The document discusses the finding that the equality between the Riemann zeta function and the Euler product does not seem to hold. Through examples calculating finite zeta functions up to k=8, an error term is found that gets larger as k increases. Testing on larger primes up to 2147321881 confirms the error increases and the functions do not appear to be equal. The equality only seems to hold for the Riemann zeta function when s>1, but not for the interesting case of s=1/2 or when using the eta function for 0<s<1.

1. Disprove of equality between Riemann zeta function
and Euler product
Chris De Corte
chrisdecorte@yahoo.com
KAIZY BVBA
Beekveldstraat 22
9300 Aalst, Belgium
November 18, 2014
1

2. The goal of this document is to share with the mathematical community the
finding that the equality between the Riemann z`eta function and the Euler
product does not seem to hold.
2

3. Contents
1 Key-Words 4
2 Introduction 4
3 De

4. nition and examples 4
3.1 Definition of finite zeta function . . . . . . . . . . . . . . . . . 4
3.2 example k=2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3.3 example k=4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.4 example k=8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.5 Generalized error formula . . . . . . . . . . . . . . . . . . . . 6
3.6 Intermediate Summary . . . . . . . . . . . . . . . . . . . . . . 6
4 Testing the formula's 6
4.1 Calculating absolute error . . . . . . . . . . . . . . . . . . . . 6
4.2 Calculating the relative error . . . . . . . . . . . . . . . . . . 7
4.3 Relationship between absolute and relative error . . . . . . . 8
4.4 Some other calculation results . . . . . . . . . . . . . . . . . . 8
4.5 Using the `eta function for 0 s 1 . . . . . . . . . . . . . . 9
5 Discussions 10
6 References 10
3

5. 1 Key-Words
Riemann, Euler z`eta function, Euler product, Prime counting.
2 Introduction
The following document originated out of our interest for primes.
We remain surprised by the fact that so much attention goes to the investi-gation
of the non trivial zero’s of the Riemann zeta function in the critical
strip while we could not find an exact link with the distribution of the primes
[1].
For this reason, we started to investigate the fundamentals all over again.
The esteemed proof of the equality between the Riemann zeta function and
the Euler product can be found in [2].
This document approaches this proof from a different angle.
3 De

6. nition and examples
3.1 De

7. nition of

8. nite zeta function
We define the finite z`eta function as follows:
k =
Σk
i=1
(1=i) (1)
We will first redo the proofs found in [2] for a few examples of this limited
zeta function and then discuss k →∞ once we have clarified our thoughts.
3.2 example k=2
2 = 1 + 1=2 (2)
1=22 = 1=2 + 1=2(1=2) (3)
4

9. Subtracting both equations, we get:
(1 − 1=2)2 = 1 − 1=2(1=2) = 1 − errA(2) (4)
We notice that when redoing the proof found in [2], an error term appears
which for k = 2 equals to errA(2) = 0:25.
We call this type of error the absolute error as we will later also introduce
a relative error.
3.3 example k=4
4 = 1 + 1=2 + 1=3 + 1=4 (5)
1=24 = 1=2 + 1=4 + 1=2(1=3 + 1=4) (6)
Subtracting equation (6) from (5), we get:
(1 − 1=2)4 = 1 + 1=3 − 1=2(1=3 + 1=4) (7)
Preparing to eliminate prime number 3 from the right side of the equation:
1=3(1 − 1=2)4 = 1=3 + 1=3:1=3 − 1=2:1=3(1=3 + 1=4) (8)
Subtracting equation (8) from (7), we get:
(1−1=3)(1−1=2)4 = 1−1=2(1=3+1=4)(1−1=3)−1=3:1=3 = 1−errA(4) (9)
We notice that when redoing the proof for k = 4, a bigger error appears of
errA(4) = 0:3055
3.4 example k=8
4 = 1 + 1=2 + 1=3 + 1=4 + 1=5 + 1=6 + 1=7 + 1=8 (10)
Trying to eliminate all the fractions from the right side of the equation
results in:
(1 − 1=7)(1 − 1=5)(1 − 1=3)(1 − 1=2)8 =
1 − 1=2(1=5 + 1=6 + 1=7 + 1=8)(1 − 1=3)(1 − 1=5)(1 − 1=7)
− 1=3(1=3 + 1=5 + 1=7)(1 − 1=5)(1 − 1=7) − 1=5(1=5 + 1=7)(1 − 1=7)
= 1 − errA(8) (11)
5

10. We notice that when redoing the proof for k = 8, an even bigger error
appears of errA(8) = 0:3584
3.5 Generalized error formula
Based on formula (11), the error term for k → ∞ can be assumed to be of
a size that can be represented as follows:
errA(∞) ≈
Σ
8pi

(1=pi)
Σ
8pjpi
(1=pj)
Π
8pkpi

 (12)
(1 − 1=pk)
Note: we will not use this formula in the remainder of this document.
3.6 Intermediate Summary
When using a limited zeta function and trying to eliminate the original
fractions on the right side of the equation, we are actually creating extra
fractions in stead and the frightening thing is that when increasing the k
of our limited zeta function, the error factor only becomes bigger. We can
actually recognize a pattern in the structure of the error factor which gives
us no reason to believe that it will become zero for k → ∞ nor do we expect
it to change signs in between. Moreover, this error factor is not of a size that
we can neglect as it will seem. The general accepted proof for the equality
[2] seems to neglect the impact of these infinite amount of extra fractions
created.
4 Testing the formula's
4.1 Calculating absolute error
Based on the above, we can now define the formula for the absolute error as
follows:
errA(k) = 1 −
Π
8pik
(1 − 1=pi)
Σk
i=1
(1=i) (13)
6

11. We calculated this formula for a set of primes in PARI/GP and the results
are demonstrated in the table 1. For the large primes, we had to rely on
data from [3] and extra calculations done in C++.
Table 1: This table shows the absolute errors calculated for a list of values
using formula (12)
k errA
2 0.2500
11 0.3725
101 0.3809
1009 0.3938
10007 0.4041
100003 0.4106
500509 0.4139
999999937 0.4229
2147321881 0.4235
4.2 Calculating the relative error
So far, we have been occupied with calculating what we called the absolute
error. Maybe, it is more interesting to calculate the relative error. We define
the relative error as follows:
errR(k) =
Π
8pik
1
(11=pi) Σk
i=1(1=i)
(14)
A summary of this can be seen in table 2. The data of table 2 has been
represented graphically in Figure 1, which can be found at the end of this
document.
7

12. Table 2: This table shows the relative errors calculated for a list of values
using formula (13).
k errR
2 1.3333
11 1.5936
101 1.6152
1009 1.6497
10007 1.6781
100003 1.6966
500509 1.7062
999999937 1.7328
2147321881 1.7345
4.3 Relationship between absolute and relative error
Out of formula (13) and (14), we deduce:
errA(k) = 1 − 1
errR(k)
(15)
4.4 Some other calculation results
In the above, we assumed that s = 1. We recalculated the relative error for
a few examples on a smaller data set for s̸= 1. The results can be found in
table 3.
Table 3: This table shows the relative errors for some s̸= 1.
k errR(s = 1) errR(s = 2) errR(s = 1:01) errR(s = 1=2)
11 1.5936 1.0323 1.5741 6
101 1.6152 1.0043 1.5820 61
1009 1.6497 1.0005 1.5994 26251
10007 1.6781 1 1.6099
100003 1.6966 1 1.6106
500509 1.7062 1 1.6080
8

13. These calculations show that: for s ≥ 2, the relative error quickly converges
to 1, for s slightly larger than 1, the relative error first goes up but then
comes back down again towards 1, for s = 1=2, the relative error quickly
becomes very big.
4.5 Using the eta function for 0 s 1
Defining the finite `eta function as follows [4]:
k(s) =
Σk
i=1
(−1)i+1
is (16)
We replace the final z`eta function as follows [4]:
k(s) =
1
1 − 21s k(s) (17)
We redefine the formula for the relative error as follows:
errR(k) =
Π
8pik
1
(11=psi
)
1
121s k(s)
(18)
Using formula 18, we calculate some examples for the relative error in the
strip 0 s 1. The results can be found in table 4.
Table 4: This table shows the relative errors for some 0 s 1.
k errR(s = 0:01) errR(s = 1=2) errR(s = 0:99) errR(s = 0:999)
11 -1.810 109 -18.52 -0.0465 -0.0045
101 -1.938 1039 -732 -0.0874 -0.0083
1009 -1087741 -0.1326 -0.0124
10007 -0.1805 -0.0165
100003 -0.2309 -0.0207
500509 -0.2677 -0.0237
These calculations show that: for s → 1, the relative error seems to become
stable. This is in line with our previous calculations. For all s 1, also
9

14. s = 1=2, the error becomes too big too soon hence we can conclude that
there seems to be no relationship between the Euler product formula and
the z`eta function.
5 Discussions
The equality between the Euler z`eta function and the Euler product does
not seem to hold (for s = 1).
The equality between the more general Riemann z`eta function and the Euler
product does not hold for the interesting case of s = 1=2 but it seems to
hold for the less interesting cases of s 1:
1 +
1
2s +
1
3s +
1
4s
· ··̸=
1
(1 − 1
2s )
· 1
(1 − 1
3s )
· 1
(1 − 1
5s )
· ·· (19)
We don’t know where the relative error defined in (14) goes to in figure 1.
So, we would have to guess. For the moment, it seems to be a bit far away
from = 1:7810292. This , which also came up in [3], can be very closely
approximated as:
= e
where
≈ 0:57721 is the Euler −Mascheroni constant (20)
Another guess could be
√
which is 1.77.
But for the moment, the relative error in our graph does not seem to be able
to reach the 1.74 ...
We conclude by stating that only for s = 1, there seems to be a relation-ship
between the Euler product formula and the z`eta function and that in
proportion of a factor 1.7 ...
6 References
1. Wikipedia, the free encyclopedia (2014). Riemann hypothesis. Page
1.
2. Wikipedia, the free encyclopedia (2014). Proof of the Euler product
formula for the Riemann zeta function. Page 1-2.
10

15. 3. Chris De Corte (2014). Probabilistic approach to prime counting.
Page 1-11.
4. Uitgeverij Bert Bakker (2012). De zeven grootste raadsels van de
wiskunde. Page 44-45.
11

16. Figure 1: This figure represents the relative error calculated in table [2]. The
drawing extents from prime 7919 with a relative error of 1.6755 to prime
2147321881 with a relative error of 1.7345
12