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An Elementary Proof of the Riemann Hypothesis
Jose Risomar Sousa
June 27, 2021
Abstract
The Riemann hypothesis is true. In this paper I present a solution for it in a very
short and condensed way, making use of one of its equivalent problems. But as Carl
Sagan once famously said, extraordinary claims require extraordinary evidence. The
evidence here is the newly discovered inversion formula for Dirichlet series.
1 Introduction
The Riemann hypothesis is the second of the 7 Millennium Problems proposed by the Clay
Mathematics Institute back in year 2000, and it has a 1 million dollar prize for whoever solves it.
It was formulated by Bernhard Riemann in 1859, and has been unsolved for more than
150 years now. The Riemann hypothesis is about the zeta function, ζ(s), a special case of
Dirichlet series:
ζ(s) =
∞
X
n=1
1
ns, if <(s)>1(1)
A Dirichlet series is any infinite sum of the type,
Fa(s) =
∞
X
n=1
a(n)
ns
where a(n) is an arithmetic function and Fa(s) is its generating function (hence, the zeta
function has a(n) = 1).
The analytic continuation of the zeta function to 0 <<(s)<1 (the so-called critical
strip) can be attained by means of the alternating zeta function (also known as the Dirichlet
eta function). The zeta function can also be continued to <(s)<0, by means of the Rie-
mann functional equation, which pairs ζ(s) up with ζ(1 −s), therefore enabling points with
<(s)>1 to complete the function for points with <(s)<0. Finally, the Riemann hypothe-
sis is the statement that all the non-trivial zeros of the function thus obtained have <(s) = 1/2.
1
Despite its very sound arguments, this paper is not meant to be a definitive proof of the
Riemann hypothesis, unless vetted or improved upon by an expert, but it definitely adds nov-
elties to the picture, besides being a really interesting exposition. At worst, the new facts
presented here represent evidence in favor of the Riemann hypothesis, even if some details
may be missing.
In this article, we try and make a distinction between the zeta function Dirichlet series and
the function itself.
2 The background
It stems from the Euler product that when <(s)>1, 1/P1/nsis a Dirichlet series that
has a(n) = µ(n), the so-called M¨obius function. The Riemann hypothesis is then equivalent
to a statement that can be made about this new Dirichlet series and the reciprocal of the zeta
function in the upper half portion of the critical strip, <(s)>1/2. But let’s go over some of
the details first.
2.1 The Euler product
In 1737, German mathematician Euler discovered an interesting relationship between the
zeta function and the primes known as Euler product, valid when <(s)>1:
ζ(s) =
∞
Y
n=1
1
1−p−s
n
=
∞
X
n=1
1
ns(2)
When (2) is inverted, it reveals a relationship between the reciprocal of the zeta function
and the square-free numbers:
1
ζ(s)=
∞
Y
n=1 1−1
ps
n=1−1
2s1−1
3s1−1
5s· · ·
The observation of this relation allows us to write 1/ζ(s) as function of µ(n), a function
introduced in 1832 by another German mathematician, August F. M¨obius, which involves the
concept of square-free numbers. That is, provided that <(s)>1.
2.2 The M¨obius function
A square-free number is a number that can’t be divided by any squared prime. In other
words, if nis square-free, p1p2· · · pkis its unique prime decomposition. Hence, we can define
a function µ(n) such that:
µ(n) =
1,if n=1
(−1)k,if nis square-free with kprime factors
0,if nis not square-free
2
This is the M¨obius function from the previous section.
With it, it’s easy to write the reciprocal of the zeta function Dirichlet series, which also
happens to be a Dirichlet series. By alternating the signs of the terms in the zeta series (and
eliminating circa 40% of them, the square-full), the result seems to converge for <(s)>1/2,
based on empirical evidence:
1/ ∞
X
n=1
1
ns!=
∞
X
n=1
µ(n)
ns(3)
At this point, this equation is still only valid if <(s)>1. It stems from the Euler product,
and the new series on the right is referred to throughout this text as F(s).
However, unlike the zeta function series, F(s) should converge even for 1/2<<(s)≤1,
and according to the literature, a certain statement about it and the zeta function is equivalent
to the Riemann hypothesis.
2.3 A restatement of the Riemann hypothesis
The Riemann hypothesis is equivalent to the statement that the equation,
1
ζ(s)=
∞
X
n=1
µ(n)
ns,(4)
is valid for every swith real part greater than 1/2.
This restatement says that when we invert the zeta function Dirichlet series, we obtain
another Dirichlet series, F(s), that is in keeping with the analytically continued values of the
zeta function (that is, values such as 1/2<<(s)≤1). It’s not clear if this restatement requires
that F(s) be defined only for <(s)>1/2 for the RH to hold.
To help lay the foundation, let’s define FN(s) as the partial sums of F(s):
FN(s) =
N
X
n=1
µ(n)
ns(5)
3 The solution
What this proof really establishes is the validity of equation (4) for every swith <(s)>1/2,
even if ζ(s) = 0 for some such s(which would mean the Riemann hypothesis is false). In that
case, however, the equation would still hold in a sense, if F(s) diverged (which seems implausi-
ble, since Dirichlet series either always converge or never converge for <(s) in a given domain).
The Riemann hypothesis therefore implies that F(s) never diverges for <(s)>1/2.
Convergence of F(s) for <(s)>1/2 is based on empirical evidence. It’s not proved in this
paper and it’s probably not trivial.
3
3.1 Inversion formula for Dirichlet series
Before laying out the points of the proof, let’s review the inversion formula for Dirichlet
series, whose proof and ubiquitous evidence for were given in paper [1]. It’s a proven theorem
that relates a Dirichlet series to its arithmetic function, and can be stated as follows.
Theorem 1 Suppose that Fa(s) is a Dirichlet series and a(n) is its associated arithmetic
function. Then for any positive integer qsuch that Fa(2q)<∞,a(n) is given by:
a(n) = −2
∞
X
i=q
(−1)i(2πn)2i
i
X
j=q
(−1)j(2π)−2jFa(2j)
(2i+ 1 −2j)!
Proof 1 Although not obvious, this is a very powerful result. The above power series
converges for all nand is the analytic continuation of:
−sin 2πn
πn
∞
X
j=q
n2jFa(2j), (6)
since they have the same Taylor series expansion and (6) only converges for |n|<1. In some
cases it’s possible to find a closed-form for a(n), though it can be challenging.
The proof is short and simple:
−2
∞
X
i=1
(−1)i(2πn)2i
i
X
j=1
(−1)j(2π)−2jFa(2j)
(2i+ 1 −2j)! ⇒
−2
∞
X
i=1
(−1)i(2πn)2i
i
X
j=1
(−1)j(2π)−2j
(2i+ 1 −2j)!
∞
X
k=1
a(k)
k2j⇒
∞
X
k=1
a(k) −2
∞
X
i=1
(−1)i(2πn)2i
i
X
j=1
(−1)j(2π)−2jk−2j
(2i+ 1 −2j)! !
The theorem then follows from the following equation:
−2
∞
X
i=1
(−1)i(2πn)2i
i
X
j=1
(−1)j(2πk)−2j
(2i+ 1 −2j)! =(1,if n=k
0,otherwise (7)
And the above equation is justified for being the convolution of µ0(n), the unit function,
and the arithmetic function of the series k−s,b(n), since from the convolution formula:
c(n) = (µ0∗b)(n) = X
d|n
µ0(d)bn
d=b(n) = (1,if n=k
0,otherwise
4
For more details on the proof, please refer to [1].
This is a real breakthrough and a surprising result. It says that every Dirichlet series has
coefficients given by a Taylor series. One of the advantages of this formula is that if you know
Fa(s) at the even integers, you know the series expansion of the a(n). Another advantage is
that it extends a(n) to the complex numbers. Perhaps this extended function even has some
deeper connection to the Dirichlet series, though that’s not currently known.
The inversion formula also has a property that’s trivial to see. For any complex s:
a(n)
ns=−2
∞
X
i=q
(−1)i(2πn)2i
i
X
j=q
(−1)j(2π)−2jFa(s+ 2j)
(2i+ 1 −2j)!
Theorem 2 Let G(s) be a function that admits a Dirichlet series representation, Fa(s), in
at least part of its domain. Then at the integers the a(n), derived from G, are finite and not
all zero.
Proof 2 Self-evident.
In other words, the inversion formula can be used to check if a function is a Dirichlet series.
It says nothing about the convergence domain of Fa(s), though.
3.2 Riemann hypothesis proof
First we note that, even though the zeta function only admits a Dirichlet series representa-
tion for <(s)>1, its reciprocal can be expressed as a Dirichlet series supposedly for <(s)>1/2.
The analytic continuation of the zeta function to 0 <<(s)<1 is achieved through the
Dirichlet eta function, η(s), that is, the below equation holds for <(s)>0:
ζ(s) = η(s)
1−21−s=1
1−21−s
∞
X
j=1
(−1)j+1
js,
With everything that’s been said, we’re now in a position to produce a short proof. Suppose
that for <(s)>1/2,
1
ζ(s)=1−21−s
η(s)=
∞
X
n=1
a(n)
ns(8)
What is a(n)?
The inversion formula tells us that a(n) is given by the below power series:
a(n) = −2
∞
X
i=1
(−1)i(2πn)2i
i
X
j=1
(−1)j(2π)−2jζ(2j)−1
(2i+ 1 −2j)! (9)
5
Despite the fact that this formula yields the coefficients of the Dirichlet series 1/ζ(s) inde-
pendently of its convergence domain, it has no known closed-form and hence in principle we
don’t know what values it assumes at the positive integers. However, since for <(s)>1 we
know from the Euler product that a(n) = µ(n), and (9) doesn’t depend on s, we conclude that
a(n) = µ(n) for every positive integer n.
Below the graph of µ(n) was plotted in the (0,11) interval for some insight into its shape
and local minima and maxima (it crosses the x-axis at the square-full and half-integers):
Prior to this discovery nobody even knew that the M¨obius function could be continuous or
analytic, or what its shape could look like. Therefore, there is no question that equation (4)
holds whenever F(s) converges (F(s) is the Dirichlet series representation of (1 −21−s)/η(s)).
3.3 Convergence of F(s)
Intuitively, it seems easy to see why F(s) should always converge when 1/2<<(s)≤1,
the sum of the square-free numbers with an odd number of primes (raised to s) should always
be an infinity more or less equal to the sum of those with an even number of primes and these
infinities should cancel out. As a matter of fact, these infinities will be exactly equal when
s= 1 (which is a pole in the zeta function), causing F(1) to zero out. There seems to be an
inflection point though, <(s) = 1/2, from which these infinities either become infinitely differ-
ent, or they are still close but their difference varies erratically as more terms are added (the
limit of FN(s) doesn’t exist, which may have to do with the spacing between large square-free
numbers).
6
We know the partial zeta function is asymptotic to the logarithm when Nis large, H(N)∼
γ+ log N1, but what about FN(1/2)?
4 Conclusion
The equivalent problem shown in this paper is a result from the literature, some mathe-
matician has found a way to prove that it implies the Riemann hypothesis. The missing piece
of the puzzle was to come up with an alternative way to prove that,
1
ζ(s)=
∞
X
n=1
a(n)
ns⇒a(n) = µ(n),
without the requirement that <(s)>1, which the inversion formula for Dirichlet series does.
References
[1] Risomar Sousa, Jose An exact formula for the prime counting function, eprint
arXiv:1905.09818, 2019.
1That is, s= 1 is the inflection point, with H(N) the harmonic numbers.
7