A Short Telescoping Proof of Hata's Formula for the Euler-Mascheroni Constant

#A90 INTEGERS 25 (2025) A SHORT TELESCOPING PROOF OF HATA’S FORMULA FOR THE EULER-MASCHERONI CONSTANT Nikita Kalinin Mathematics and Computer Science Department, Guangdong Technion Israel Institute of Technology, Shantou, China [email protected] Received:7/17/25 , Revised: 8/21/25, Accepted: 9/17/25, Published: 11/5/25 Abstract This note presents an elementary proof of a formula for Euler’s constant γ, originally due to Masayoshi Hata. We obtain Hata’s formula by a telescoping argument over Farey intervals. 1. Introduction Recall that Euler’s constant (also known as the Euler–Mascheroni constant) is defined as γ= lim n→∞ 1 + 1 2+· · · +1 n−ln n. Alternative series representations include γ= ∞ X k=2 (−1)kζ(k) k= log 4 π+ ∞ X m=2 (−1)m m2m m X k=1 m k. Euler’s constant is also related to the Riemann ζ-function through the identity γ= log 4π+X ρ 1 ρ−2, where the sum is over the non-trivial zeros ρof the Riemann zeta function. Further formulas and connections can be found in the survey [1] and in a more elaborate survey [4]. However, these and many other presentations of γoffer little hope for proving that γis an irrational number (which is widely believed), so new formulas and methods are highly desirable. In this note, we investigate a lesser-known series for γ, due to Masayoshi Hata. DOI: 10.5281/zenodo.17535156 INTEGERS: 25 (2025) 2 Definition 1. An interval a b,c dsuch that a, b, c, d ∈Z≥0,0≤a b<c d≤1,and ad −bc =−1 is called a Farey interval. In the literature, Farey intervals are also often called Stern-Brocot intervals. Let Fdenote the set of Farey intervals. Let F∗=0 1,1 n:n∈N⊂ F. Masayoshi Hata proved the following theorem. Theorem 1 ([2]).In the above notation γ=1 2+1 2X [a b,c d]∈F\F∗ 1 abcd(a+c)(b+d). Hata’s proof is very nice. He studies presentations of functions in a certain Schauder basis associated with Farey intervals. Then, using a Parseval-type identity for the function ψ(t) = t{1 t}1− {1 t}, he derives the above theorem. In subsequent work, Haynes–Vaaler [3] showed that the derivatives of these piecewiselinear functions from that Schauder basis form an orthonormal basis and a complete system of martingale differences in L2([0,1]), with applications to metric properties of continued fractions. The goal of this note is to give a short and elementary proof of the above theorem using the telescoping structure over Farey intervals. 2. A Telescoping Proof of Hata’s Theorem We begin with the following lemma, which is proved by direct computation. Lemma 1. Let I=a b,c dbe a Farey interval. Then ad +bc abcd −a(b+d) + b(a+c) ab(a+c)(b+d)−(a+c)d+ (b+d)c (a+c)(b+d)cd =(bc −ad)2 abcd(a+c)(b+d)=1 abcd(a+c)(b+d). The identity in Lemma 1 suggests to associate the term ad+bc abcd to the interval I, so denote fha b,c di=f(I) = ad +bc abcd =1 bc +1 ad. INTEGERS: 25 (2025) 3 The other two terms in Lemma 1 correspond to the values of f(Ileft) and f(Iright) for two Farey intervals Ileft and Iright from the subdivision of Iby the mediant a+c b+d of two fractions a band c d: Ileft =a b,a+c b+dand Iright =a+c b+d,c d. Hence, Lemma 1 can be reformulated as 1 abcd(a+c)(b+d)=f(I)−f(Ileft)−f(Iright).(1) The idea of the proof of Theorem 1 is to sum the Equation (1) over all I∈ F \F∗. To illustrate the telescoping property, we consider the following example, where we sum Equation (1) over I=1 2,1 1with Ileft =1 2,2 3and Iright =2 3,1 1: f1 2,1 1−f1 2,2 3−f2 3,1 1 +f1 2,2 3−f1 2,3 5−f3 5,2 3 +f2 3,1 1−f2 3,3 4−f3 4,1 1 =f1 2,1 1−f1 2,3 5−f3 5,2 3−f2 3,3 4−f3 4,1 1, which is the seed term f1 2,1 1minus the sum of boundary terms corresponding to the partition 1 2,3 5,2 3,3 4,1 1of 1 2,1 1. To compute the limit of finite sums as above, in the following lemma we show that the sums of boundary terms converge to the integral of 2 tas the mesh of partition tends to zero. Lemma 2. Let I=a b,c dbe a Farey interval with a partition a b=a1 b1 =x1<a2 b2 =x2<· · · <am bm =c d=xm, such that [ak bk,ak+1 bk+1 ]is a Farey interval for each k= 1, . . . , m −1and a > 0. Then  m−1 X k=1 fak bk ,ak+1 bk+1 − c/d Z a/b 2 tdt ≤ c d−a b·b2 a2·max k=1,...,m−1 1 bkbk+1 .(2) Proof. The bound on the right of Inequality (2) will follow from the standard estimate  (y−x)g(x) + g(y) 2− y Z x g(t)dt ≤|y−x|2 2max t∈[x,y]|g′(t)|,(3) INTEGERS: 25 (2025) 4 for a C1-function gon an interval [x, y]. We consider g(t) = 1/t. Each Farey interval [xk, xk+1] = hak bk,ak+1 bk+1 ihas length xk+1 −xk=1 bkbk+1 . The function fcan be rewritten as fak bk ,ak+1 bk+1 =1 bkbk+1 bk+1 ak+1 +bk ak= (xk+1 −xk)·(g(xk) + g(xk+1)). Therefore m−1 X k=1 fak bk ,ak+1 bk+1 = m−1 X k=1 (xk+1 −xk)·(g(xk) + g(xk+1)),(4) which converges to 2 c/d R a/b g(t)dt as maxk|xk+1 −xk| → 0, since it is a Riemann sum. Applying Inequality (3) with g(t)=1/t on each subinterval [xk, xk+1] and summing over k, then using Equation (4), we obtain  m−1 X k=1 fak bk ,ak+1 bk+1 − c/d Z a/b 2 tdt ≤ m−1 X k=1  ak+1 bk+1 −ak bk 2 ·max t∈[a b,c d]|g′(t)|. Finally, to establish Inequality (2), we compute maxt∈[a b,c d]|g′(t)|=b2 a2and m−1 X k=1  ak+1 bk+1 −ak bk 2 ≤max k=1,...,m−1 ak+1 bk+1 −ak bk · c d−a b= max k=1,...,m−1 1 bkbk+1 · c d−a b. We recall some classical properties of Farey intervals. Let F0=0 1,1 1,F1=0 1,1 2,1 2,1 1, F2=0 1,1 3,1 3,1 2,1 2,2 3,2 3,1 1, . . . where Fn+1 is obtained from Fnby replacing each interval I= [a b,c d]∈ Fnby two intervals [a b,a+c b+d] and [a+c b+d,c d]. Using induction by n, we see that for each [a b,c d]∈ Fnwe have max(b, d)≥n+ 1.(5) Since [0 1,1 1] is a Farey interval and a(b+d)−b(a+c) = ad −bc, it follows that Fn consists of Farey intervals for each n≥0. Since the length of I= [a b,c d]∈ Fnis 1 bd , using Inequality (5) we get the following corollary. Corollary 1. The length of each interval I∈ Fnis at most 1 n+1 . INTEGERS: 25 (2025) 5 We recall the proof that S∞ n=0 Fn=F. It follows from the construction that all Fnare disjoint. It remains to show that each Farey interval [a b,c d] belongs to a certain Fn. We proceed by induction on max(b, d). If max(b, d) = 1 then b=d= 1, a = 0, and c= 1, so [a b,c d] = [0 1,1 1]∈ F0. Assume that the claim holds for max(b, d)< k. Consider a Farey interval [a b,c d] with max(b, d) = k. Since ad −bc =−1, it is not possible that b=d=k > 1. Without loss of generality, suppose that b < d. Then a<c. Thus c dis the mediant of fractions a band c−a b−d; also a(b−d)−b(c−a) = ad −bc =−1; so ha b,c−a b−diis a Farey interval. Since max(b, b −d)< k, by induction hypothesis we have that [a b,c−a b−d]∈ Fn−1for a certain n≥1. Hence [a b,c d]∈ Fnby the definition. Corollary 2. For each interval I= [a b,c d]∈ F \ F∗there exists n≥1such that I⊂[1 n+1 ,1 n]. Note that Icannot be [0 1,1 1], so Imust be contained in [0 1,1 2] or in [1 2,1 1] = [1 n+1 ,1 n] for n= 1. If Iis contained in [0 1,1 2], then, since Icannot be equal to [0 1,1 2], Imust be contained in [0 1,1 3] or [1 3,1 2]=[ 1 n+1 ,1 n] for n= 2, etc. Lemma 3. Fix n≥1. Consider the interval Jn=h1 n+1 ,1 ni∈ Fn. Then the sum of expressions in Equation (1) over all Farey intervals I⊂Jnequals 1 n+1 n+ 1 −2 ln(n+ 1 n). Proof. Sum Equation (1) over the intervals I=h1 n+1 ,1 ni,h1 n+1 ,2 2n+1 i,h2 2n+1 ,1 ni, etc., each time subdividing all intervals by the mediant of its endpoints. After N steps we have a partition of Jnas in Lemma 2. Therefore X I∈SN i=0 Fn+i, I⊂Jnf(I)−f(Ileft)−f(Iright)=f 1 n+ 1,1 n− 2N X k=1 fak bk ,ak+1 bk+1 , (6) where {hak bk,ak+1 bk+1 i}2N k=0 ={I⊂Fn, I ∈ Fn+k}. Due to Inequality (2) and Corollary 1, as N→ ∞, Equation (6) converges to f 1 n+ 1,1 n− 1 n Z1 n+1 2 tdt =1 n+1 n+ 1 −2 ln(n+ 1 n). Proof of Theorem 1. Since each interval in F \F∗is contained in one of the intervals Jn= [ 1 n+1 ,1 n], using the telescoping identity from Lemma 1 and Lemma 3 for each INTEGERS: 25 (2025) 6 of the intervals Jn, we can rearrange the summands so that for each k, we group all terms corresponding to the intervals contained in [ 1 k+1 ,1], compute their sum, and then let k→ ∞. Therefore, we split the sum from Theorem 1 as 1 2+1 2lim k→∞ k X n=1 1 n+1 n+ 1−2 ln n+ 1 n!, which evaluates to 1 2·2 lim k→∞ k+1 X n=1 1 n−ln(k+ 1) −1 2(k+ 1)!=γ. Acknowledgement. I would like to thank the reviewer whose suggestions improved this article. References [1] T. P. Dence and J. B. Dence, A survey of Euler’s constant, Math. Mag. 82 (4) (2009), 255–265. [2] M. Hata, Farey fractions and sums over coprime pairs, Acta Arith. 70 (2) (1995), 149–159. [3] A. K. Haynes and J. D. Vaaler, Martingale differences and the metric theory of continued fractions, Illinois J. Math. 52 (1) (2008), 213–242. [4] J. Lagarias, Euler’s constant: Euler’s work and modern developments, Bull. Amer. Math. Soc. (N.S.) 50 (4) 2013, 527–628.