ(5) to ζ(25) is irrational. by elementary means |
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Інститут математики НАН України |
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One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means / W. Zudilin // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 10 назв. — англ. |
| series |
Symmetry, Integrability and Geometry: Methods and Applications |
| work_keys_str_mv |
AT zudilinw oneoftheoddzetavaluesfromz5toz25isirrationalbyelementarymeans |
| first_indexed |
2025-11-24T07:13:48Z |
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2025-11-24T07:13:48Z |
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1849654958538883072 |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 14 (2018), 028, 8 pages
One of the Odd Zeta Values
from ζ(5) to ζ(25) Is Irrational.
By Elementary Means
Wadim ZUDILIN
Department of Mathematics, IMAPP, Radboud University,
PO Box 9010, 6500 GL Nijmegen, The Netherlands
E-mail: w.zudilin@math.ru.nl
URL: http://www.math.ru.nl/~wzudilin/
Received January 31, 2018, in final form March 26, 2018; Published online March 29, 2018
https://doi.org/10.3842/SIGMA.2018.028
Abstract. Available proofs of result of the type ‘at least one of the odd zeta values
ζ(5), ζ(7), . . . , ζ(s) is irrational’ make use of the saddle-point method or of linear indepen-
dence criteria, or both. These two remarkable techniques are however counted as highly non-
elementary, therefore leaving the partial irrationality result inaccessible to general mathe-
matics audience in all its glory. Here we modify the original construction of linear forms
in odd zeta values to produce, for the first time, an elementary proof of such a result —
a proof whose technical ingredients are limited to the prime number theorem and Stirling’s
approximation formula for the factorial.
Key words: irrationality; zeta value; hypergeometric series
2010 Mathematics Subject Classification: 11J72; 11M06; 33C20
1 Introduction
Without touching deeply a history of the question (see [3] for an excellent account of this), we
notice that the irrationality of the zeta values — values of Riemann’s zeta function
ζ(s) =
∞∑
n=1
1
ns
at integers s = 2, 3, . . . , is known for even s and also for s = 3, while there are only partial
results in this direction for odd s ≥ 5. A starting point here has been set in the work [1] with
further development, particularly focusing on ζ(5), in [7] and [10].
We fix an odd integer s ≥ 7. Our strategy is constructing two sequences of linear forms rn
and r̂n living in the Q-space Q+Qζ(3) +Qζ(5) + · · ·+Qζ(s), for which we have a control of the
common denominators λn of rational coefficients and an elementary access to their asymptotic
behaviour as n → ∞; more importantly, the two coefficients of ζ(3) in these forms are propor-
tional (with factor 7), so that 7rn − r̂n belongs to the space Q + Qζ(5) + · · · + Qζ(s). Finally,
using 7rn − r̂n > 0 and the asymptotics λn(7rn − r̂n)→ 0 as n→∞ of the linear forms
λn(7rn − r̂n) ∈ Z + Zζ(5) + Zζ(7) + · · ·+ Zζ(s)
when s = 25, we conclude that it cannot happen that all the quantities ζ(5), ζ(7), . . . , ζ(25) are
rational.
This paper is a contribution to the Special Issue on Orthogonal Polynomials, Special Functions and Applica-
tions (OPSFA14). The full collection is available at https://www.emis.de/journals/SIGMA/OPSFA2017.html
mailto:w.zudilin@math.ru.nl
http://www.math.ru.nl/~wzudilin/
https://doi.org/10.3842/SIGMA.2018.028
https://www.emis.de/journals/SIGMA/OPSFA2017.html
2 W. Zudilin
The original idea of using the so-called well-poised hypergeometric series to construct linear
forms in zeta values of a given parity is due to Ball and Rivoal [1]; our new ingredient here
is using simultaneously such series and their ‘twists by half’, for an appropriate choice of the
parameters. More precisely, our hypergeometric series assume the form
rn =
∞∑
ν=1
Rn(ν) and r̂n =
∞∑
ν=1
Rn
(
ν − 1
2
)
, (1)
where the rational-function summand Rn(t) is defined as follows:
R(t) = Rn(t) =
n!s−5
∏n
j=1(t− j) ·
∏n
j=1(t+ n+ j) · 26n
∏3n
j=1
(
t− n− 1
2 + j
)∏n
j=0(t+ j)s
=
26nn!s−5
∏6n
j=0
(
t− n+ 1
2j
)∏n
j=0(t+ j)s+1
. (2)
The following Sections 2 and 3 discuss, respectively, the arithmetic and analysis of the
forms (1). In Section 4 we use this information to conclude with the proof of the claimed
result and make some relevant comments.
2 Arithmetic ingredients
The notation dn will be used for the least common multiple of 1, 2, . . . , n. Recall that the prime
number theorem is equivalent to the asymptotics
lim
n→∞
d1/nn = e. (3)
A rational function S(t) of the form
S(t) =
P (t)
(t− t1)s1(t− t2)s2 · · · (t− tq)sq
,
whose denominator has degree larger than its numerator, possesses a unique partial-fraction
decomposition
S(t) =
q∑
j=1
sj∑
i=1
bi,j
(t− tj)i
.
The coefficients here can be computed on the basis of explicit formula
bi,j =
1
(sj − i)!
(
S(t)(t− tj)sj
)(sj−i)∣∣∣
t=tj
for all i, j in question. This procedure can be illustrated on the following examples, in which
all the exponents sj are equal to 1:
n!∏n
j=0(t+ j)
=
n∑
k=0
(−1)k
(
n
k
)
t+ k
,∏n
j=1(t− j)∏n
j=0(t+ j)
=
n∑
k=0
(−1)n+k
(
n+k
n
)(
n
k
)
t+ k
,∏n
j=1(t+ n+ j)∏n
j=0(t+ j)
=
n∑
k=0
(−1)k
(
2n−k
n
)(
n
k
)
t+ k
,
One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational 3
22n
∏n
j=1(t+ 1
2 − j)∏n
j=0(t+ j)
=
n∑
k=0
(−1)n+k
(
2n+2k
2n
)(
2n
n+k
)
t+ k
,
22n
∏n
j=1(t−
1
2 + j)∏n
j=0(t+ j)
=
n∑
k=0
(
2k
k
)(
2n−2k
n−k
)
t+ k
,
22n
∏n
j=1(t+ n− 1
2 + j)∏n
j=0(t+ j)
=
n∑
k=0
(−1)k
(
4n−2k
2n
)(
2n
k
)
t+ k
.
It also means that the function R(t) in (2) can be written as
R(t) =
s∑
i=1
n∑
k=0
ai,k
(t+ k)i
(4)
with the recipe to compute the coefficients ai,k in its partial-fraction decomposition. At the
same time, the function R(t) is a product of ‘simpler’ rational functions given above, with all
coefficients of their partial fractions being integral.
Lemma 1. Let k1, . . . , kq be pairwise distinct numbers from the set {0, 1, . . . , n} and s1, . . . , sq
positive integers. Then the coefficients in the expansion
1∏q
j=1(t+ kj)sj
=
q∑
j=1
sj∑
i=1
bi,j
(t+ kj)i
satisfy
ds−in bi,j ∈ Z, where i = 1, . . . , sj and j = 1, . . . , q, (5)
where s = s1 + · · ·+ sq.
In particular,
ds−in ai,k ∈ Z, where i = 1, . . . , s and k = 0, 1, . . . , n, (6)
for the coefficients in (4).
Proof. Denote the rational function in question by S(t). The statement is trivially true when
q = 1, therefore we assume that q ≥ 2. In view of the symmetry of the data, it is sufficient
to demonstrate the inclusions (5) for j = 1. Differentiating a related product m times, for any
m ≥ 0, we obtain
1
m!
(
S(t)(t+ k1)
s1
)(m)
=
1
m!
q∏
j=2
(t+ kj)
−sj
(m)
=
∑
`2,...,`q≥0
`2+···+`q=m
q∏
j=2
1
`j !
(
(t+ kj)
−sj
)(`j)
=
∑
`2,...,`q≥0
`2+···+`q=m
q∏
j=2
(−1)`j
(
sj + `j − 1
`j
)
(t+ kj)
−(sj+`j).
This implies that
bi,1 =
∑
`2,...,`q≥0
`2+···+`q=s1−i
q∏
j=2
(−1)`j
(
sj + `j − 1
`j
)
1
(kj − k1)sj+`j
4 W. Zudilin
for i = 1, . . . , s1. Using dn/(kj − k1) ∈ Z for j = 2, . . . , q and
∑q
j=2(sj + `j) = s − i for each
individual summand, we deduce the desired inclusion in (5) for j = 1, hence for any j.
The second claim in the lemma follows from considering R(t) as a product of the ‘simpler’
rational functions. �
Lemma 2. For the coefficients ai,k in (4), we have
ai,k = (−1)i−1ai,n−k for k = 0, 1, . . . , n and i = 1, . . . , s,
so that
n∑
k=0
ai,k = 0 for i even.
Proof. Since s is odd, the function (2) possesses the following (well-poised) symmetry: R(−t−
n) = −R(t). Substitution of the relation into (4) results in
−
s∑
i=1
n∑
k=0
ai,k
(t+ k)i
=
s∑
i=1
n∑
k=0
ai,k
(−t− n+ k)i
=
s∑
i=1
(−1)i
n∑
k=0
ai,k
(t+ n− k)i
=
s∑
i=1
(−1)i
n∑
k=0
ai,n−k
(t+ k)i
,
and the identities in the lemma follow from the uniqueness of decomposition into partial fractions.
The second statement follows from
n∑
k=0
ai,k = (−1)i−1
n∑
k=0
ai,n−k = (−1)i−1
n∑
k=0
ai,k. �
Lemma 3. For each n,
rn =
s∑
i=2
i odd
aiζ(i) + a0 and r̂n =
s∑
i=2
i odd
ai
(
2i − 1
)
ζ(i) + â0,
with the following inclusions available:
ds−in ai ∈ Z for i = 3, 5, . . . , s, and dsna0, d
s
nâ0 ∈ Z.
Notice that(
2i − 1
)
ζ(i) =
∞∑
`=1
1(
`− 1
2
)i
for i ≥ 2.
Proof. Our strategy here is to write the series in (1) using the partial-fraction decomposition (4)
of R(t). To treat the first sum rn we additionally introduce an auxiliary parameter z > 0, which
we later specialise to z = 1:
rn(z) =
∞∑
ν=1
Rn(ν)zν =
∞∑
ν=1
s∑
i=1
n∑
k=0
ai,kz
ν
(ν + k)i
=
s∑
i=1
n∑
k=0
ai,kz
−k
∞∑
ν=1
zν+k
(ν + k)i
=
s∑
i=1
n∑
k=0
ai,kz
−k
(
Lii(z)−
k∑
`=1
z`
`i
)
=
s∑
i=1
Lii(z)
n∑
k=0
ai,kz
−k −
s∑
i=1
n∑
k=0
k∑
`=1
ai,kz
−(k−`)
`i
,
One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational 5
where
Lii(z) =
∞∑
`=1
z`
`i
for i = 1, . . . , s are the polylogarithmic functions. The latter are well defined at z = 1 for i ≥ 2,
where Lii(1) = ζ(i), while Li1(z) = − log(1− z) does not have a limit as z → 1−. By taking the
limit as z → 1− in the above derivation and using Rn(ν) = O
(
ν−2
)
as ν →∞, we conclude that
n∑
k=0
a1,k = lim
z→1−
n∑
k=0
a1,kz
−k = 0,
and
rn =
s∑
i=2
ζ(i)
n∑
k=0
ai,k −
s∑
i=1
n∑
k=0
ai,k
k∑
`=1
1
`i
. (7)
We proceed similarly for r̂n, omitting introduction of the auxiliary parameter z. Since R(t)
in (2) vanishes at t = −1
2 ,−
3
2 , . . . ,−n+ 1
2 , we can shift the starting point of summation for r̂n
to t = −m− 1
2 , where m =
⌊
n−1
2
⌋
, so that
r̂n =
∞∑
ν=−m
Rn
(
ν − 1
2
)
=
∞∑
ν=−m
s∑
i=1
n∑
k=0
ai,k(
ν + k − 1
2
)i
=
s∑
i=1
n∑
k=0
ai,k
∞∑
ν=−m
1(
ν + k − 1
2
)i
=
s∑
i=1
m∑
k=0
ai,k
∞∑
ν=−m
1(
ν + k − 1
2
)i +
s∑
i=1
n∑
k=m+1
ai,k
∞∑
ν=−m
1(
ν + k − 1
2
)i
=
s∑
i=1
m∑
k=0
ai,k
(
0∑
`=k−m
1(
`− 1
2
)i +
∞∑
`=1
1(
`− 1
2
)i
)
+
s∑
i=1
n∑
k=m+1
ai,k
( ∞∑
`=1
1(
`− 1
2
)i − k−m−1∑
`=1
1(
`− 1
2
)i
)
=
s∑
i=2
(2i − 1)ζ(i)
n∑
k=0
ai,k +
s∑
i=1
m∑
k=0
ai,k
m−k∑
`=0
(−1)i(
`+ 1
2
)i
−
s∑
i=1
n∑
k=m+1
ai,k
k−m−1∑
`=1
1(
`− 1
2
)i . (8)
Now the statement of the lemma follows from the representations in (7) and (8), Lemma 2, the
inclusions (6) of Lemma 1 and
din
k∑
`=1
1
`i
∈ Z for 0 ≤ k ≤ n and i ≥ 1,
din
m−k∑
`=0
(−1)i
(`+ 1
2)i
∈ Z for 0 ≤ k ≤ m and i ≥ 1,
din−1
k−m−1∑
`=1
1(
`− 1
2
)i ∈ Z for m+ 1 ≤ k ≤ n and i ≥ 1. �
6 W. Zudilin
3 Asymptotic behaviour
In this section we make frequent use of Stirling’s asymptotic formula
n! ∼
√
2πn
(n
e
)n
as n→∞,
and its corollary(
2n
n
)
∼ 22n√
πn
as n→∞
for the central binomial coefficients. (One may also use somewhat weaker but ‘more elementary’
lower and upper bounds∫ n
1
log x dx ≤ log(n!) ≤
∫ n+1
2
log x dx
for the factorial coming out from estimating integral sums of the logarithm function, with
a nemesis of running into more sophisticated versions for the asymptotics and inequalities below.)
Because the rational function Rn(t) in (2) vanishes at 1, 2, . . . , n and at 1
2 ,
3
2 , . . . , n −
1
2 , the
hypergeometric series (1) can be alternatively written as
rn =
∞∑
ν=n+1
Rn(ν) =
∞∑
k=0
ck and r̂n =
∞∑
ν=n+1
Rn(ν − 1
2) =
∞∑
k=0
ĉk,
with the involved summands
ck = Rn(n+ 1 + k) =
26nn!s−5
∏6n
j=0
(
k + 1 + 1
2j
)∏n
j=0(n+ k + 1 + j)s+1
=
n!s−5(6n+ 2k + 2)!(n+ k)!s+1
2(2k + 1)!(2n+ k + 1)!s+1
(9)
and
ĉk = Rn
(
n+ 1
2 + k
)
=
26nn!s−5
∏6n
j=0
(
k + 1
2 + 1
2j
)∏n
j=0
(
n+ k + 1
2 + j
)s+1
strictly positive. Observe that
ck
ĉk
=
∏6n
j=0(2k + 2 + j)∏6n
j=0(2k + 1 + j)
·
n∏
j=0
n+ k + 1
2 + j
n+ k + 1 + j
s+1
=
6n+ 2k + 2
2k + 1
·
(
2−2(n+1)
(
4n+2k+2
2n+k+1
)(
2n+2k
n+k
) )s+1
∼ 6n+ 2k + 2
2k + 1
(
n+ k
2n+ k + 1
)(s+1)/2
as n+ k →∞. (10)
Lemma 4. For s ≥ 7 odd,
lim
n→∞
r1/nn = lim
n→∞
r̂1/nn = g(x0) and lim
n→∞
rn
r̂n
= 1
where
g(x) =
26(x+ 3)6(x+ 1)s+1
(x+ 2)2(s+1)
and x0 is the unique positive zero of the polynomial
x(x+ 2)(s+1)/2 − (x+ 3)(x+ 1)(s+1)/2.
One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational 7
Proof. We have
ck+1
ck
=
(
k + 3n+ 3
2
)
(k + 3n+ 2)
(k + 1)
(
k + 3
2
) (
k + n+ 1
k + 2n+ 2
)s+1
∼ f
(
k
n
)2
as n+ k →∞, (11)
where
f(x) =
x+ 3
x
(
x+ 1
x+ 2
)(s+1)/2
.
For an ease of notation write q = (s+ 1)/2 ≥ 4. Since
f ′(x)
f(x)
=
1
x+ 3
− 1
x
+ q
(
1
x+ 1
− 1
x+ 2
)
=
(q − 3)x2 + 3(q − 3)x− 6
x(x+ 1)(x+ 2)(x+ 3)
and the quadratic polynomial in the latter numerator has a unique positive zero x1, the func-
tion f(x) monotone decreases from +∞ to f(x1) when x ranges from 0 to x1 and then monotone
increases from f(x1) to f(+∞) = 1 (not attaining the value!) when x ranges from x1 to +∞.
In particular, there is exactly one positive solution x0 to f(x) = 1. Notice that 0 < x0 < 1,
because f(1) = 4 · (2/3)q < 1.
The information gained and asymptotics in (11) imply that ck+1/ck > 1 for the indices
k < x0n− γ
√
n and ck+1/ck < 1 for k > x0n+ γ
√
n for an appropriate choice of γ > 0 dictated
by application of Stirling’s formula to the factorials defining ck in (9) (see [2, Section 3.4] as
well as the second proof of Lemma 3 in [1]). This means that the asymptotic behaviour of
the sum rn =
∑∞
k=0 ck is determined by the asymptotics of ck0 and its neighbours ck, where
k0 = k0(n) ∼ x0n and |k − k0| ≤ γ
√
n, so that
lim
n→∞
r1/nn = lim
n→∞
c
1/n
k0(n)
= lim
n→∞
((n
e
)(s−5)n(6n+ 2k0 + 2
e
)6n+2k0+2( e
2k0 + 1
)2k0+1
×
(
n+ k0
e
)(s+1)(n+k0)( e
2n+ k0 + 1
)(s+1)(2n+k0+1)
)1/n
=
(2x0 + 6)2x0+6(x0 + 1)(s+1)(x0+1)
(2x0)2x0(x0 + 2)(s+1)(x0+2)
=
26(x0 + 3)6(x0 + 1)s+1
(x0 + 2)2(s+1)
· f(x0)
2x0 = g(x0).
It now follows from (10) that
ĉk+1
ĉk
∼ ck+1
ck
as n+ k →∞, (12)
so that the above analysis applies to the sum r̂n =
∑∞
k=0 ĉk as well, and its asymptotic behaviour
is determined by the asymptotics of ĉk0 and its neighbours ĉk, where k0 = k0(n) ∼ x0n and
|k − k0| ≤ γ̂
√
n. From (12) we deduce that the limits of ĉ
1/n
k0(n)
and c
1/n
k0(n)
as n → ∞ coincide,
hence r̂
1/n
n → g(x0) as n→∞. In addition to this, we also get
lim
n→∞
rn
r̂n
= lim
n→∞
ck0(n)
ĉk0(n)
= lim
n→∞
6n+ 2k0 + 2
2k0 + 1
(
n+ k0
2n+ k0 + 1
)(s+1)/2
= f(x0),
which leads to the remaining limiting relation. �
8 W. Zudilin
4 Conclusion
We choose s = 25 and apply Lemma 4 to find out that 7rn− r̂n > 0 for n sufficiently large, and
lim
n→∞
(7rn − r̂n)1/n = g(x0) = exp(−25.292363 . . . ),
where x0 = 0.00036713 . . . . Assuming that the odd zeta values from ζ(5) to ζ(25) are all rational
and denoting by a their common denominator, we use Lemma 3 and the asymptotics (3) to
conclude that the sequence of positive integers
ad25n (7rn − r̂n)
tends to 0 as n → ∞; contradiction. Thus, at least one of the numbers ζ(5), ζ(7), . . . , ζ(25) is
irrational.
Those who count the prime number theorem as insufficiently elementary may use weaker
versions of (3), for example, dn < 3n from [5] and the choice s = 33 instead, to arrive at the
same conclusion (for the larger value of s, of course).
Finally, we remark that the novelty of eliminating an ‘unwanted’ term of ζ(3) in linear forms
in odd zeta values can be further used with the arithmetic method in [10] to significantly reduce
the size of s. Since this does not let s be down to s = 9, hence leaving the achievement ‘at
least one of the four numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational’ unchanged, we do not discuss
this generalisation in greater details. We point out, however, that there are other applications
of the hypergeometric ‘twist-by-half’ idea, some discussed in the joint papers [6, 8], and that
a far-going extension to general ‘twists’ introduced by J. Sprang in [9] leads to an elementary
proof of a version of the Ball–Rivoal theorem from [1] as well as to a significant improvement of
the latter — see [4] for details.
Acknowledgements
I thank Stéphane Fischler, Tanguy Rivoal, Johannes Sprang and the anonymous referees for
their feedback on the manuscript.
References
[1] Ball K., Rivoal T., Irrationalité d’une infinité de valeurs de la fonction zêta aux entiers impairs, Invent.
Math. 146 (2001), 193–207.
[2] de Bruijn N.G., Asymptotic methods in analysis, Bibliotheca Mathematica, Vol. 4, North-Holland Publishing
Co., Amsterdam, P. Noordhoff Ltd., Groningen, Interscience Publishers Inc., New York, 1958.
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math.NT/0303066.
[4] Fischler S., Sprang J., Zudilin W., Many odd zeta values are irrational, arXiv:1803.08905.
[5] Hanson D., On the product of the primes, Canad. Math. Bull. 15 (1972), 33–37.
[6] Krattenthaler C., Zudilin W., Hypergeometry inspired by irrationality questions, arXiv:1802.08856.
[7] Rivoal T., Irrationalité d’au moins un des neuf nombres ζ(5), ζ(7), . . . , ζ(21), Acta Arith. 103 (2002), 157–
167, math.NT/0104221.
[8] Rivoal T., Zudilin W., A note on odd zeta values, arXiv:1803.03160.
[9] Sprang J., Infinitely many odd zeta values are irrational. By elementary means, arXiv:1802.09410.
[10] Zudilin W., Arithmetic of linear forms involving odd zeta values, J. Théor. Nombres Bordeaux 16 (2004),
251–291, math.NT/0206176.
https://doi.org/10.1007/s002220100168
https://doi.org/10.1007/s002220100168
https://arxiv.org/abs/math.NT/0303066
https://arxiv.org/abs/1803.08905
https://doi.org/10.4153/CMB-1972-007-7
https://arxiv.org/abs/1802.08856
https://doi.org/10.4064/aa103-2-5
https://arxiv.org/abs/math.NT/0104221
https://arxiv.org/abs/1803.03160
https://arxiv.org/abs/1802.09410
https://doi.org/10.5802/jtnb.447
https://arxiv.org/abs/math.NT/0206176
1 Introduction
2 Arithmetic ingredients
3 Asymptotic behaviour
4 Conclusion
References
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