On the divergence of the sum of prime reciprocals
PIERRE DUSART
XLIM - UMR CNRS 7252, Faculté des Sciences et Techniques
Université de Limoges
123 avenue Albert THOMAS, 87060 LIMOGES CEDEX
FRANCE
Abstract: The sum of the reciprocals of all prime numbers diverges but the divergence is very slow. We propose
effective lower and upper bounds for partial sums under the Riemann hypothesis. We give an explicit error term
for all Mertens’ theorems.
Key-Words: Number Theory, arithmetic functions, Prime numbers, Mertens’ theorems
Received: November 28, 2022. Revised: May 22, 2023. Accepted: June 15, 2023. Published: July 14, 2023.
1 Introduction
In 14th-century, the study, [1], was considering the
harmonic series
∞
X
n=1
1
n= 1 + 1
2+1
3+1
4+···
and proves for the first time the divergence of the
sum of the reciprocals of the integers. Three centuries
later, [2], would also consider this sum and used the
relation
∞
X
n=1
1
n=Y
p∈P 1 + 1
p+1
p2+···=Y
p∈P
1
1−p−1,
where the product is taken over the set Pof all primes,
to show once again the existence of infinitely many
primes, the one appearing in Euclid’s elements.
We can ask ourselves the question if you reduce the
summation set only to the prime set, also an infinite
set but a smaller set, the sum remains divergent or not?
The question was resolved by [2], who proved in
1744 the divergence of the sum of the reciprocals of
all prime numbers
X
pprime
1
p=1
2+1
3+1
5+1
7+1
11 +1
13 +1
17 +···
= +∞
The next step is to exhibit the growth of the series.
You have to count the terms of the series up to a bound
x
F(x) = X
pprime≤x
1
p,(1)
sum which becomes a function of x. By [3], the
growth of this function is asymptotic to the “ln ln”
function (iterate of ln function, the natural logarithm).
Step by step, we can find the asymptotic expansion by
studying the difference
lim
x→+∞
X
pprime≤x
1
p−ln ln x
,(2)
following the same idea on Euler’s constant, [4],
γ=lim
n→∞ n
X
k=1
1
k−ln n!.
One can establish that the limit of (2) exists (Mertens’
second theorem, [5]) and converges to a constant M,
the well-known Meissel-Lehmer constant. The value
of Mis approximately (sequence A077761 in the
OEIS1)
M≈0.2614972128476427837554268386...
Next, we can express completely the asymptotic ex-
pansion which expresses the behavior of the sum up
to xand define
X
pprime≤x
1
p=ln ln x+M+O1
ln x.(3)
(See proof of [6])
1https://oeis.org/A077761
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If you want to compute F(x)for high values of x, an
algorithm for fast computation of (1) is presented in
[7].
The aim of this paper is to have a practical approxi-
mation for this sum, which follows the asymptotic ex-
pansion. We place ourselves under the Riemann hy-
pothesis to have the best possible results.
We also put practical approximations for Mertens’
first theorem about Pln p/pand Mertens’ third the-
orem about Q(1 −1/p).
2 Sum of the reciprocals of primes
Effective estimates are very useful to show other re-
sults. For example, if you use a classical upper bound
for pn(the n-th prime number)
pn< n ln n+nln ln nfor n≥6
then showing the divergence of F(x)is easy:
∞
X
n=1
1
pn≥∞
X
n=6
1
pn
≥∞
X
n=6
1
nln n+nln ln n
≥∞
X
n=6
1
2nln n=∞
by the integral test for convergence. We found, [1],
result by another way (but not with the same tools).
The best-known results about effective estimates of
the sum (3) are from [8], [9], with an error term
Oln x
√xinstead of O1
ln x.
2.1 First lemmas
Let’s introduce ϑand ψas the first and second Cheby-
shev functions respectively.
Let’s start with two results found by [10]:
Lemma 2.1 (Lemma 1, [10]).Let, for n∈N∗and
ρ∈C,
Fρ,n(x) = Z∞
x
xρ
tn+1
n(ln t+ 1)
ln2(t)dt.
If Re(ρ) = 1/2 then
Fρ,n(x) = n
n−ρ
xρ−n
ln n+rρ,n(x)
with
|rρ,n(x)|⩽1
|ρ|
1
xn−1/2 ln2x1 + 4
(2n−1) ln x.
Proof. By integration by parts,
Z∞
x
xρ
tn+1
n(ln t+ 1)
ln2(t)dt=n
n−ρ
xρ−n
ln n
+ρ
(ρ−n)2−xρ−n
ln2x+ 2 R∞
x
tρ−n−1
ln3tdt
Lemma 2.2 (Lemma 4, [10]).Under the Riemann hy-
pothesis,
Z∞
x
(ϑ(t)−t)(ln t+ 1)
(tln t)2dt
⩽0.0462
x1/2 ln x1 + 1
ln x+4
ln2x+ln 2π
xln x
+2
√xln x1−1
ln x+4
ln2(x)+1
x1/6 .
Proof. Let ϑ(x) = x+ε1(x)and ψ(x) = x+ε2(x).
Since ε1(x) = ϑ(x)−x=ϑ(x)−(ψ(x)−ε2(x)),
this last integral can be expressed by
Z∞
x
ε1(t)(t+ 1)
(tln t)2dt=I1(x)−I2(x)
where
I1(x) = Z∞
x
ε2(t)(ln t+ 1)
(tln t)2dt,
and
I2(x) = Z∞
x
(ψ(t)−ϑ(t))(ln t+ 1)
(tln t)2dt.
To obtain an upper bound for I1, we use a classical
explicit formula in number theory (Chapter 17 of [11])
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ψ(x) = x−ln 2π−Pρ
xρ
ρ−1
2ln(1 −1/x2)in the
form
Z∞
x
fn(t)(ψ(t)−t+ln 2π+1
2ln(1 −1/t2))dt
=−Z∞
xX
ρ
fn(t)tρ
ρdt
and taking fn(t) = n(ln t+1)
tn+1 ln2t(derivative of
−1/(tnln t)), we have f′
n(t) = O(1
t3ln t)for n⩾1
and thus inverting the sum and integral signs by the
dominated convergence theorem. Thus
I1(x) = − X
ρZ∞
x
f1(t)tρ
ρdt!
−R∞
xf1(t)(ln 2π+1
2ln(1 −1/t2))dt.
Moreover1
2ln(1 −1/t2)<0and |1
2ln(1 −1/t2)|<
ln 2πhence
−Z∞
x
fn(t)(ln 2π)dt
<−Z∞
x
fn(t)(ln 2π+1
2ln(1 −1/t2))dt
<0
Thus
0<Z∞
x
fn(t)(ln 2π+1
2ln(1 −1/t2))dt⩽ln 2π
xnln x.
On the other hand, according to Lemma 2.1,
− X
ρZ∞
x
fn(t)tρ
ρdt!
=X
ρ
n
ρ(ρ−n)
xρ−n
ln x−X
ρ
rρ(x)
ρ
Then
n
xn−1/2 ln xX
ρ
xiℑ(ρ)
ρ(ρ−n)−X
ρ
rρ(x)
ρ
and |Pρ
xiℑ(ρ)
ρ(ρ−n)|⩽Pρ1
|ρ|2=γ+ 2 −ln 4π⩽
0.0462. Thus, the formula applied for n= 1,
|I1(x)|⩽0.0462
x1/2 ln x1 + 1
ln x+4
ln2x+ln 2π
xln x.
To obtain an upper bound for I2, a frame of the
difference ψ−ϑis used: according to Lemma 3 of
[10], under the Riemann hypothesis, for x⩾121,
√x < ψ(x)−ϑ(x)<√x+4
3x1/3.
Thus, with the notations of Lemma 2.1,
I2(x)⩽F1/2,1(x) + 4
3F1/3,1(x).
Moreover, F1/2,1(x)⩽2
√xln x−2
√xln2x+8
√xln3xand
F1/3,1(x)⩽3
2x2/3 ln x. As a result,
I2(x)⩽2
√xln x1−1
ln x+4
ln2(x)+1
x1/6 .
2.2 Improvement of the error estimate.
In this section, we update the results, [8], [9] (Theo-
rem 4.1 & 4.4).
Here and throughout the rest of the paper, f(x) =
O∗(g(x)) means |f(x)|⩽g(x).
Theorem 2.3. Let M= 0.261497 ···, the Meissel-
Mertens constant (sequence A077761 of OEIS). If the
Riemann hypothesis is true, then we have for x⩾
1628.6,
X
p⩽x
1
p=ln ln x+M+O∗ln(x/ln x)
8π√x.
Proof. The sum of the reciprocals of the primes is re-
lated to ϑ(x)by (4.20) of [12],
X
p⩽x
1
p=ln2x+M+ϑ(x)−x
xln x
−R∞
x
(ϑ(y)−y)(1+ln y)
y2ln2ydy.
Let’s define Z1by
Z1=X
p⩽x
1
p−ln2x−M
.(4)
As
Z1⩽|ϑ(x)−x|
xln x+Z∞
x
|ϑ(y)−y|(1 + ln y)
y2ln2ydy.
(5)
we have for x≥1011 by Proposition 2.5 of [9], and
Lemma 2.2,
Z1≤1
8π
ln x−ln ln x−2
√x+I1(x) + I2(x)
≤1
8π
ln(x/ln x)
√x.
with I1(x)and I2(x)defined in the proof of
Lemma 2.2.
We check by direct calculation up to 1011 that the
bounds remain valid for Z1.
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3 Others sums or products of primes
We need to evaluate another integral for the other
functions of primes.
Proposition 3.1. Under the Riemann hypothesis,
Z∞
x
ϑ(t)−t
t2dt⩽2
√x+0.0462
x1/2 +2
x2/3 +ln 2π
x.
Proof. We need to evaluate
Z∞
x
ϑ(t)−t
t2dt=Z∞
x
ϑ(t)−ψ(t) + ψ(t)−t
t2dt
=I3(x)−I4(x)
where
I3(x) = Z∞
x
ψ(t)−t
t2dt,
and
I4(x) = Z∞
x
ψ(t)−ϑ(t)
t2dt.
To obtain an upper bound for I3, we use a classical
explicit formula in number theory (Chapter 17 of [11])
ψ(x) = x−ln 2π−Pρ
xρ
ρ−1
2ln(1 −1/x2)in
the form of
Z∞
x
f(t)(ψ(t)−t+ln 2π+1
2ln(1 −1/t2))dt
=−Z∞
xX
ρ
f(t)tρ
ρdt
and taking f(t) = 1
t2, we have f′(t) = O(1
t3)for
n⩾1and thus invert the sum and integral signs by
the dominated convergence theorem. Thus
I3(x) = − X
ρZ∞
x
f(t)tρ
ρdt!
−R∞
xf(t)(ln 2π+1
2ln(1 −1/t2))dt.
For t⩾2,0<−1
2ln(1 −1/t2)<ln 2πso for x⩾2,
−Z∞
x
f(t)(ln 2π)dt
<−Z∞
x
f(t)(ln 2π+1
2ln(1 −1/t2))dt < 0
Thus
0<Z∞
x
f(t)(ln 2π+1
2ln(1 −1/t2))dt⩽ln 2π
x.
On the other hand,
− X
ρZ∞
x
f(t)tρ
ρdt!=X
ρ
xρ−1
ρ(ρ−1)
Let, under the Riemann hypothesis
1
x1/2 X
ρ
xiℑ(ρ)
ρ(ρ−1)
and |Pρ
xiℑ(ρ)
ρ(ρ−1) |⩽Pρ1
|ρ|2=γ+ 2 −ln 4π⩽
0.0462. Thus we have,
|I3(x)|⩽0.0462
x1/2 +ln 2π
x.
To obtain an upper bound for I4, bounds of the dif-
ference ψ−ϑare used: according to Lemma 3 of [10],
under the Riemann hypothesis, for x⩾121,
√x < ψ(x)−ϑ(x)<√x+4
3x1/3.
Thus,
Z∞
x
√t+ 4/3 3
√t
t2dt=−2
√t−2
t2/3 ∞
x
.
As a result,
I4(x)⩽2
√x+2
x2/3 .
Theorem 3.2. Let B3(sequence A083343 in OEIS)
the constant given by the infinite sum
B3=γ+∞
X
n=2 X
p
(ln p)/pn≈1.33258 22757 33221.
Assuming the Riemann hypothesis, we have for x⩾
1674.5,
X
p⩽x
ln p
p=ln x−B3+O∗ln(x/ln x)
8π√xln x.(6)
Proof. By [12], (4.21), which refines Mertens’ first
theorem,
X
p≤x
ln p
p=ln x−B3+ϑ(x)−x
x−Z∞
x
ϑ(y)−y
y2dy.
Let’s define Z2by
Z2=|X
p≤x
ln p
p−ln x+B3|.(7)
Hence, the remainder term can be bounded using the
theta function
Z2⩽|ϑ(x)−x|
x+Z∞
x
ϑ(y)−y
y2dy
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We use Proposition 2.5 of [9], for the first term, and
Proposition 3.1 for the second, so we obtain
Z2⩽ln x
8π√x(ln(x/ln x)−2) + Z∞
x
ϑ(y)−y
y2dy
⩽ln x
8π√x(ln(x/ln x)−2) + 2
√x+0.0462
x1/2
+2
x2/3 +ln 2π
x≤ln x
8π√xln(x/ln x),
where the last inequality is only valid for x≥2.04 ·
1011.
We check (6) for x≤2.1·1011 by computer.
Lemma 3.3. For x≥2, we have
1
ζ(2) ≤Y
p≤x1−1
p2<1
ζ(2) 1 + 1
x.
Proof. We have by [6], (Proof of Theorem 280)
ζ(s)−Y
p≤q1
1−p−s
<∞
X
n=q+1
1
nσ,
where σ=ℜ(s). Hence
1
ζ(2) + Pn>x 1
n2
<Y
p≤x1−1
p2
and
Y
p≤x1−1
p2<1
ζ(2) −Pn>x 1
n2
=1
ζ(2)
1
1−1
ζ(2) Pn>x 1
n2
.
Let’s take
u=1
ζ(2) X
n>x
1
n2.(8)
For x≥2, we have u < 1
ζ(2) (ζ(2) −1−1/4) <0.25
and
1
1−u<1 + u+4
3u2for u < 0.25.(9)
We have also Pn>x 1
n2<R∞
x−1
dt
t2=1
x−1. By com-
bining definition (8) with inequation (9),
1
1−u<1+ 1
ζ(2)
1
x−1+4
31
ζ(2)
1
x−12
<1+ 1
x
for x≥5.
Theorem 3.4. If the Riemann hypothesis is satisfied,
we have
for x≥1628.0,
Y
p≤x1−1
p=e−γ
ln x1 + O∗ln(x/ln x)
8π√x
for x≥1628.4,
Y
p≤xp
p−1=eγln x1 + O∗ln(x/ln x)
8π√x
for x≥1629.2,
Y
p≤x
p+ 1
p=6
π2eγln x1 + O∗ln(x/ln x)
8π√x.
Proof. Let S=Pp>x (ln(1 −1/p) + 1/p). By [12,
(8.10)], we have
0> S > S0=−1.02
(x−1) ln xif x > 1.(10)
Then
P1=Y
p≤x1−1
p=exp
X
p≤x
ln 1−1
p
=exp
M−γ−X
p≤x
1
p−S
Proceeding in the same way as in the proof of Theo-
rem 2.3, we obtain
Z1≤t(x) = 1
8π√xln(x/ln x)−2ln ln x
8π√xln x(11)
Let z(x) = 1
8π√xln(x/ln x). We have exp(t(x) +
1.02
(x−1) ln x)≤1 + z(x)and (1 + 1/x)exp(t(x)) ≤
1 + z(x)for x≥1011.
Combining (11) and (10) with the previous result,
we get for x≥1011,
P1≤e−γ
ln xexp(Z1−S)≤e−γ
ln xexp(t(x)−S0)
≤e−γ
ln xln x(1 + z(x)).
Similarly, P2=1
P1=Qp≤xp
p−1≤
eγ(ln x)exp t(x)≤eγln x(1 + z(x)).
Moreover, P1≥1
eγln x(1+z(x)) =e−γ
ln x
1
1+z(x)≥
e−γ
ln x(1 −z(x)).We also have P2= 1/P1≥
1
e−γ(1+z(x))/ ln x≥eγln x1
1+z(x)> eγln x(1 −z(x)).
The last product is closely related to the others.
Since 1 + 1/p= (1 −1/p2)/(1 −1/p), we write
P3=Y
p≤x
p+ 1
p=P2·Y
p≤x1−1/p2.
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Hence, by Lemma 3.3, P3< P2·1
ζ(2) (1 + 1/x)<
eγln x
ζ(2) (1+1/x)exp t(x)<eγln x
ζ(2) (1+z(x)) and P3>
P2/ζ(2) >eγln x
ζ(2) (1 −z(x)).
We check by computer for x≤1011 that the in-
equalities are still valid.
4 Conclusion
Analytic number theory, [13], studies the properties
of functions on prime numbers using analytic objects
(for example, here, the use of Riemann’s zeta func-
tion). Under Riemann’s hypothesis concerning the
zeros of the zeta function, we obtain, in Section 2,
an accurate estimate of the sum of the reciprocals of
the primes. The result can be applied to other sums,
as we saw in Section 3. This study can be developed
for other functions using prime numbers.
Acknowledgements: We would like to thank the
anonymous reviewers whose comments helped me to
improve the quality of this article.
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