Acta Applicandae Mathematicae

, Volume 121, Issue 1, pp 1–3

Rebuttal of Kowalenko’s Paper As Concerns the Irrationality of Euler’s Constant γ


  • Mark W. Coffey
    • Department of PhysicsColorado School of Mines

DOI: 10.1007/s10440-012-9697-z

Cite this article as:
Coffey, M.W. & Sondow, J. Acta Appl Math (2012) 121: 1. doi:10.1007/s10440-012-9697-z


We rebut Kowalenko’s claims in this journal that he proved the irrationality of Euler’s constant γ, and that his rational series for γ is new.


Euler’s constantIrrationalityKluyver’s numbersRebuttal

1 Introduction

The irrationality of Euler’s constant
$$\gamma=\lim_{n \to\infty} \Biggl(\sum_{k=1}^n\frac{1}{k}-\log n \Biggr)=0.577215664901532860606512090082402431042\ldots $$
has been long conjectured. However, it remains an open problem.

Recently in this journal Kowalenko claimed that simple arguments suffice to settle this matter [4]. As he offered no general framework or new mathematical principle, we believe that the following illustrations are sufficient to describe the flaws in his very limited approach.

2 A Faulty Irrationality Argument

Kowalenko derives the following formula for Euler’s constant in (65) on p. 428:
$$\gamma= \sum_{k=1}^{\infty}\frac{A_0}{k(k + 1)} -\sum_{k=2}^{\infty }\frac{A_1}{k^2} + \sum _{k=3}^{\infty}\frac{A_2}{k(k - 1)} - \sum _{k=4}^{\infty}\frac{A_3}{k(k - 2)}+\dotsb.$$
Here \(A_{0},A_{1},A_{2},A_{3},\dotso\) are certain rational numbers. He writes:

With the exception of the second series on the rhs, all the series in (65) can be easily evaluated by decomposing them into partial fractions. On the other hand, the second series on the rhs is virtually equal to ζ(2).

He then transforms formula (65) into the following series in (69):
$$\gamma= \frac{3}{2}-\frac{1}{2} \biggl(\frac{\pi^2}{6} + \frac{1}{12} +\frac {5}{144} + \frac{247}{12960} +\frac{ 77}{6400} +\frac {25027}{3024000}+\dotsb \biggr).$$
Kowalenko states:

For γ to be rational the term involving π2/6, which arises solely from the summation over 1/k2 or ζ(2) in (65), has to be cancelled by the remaining sum. This means that we need to examine the methods for converting an irrational number into a rational number by the process of addition. There are only two possible methods for achieving this, which are best understood if we regard an irrational number as an infinite random distribution of decimal digits. \(\dotso\) Therefore, for γ to be rational, we need to convert a random distribution into a non-random one.

But, for example, the distribution of digits in Liouville’s irrational number
$$\sum_{n=1}^\infty\frac{1}{10^{n!}}=0.1100010000000000000000010 \ldots $$
is not random, as the sum formula shows. Thus Kowalenko’s understanding of irrationality is lacking.
He continues:

The first method by which an irrational number can be converted to a rational number is to add another number, which at some stage possesses the opposite random distribution to the original irrational number. This represents the situation whereby the second number can be expressed as Cπ2/6, with C a rational number. Such a situation, however, cannot occur with (69). First, we note that if we are to obtain Cπ2/6 from the remaining terms after the π2/6 term in the parenthesis of (69), then these terms would have to yield a summation involving 1/k2. This is simply not possible as all the 1/k2 terms have already been removed as mentioned above.

Here Kowalenko apparently assumes that Euler’s series \(\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) is the only way to represent π2/6 as the sum of a series of rational numbers. Of course, that is not true; there are infinitely many such representations of π2/6, and hence of Cπ2/6, for any rational number C>π2/6.

Kowalenko concludes the paragraph:

Furthermore, the summation would have to be negative. Yet all the terms in the parenthesis in (69) are positive definite. Therefore, it is simply impossible for all the terms in (69) to yield Cπ2/6.

Here he claims that the sum of a series of positive rational numbers cannot be equal to Cπ2/6. But, for example, decimal expansion does give such a series:
$$C-\frac{\pi^2}{6} = n + 0.d_1d_2d_3\dotso= n+ \sum_{k=1}^{\infty}\frac {d_k}{10^k}.$$

In view of his misconceptions, Kowalenko has not proven that the “first method by which an irrational number can be converted to a rational number” does not lead to the rationality of Euler’s constant. Thus its irrationality remains an open problem.

3 A Known Representation

Finally, we point out that Kowalenko’s claim to have found “a new representation for Euler’s constant” [4, p. 143] is also incorrect. Namely, what he calls “Hurst’s formula,” which is \(\gamma= \sum_{k=1}^{\infty}\frac{\lvert A_{k}\rvert}{k}\) in (60) of [4], is known. According to Gourdon and Sebah [1, p. 6], the formula was discovered in 1924 by Kluyver [2, 3], who wrote it as \(\gamma= \sum_{k\ge1} \frac{a_{k}}{k}\). By comparing the values of Kluyver’s numbers ak for 1≤k≤7 (see the end of Sect. 2.3 in [1, p. 6]) with those of Kowalenko’s numbers Ak (see Table 1 on p. 418), one readily observes that ak=|Ak|.

Indeed, letting
$$(z)_n=\frac{\Gamma(z+n)}{\Gamma(z)} = z(z+1)\dotsb(z+n-1)$$
denote the Pochhammer symbol, with Γ the Gamma function, we have the following relations. From (18) of [4],
$$A_k=\frac{(-1)^k}{k!} \int_0^1(-x)_k dx,$$
while from [2] (see also [3, p. 143]),
$$a_k=\frac{1}{k!}\int_0^1x(1-x)_{k-1}dx=-\frac{1}{k!}\int_0^1(-x)_k dx,$$
and we therefore conclude that ak=(−1)k+1Ak.

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© Springer Science+Business Media B.V. 2012