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Lemma 1.1. There exist the constants ∞ = ζ(ν)=2.2948565916... , C1 ν=2 Y∞ = ζ(2ν)=1.8210174514... , C2 ν=1 Y∞ = ζ(2ν +1)=1.2602057107... . C3 ν=1 Y Proof. We have log(1 + x) < x for real x> 0. Then ∞ ∞ ∞ 3 log ζ(2ν)= log ζ(2ν) < (ζ(2ν) 1) = . (1.3) − 4 ν=1 ν=1 ν=1 Y X X The last sum of (1.3) is well known and follows by rearranging in geometric , since 2 3/4 we have . We then obtain that π /6 < 2 < e , ζ(3) < 3 < 2, and = . C C C  C1 C2C3 To compute the infinite products above within a given precision, one can use the following arguments. A standard estimate for the partial sum of ζ(s) is given by N N 1−s ζ(s) ν−s < , s R, s> 1. − s 1 ∈ ν=1 X − This follows by comparing the sum of ν−s and the of x−s in the interval (N, ). Now, one can estimate the number N depending on s and the needed precision. However,∞ we use a , that computes ζ(s) to a given precision with already accelerated built-in . Since ζ(s) 1 monotonically as s , we next have to determine a finite product that suffices the→ precision. From above, →∞we obtain 2 ζ(s) 1 < 2−s 1+ , s R, s> 1. (1.4) − s 1 ∈  −  According to (1.3) and (1.4), we then get an estimate for the remainder of the infinite product by ′ log ζ(ν) < 2−N +ε ′ ν>NY where we can take ε =3/N ′; the choice of ε follows by 2x 1+ x log 2 and (1.4). ≥ We give the following example where the constant plays an important role; see Finch C1 [8]. Let a(n) be the number of non-isomorphic abelian groups of order n. The constant 1 equals the average of the numbers a(n) by taking the . Thus, we have C 1 N 1 = lim a(n). C N→∞ N n=1 X ON ASYMPTOTIC CONSTANTS ... 3

By definition the constant 2 is connected with values of the on the positive real axis. Moreover,C this constant is also connected with values of the Dedekind eta function ∞ η(τ)= eπiτ/12 (1 e2πiντ ), τ C, Im τ > 0, − ∈ ν=1 on the upper imaginary axis. Y Lemma 1.2. The constant is given by C2 1 log p 1/ = p 12 η i C2 π p Y   where the product runs over all primes. Proof. By Lemma 1.1 and the Euler product (1.1) of ζ(s), we obtain ∞ ∞ = (1 p−2ν)−1 = (1 p−2ν )−1 C2 − − ν=1 p p ν=1 Y Y Y Y where we can change the order of the products because of absolute convergence. Rewriting p−2ν = e2πiντ with τ = i log p /π yields the result.  We used Mathematica [17] to compute all numerical values in this paper. The values were checked again by increasing the needed precision to 10 more digits. 2. Preliminaries

We use the notation f g for real-valued functions when limx→∞ f(x)/g(x) = 1. As usual, O( ) denotes Landau’s∼ symbol. We write log f for log(f(x)). · Definition 2.1. Define the spaces ν ν Ωn = span x , x log x , n 0, 0≤ν≤n{ } ≥ over R where f Ω is a function f : R+ R. Let ∈ n → Ω∞ = Ωn. n≥0 [ Define the linear map ψ : Ω∞ R which gives the constant term of any f Ω∞. For the class of functions → ∈ F (x)= f(x)+ O(x−δ), f Ω , n 0, δ> 0, (2.1) ∈ n ≥ + define the linear operator [ ]: C(R ; R) Ω∞ such that [F ] = f and [F ] Ωn. Then ψ([F ]) is defined to be the asymptotic· constant→ of F . ∈ We shall examine functions h : N R which grow exponentially; in particular these functions are represented by certain→ products. Our problem is to find an asymptotic function h˜ : R+ R where h h˜. If F = log h˜ satisfies (2.1), then we have [log h˜] Ω → ∼ ∈ n for a suitable n and we identify [log h˜] = [log h] Ω in that case. ∈ n 4 BERND C. KELLNER

Lemma 2.2. Let f Ωn where ∈ n ν ν f(x)= (αν x + βν x log x) ν=0 X with coefficients α , β R. Let g(x) = f(λx) with a fixed λ R+. Then g Ω and ν ν ∈ ∈ ∈ n ψ(g)= ψ(f)+ β0 log λ. Proof. Since g(x)= f(λx) we obtain n ν ν g(x)= (αν (λx) + βν (λx) (log λ + log x)). ν=0 X This shows that g Ω . The constant terms are α and β log λ, thus ψ(g) = ψ(f)+ ∈ n 0 0 β0 log λ.  Definition 2.3. For a function f : R+ R we introduce the notation → ′ f(x)= fν(x) ν≥1 X with functions f : R+ R in case f has a expansion such that ν → m−1 f(x)= f (x)+ θ (x)f (x), θ (x) (0, 1), m N , ν m m m ∈ ≥ f ν=1 X where Nf is a suitable constant depending on f. Next we need some well known facts which we state without proof, cf. [10]. Proposition 2.4. Let n 1 H =0, H = , n 1, 0 n ν ≥ ν=1 X −1 be the nth . These numbers satisfy Hn = γ + log n + O(n ) for n 1, where γ =0.5772156649... is Euler’s constant. ≥ Proposition 2.5 (Stirling’s series). The Γ(x) has the divergent series expansion 1 1 B log Γ(x +1)= log(2π)+ x + log x x + ′ 2ν x−(2ν−1), x> 0. 2 2 − 2ν(2ν 1) ν≥1   X − Remark 2.6. When evaluating the divergent series given above, we have to choose a suitable index m such that B m−1 B ′ 2ν x−(2ν−1) = 2ν x−(2ν−1) + θ (x)R (x) 2ν(2ν 1) 2ν(2ν 1) m m ν≥1 ν=1 X − X − and the remainder θm(x)Rm(x) is as small as possible. Since θm(x) (0, 1) is not effec- tively computable in| general, we| have to use R (x) instead as an error∈ bound. Sch¨afke | m | ON ASYMPTOTIC CONSTANTS ... 5 and Finsterer [15], among others, showed that the so-called Lindel¨of error bound L = 1 for the estimate L θ (x) is best possible for positive real x. ≥ m Proposition 2.7. If α R with 0 α< 1, then n ∈ ≤ Γ(n +1 α) √2π n n 1 (ν α)= − n 2 −α as n . − Γ(1 α) ∼ Γ(1 α) e →∞ ν=1 Y − −   Proposition 2.8 (Euler). Let Γ(x) be the Gamma function. Then n−1 n−1 ν (2π) 2 Γ = . n √n ν=1 Y   Proposition 2.9 (Glaisher [9], Kinkelin [11]). Asymptotically, we have n 1 1 n2 νν n 2 n(n+1)+ 12 e− 4 as n ∼ A →∞ ν=1 Y where =1.2824271291... is the Glaisher–Kinkelin constant, which is given by A 1 γ 1 ζ′(2) log = ζ′( 1) = + log(2π) . A 12 − − 12 12 − 2π2 Numerous digits of the decimal expansion of the Glaisher–Kinkelin constant are recorded as A074962 in OEIS [16]. A

3. Products of factorials In this section we consider products of factorials and determine their asymptotic expan- sions and constants. For these asymptotic constants we derive a divergent series represen- tation as well as a closed formula. Theorem 3.1. Let k be a positive integer. Asymptotically, we have k n 2 n(n+1) 1 n 1 k 1 k k n k/2−1 2 + + (kν)! (2π) 4 2πke n n 4 12 12k as n ∼ Fk A e3/2 →∞ ν=1   Y  with certain constants which satisfy Fk γ B ζ(2j 1) log = + ′ 2j − . Fk 12k 2j(2j 1) k2j−1 j≥2 X − Moreover, the constants have the asymptotic behavior that n k γ/12 γ/12 lim k =1, lim k = e , and k ∞ n as n k→∞ F k→∞ F F ∼ F →∞ Yk=1 with γ2 B ζ(2j 1)2 log = + ′ 2j − . F∞ 12 2j(2j 1) j≥2 X − 6 BERND C. KELLNER

Theorem 3.2. If k is a positive integer, then

1 1 1 k k−1 ν ν log = k + log + log k + log(2π) log Γ . Fk − k A 12k − 12k 4 − k k ν=1   X   We will prove Theorem 3.2 later, since we shall need several preparations.

Proof of Theorem 3.1. Let k 1 be fixed. By Stirling’s approximation, see Proposition 2.5, we have ≥ 1 1 log(kν)! = log(2π)+ kν + log(kν) kν + f(kν) (3.1) 2 2 −   where we can write the remaining divergent sum as

1 B f(kν)= + ′ 2j . 12kν 2j(2j 1)(kν)2j−1 j≥2 X − Define S(n)=1+ + n = n(n + 1)/2. By we obtain ··· n n 1 log(kν)! = log(2πk)+ log n! kS(n)+ kS(n) log k 2 2 − ν=1 X n n + k ν log ν + f(kν). ν=1 ν=1 X X 1 The term 2 log n! is evaluated again by (3.1). Proposition 2.9 provides that

n k k n k ν log ν = k log + kS(n) log n + log n S(n) + O(n−δ) A 12 − 2 − 2 ν=1 X   with some δ > 0. Since limn→∞ Hn log n = γ, we asymptotically obtain for the remaining sum that −

n 1 γ ′ B2j ζ(2j 1) lim f(kν) log n = + − =: log k. (3.2) n→∞ − 12k 12k 2j(2j 1) k2j−1 F ν=1 ! j≥2 X X − Here we have used the following arguments. We choose a fixed index m > 2 for the remainder of the divergent sum. Then

n B2m B2m ζ(2m 1) lim θm(kν) = ηm − (3.3) n→∞ 2m(2m 1)(kν)2m−1 2m(2m 1) k2m−1 ν=1 X − − ON ASYMPTOTIC CONSTANTS ... 7 with some ηm (0, 1), since θm(kν) (0, 1) for all ν 1. Thus, we can write (3.2) as an asymptotic series∈ again. Collecting all∈ terms, we finally≥ get the asymptotic formula n 1 3 log(kν)! = log + k log + log(2π)+ kS(n) + log(kn) Fk A 4 −2 ν=1 X   n k + log(2πk)+ 1 + log n 2 2 −   1 k 1 ′ + + + log n + O(n−δ ) 4 12 12k   with some δ′ > 0. Note that the exact value of δ′ does not play a role here. Now, let k be an arbitrary positive integer. From (3.2) we deduce that γ γ log = + O(k−3) and k log = + O(k−2). (3.4) Fk 12k Fk 12 The summation of (3.2) yields n γ n B ζ(2j 1) log = H + ′ 2j − . (3.5) Fk 12 n 2j(2j 1) k2j−1 j≥2 Xk=1 Xk=1 X − Similar to (3.2) and (3.3), we can write again:

n 2 2 γ γ ′ B2j ζ(2j 1) lim log k log n = + − =: log ∞. (3.6) n→∞ F − 12 12 2j(2j 1) F ! j≥2  Xk=1 X − The case k = 1 of Theorem 3.1 is related to the so-called Barnes G-function, cf. [2]. Now we shall determine exact expressions for the constants k. For k 2 this is more complicated. F ≥

1 1 Lemma 3.3. We have = (2π) 4 e 12 / 2. F1 A Proof. Writing down the product of n! repeatedly in n + 1 rows, one observes by counting in rows and columns that n n n!n+1 = ν! νν . (3.7) ν=1 ν=1 Y Y From Proposition 2.5 we have n +1 1 1 (n + 1) log n!= log(2π) n(n +1)+(n + 1) n + log n + + O(n−1). 2 − 2 12   Comparing the asymptotic constants of both sides of (3.7) when n , we obtain →∞ 1 1 1 (2π) 2 e 12 = (2π) 4 F1 A ·A where the right side follows by Theorem 3.1 and Proposition 2.9.  8 BERND C. KELLNER

Proposition 3.4. Let k,l be integers with k 1. Define ≥ n F (n) := (kν l)! for 0 l < k. k,l − ≤ ν=1 Y Then [log F ] Ω and F (n) F (n)= F (kn). Moreover k,l ∈ 2 k,0 ··· k,k−1 1,0 n l F (n)/F (n)= kn ν for 0 l

after some rearranging of terms. By Euler’s formula, see Proposition 2.8, we have 1 k−1 ν 1 1 1 log Γ = log(2π) log k. (3.11) k k 2 − 2k − 2k ν=1 X     Finally, substituting (3.11) into (3.10) yields the result. 

Remark 3.6. Although the formula for k has an elegant short form, one might also use (3.10) instead, since this formula omits theF value Γ(1/k). Thus we easily obtain the value 1 5 1 5 of from (3.10) at once: = (2π) 4 2 24 e 24 / 2 . F2 F2 A Corollary 3.7. Asymptotically, we have

2 n−1 1−γ 1 n ν ν e 12 (2π) 4 1 Γ n 12 as n n ∼ →∞ ν=1 ! Y   A A . 1−γ 1 with the constants e 12 / =0.8077340270... and (2π) 4 / =1.2345601953.... A A Proof. On the one hand, we have by (3.4) that γ n log = + O(n−2). Fn 12 On the other hand, Theorem 3.2 provides that 1 1 n2 n−1 ν n log = n2 +1 log + log n + log(2π) ν log Γ . Fn − A 12 − 12 4 − n ν=1  X   Combining both formulas easily gives the result.  Since we have derived exact expressions for the constants , we can improve the calcu- Fk lation of ∞. The divergent sum of ∞, given in Theorem 3.1, is not suitable to determine a value withinF a given precision, butF we can use this sum in a modified way. Note that we cannot use the limit formula n γ log ∞ = lim log k log n F n→∞ F − 12 ! Xk=1 without a very extensive calculation, because the sequence γ = H log n converges too n n − slowly. Moreover, the computation of k involves the computation of the values Γ(ν/k). This becomes more difficult for larger kF. Proposition 3.8. Let m, n be positive integers. Assume that m> 2 and the constants Fk are given by exact expressions for k = 1,...,n. Define the computable values ηk (0, 1) implicitly by ∈

γ m−1 B ζ(2j 1) B ζ(2m 1) log = + 2j − + η 2m − . Fk 12k 2j(2j 1) k2j−1 k 2m(2m 1) k2m−1 j=2 X − − ON ASYMPTOTIC CONSTANTS ... 11

Then γ2 m−1 B ζ(2j 1)2 B ζ(2m 1)2 log = + 2j − + θ 2m − F∞ 12 2j(2j 1) n,m 2m(2m 1) j=2 X − − with θ (θmin, θmax) (0, 1) where n,m ∈ n,m n,m ⊂ n η n η 1 θmin = ζ(2m 1)−1 k , θmax =1+ ζ(2m 1)−1 k − . n,m − k2m−1 n,m − k2m−1 Xk=1 Xk=1 The error bound for the remainder of the divergent sum of log is given by F∞ n 2 err −1 1 B2m ζ(2m 1) θn,m = 1 ζ(2m 1) 2m−1 | | − . − − k ! 2m(2m 1) Xk=1 − Proof. Let n 1 and m> 2 be fixed integers. The divergent sums for log k and log ∞ are given by Theorem≥ 3.1. Since we require exact expressions for , we canF computeF the Fk values ηk for k =1,...,n. We define ′ ηm,k = ηm,k = ηk for k =1,...,n and ′ ηm,k =0, ηm,k =1 for k > n. We use (3.5) and (3.6) to derive the bounds:

∞ η ∞ η′ θmin = ζ(2m 1)−1 m,k < θ < ζ(2m 1)−1 m,k = θmax. n,m − k2m−1 n,m − k2m−1 n,m Xk=1 Xk=1 min max We obtain the suggested formulas for θn,m and θn,m by evaluating the sums with ηm,k = 0, ′ resp. ηm,k = 1, for k > n. The error bound is given by the difference of the absolute values of the minimal and maximal remainder. Therefore n err max min −1 1 θn,m =(θn,m θn,m)R = 1 ζ(2m 1) 2m−1 R − − − k ! Xk=1 with R = B ζ(2m 1)2/2m(2m 1).  | 2m| − − Result 3.9. Exact expressions for : Fk 1 1 1 5 1 5 = (2π) 4 e 12 / 2, = (2π) 4 2 24 e 24 / 2 , F1 A F2 A 5 5 1 10 1 1 1 1 17 1 = (2π) 12 3 36 e 36 / 3 Γ 2 3 , = (2π) 2 2 3 e 48 / 4 Γ 3 2 . F3 A 3 F4 A 4 We have computed the constants k by their exact expression. Moreover,  we have determined the index m of the smallestF remainder of their asymptotic divergent series and the resulting error bound given by Theorem 3.1. 12 BERND C. KELLNER

Constant Value m Error bound −4 1 1.04633506677050318098... 4 6.000 10 F · −7 2 1.02393741163711840157... 7 7.826 10 F · −9 3 1.01604053706462099128... 10 1.198 10 F · −12 4 1.01204589802394464624... 13 1.948 10 F · −15 5 1.00963997283647705086... 16 3.272 10 F 1.00803362724207326544... 20 5.552 · 10−18 F6 · The weak interval of is given by Theorem 3.1. The second value is derived by F∞ Proposition 3.8 with parameters m = 17 and n = 7. Thus, exact expressions of 1,..., 7 are needed to compute within the given precision. F F F∞ Constant Value / Interval m Error bound −4 ∞ (1.02428, 1.02491) 4 6.050 10 F 1.02460688265559721480... 17 6.321 · 10−22 F∞ · 4. Products of Bernoulli numbers Using results of the previous sections, we are now able to consider several products of Bernoulli numbers and to derive their asymptotic expansions and constants. Theorem 4.1. Asymptotically, we have n n n(n+1) n 11 B (16πn) 2 n 24 as n , | 2ν| ∼ B1 πe3/2 →∞ ν=1   Yn n n2 2 B2ν n 4n 1 | | n 24 as n 2ν ∼ B2 πe3/2 πe →∞ ν=1 Y     . with the constants 2 1 1 5 1 1 1 = 2 2 (2π) 4 = 2 (2π) 2 2 24 e 24 / 2 , B C F A 1 C 5 1 1 A = 2/(2π) 4 = 2 24 e 24 / 2 . B2 C2F2A C2 A Proof. By Euler’s formula (1.2) for ζ(2ν) and Lemma 1.1 we obtain n n 2 (2ν)! n B · 2n(2π)−n(n+1) (2ν)! as n . | 2ν | ∼ C2 (2π)2ν ∼ C2 →∞ ν=1 ν=1 ν=1 Y Y Y Theorem 3.1 states for k = 2 that n n(n+1) 1 2n n 11 (2ν)! 2 (2π) 4 (4πn) 2 n 24 as n . ∼ F2 A e3/2 →∞ ν=1 Y   The expression for 2 is given in Remark 3.6. Combining both asymptotic formulas above gives the first suggestedF formula. It remains to evaluate the following product: n n 1 2n 1 (2ν)=2n n! (2π) 2 n 2 as n . ∼ e →∞ ν=1 Y   ON ASYMPTOTIC CONSTANTS ... 13

After some rearranging of terms we then obtain the second suggested formula.  Remark 4.2. Milnor and Husemoller [14, pp. 49–50] give the following asymptotic formula without proof: n B ′ n!2n+1 F (2n +1) as n (4.1) | 2ν| ∼ B →∞ ν=1 Y where 2 n n 4 n 4 8πe 1 F (n)= n 24 (4.2) 2πe3/2 n     . and ′ 0.705 is a certain constant. This constant is related to the constant . B ≈ B2 Proposition 4.3. The constant ′ is given by B 1 3 1 5 1 ′ =2 24 2− 2 = e 24 /2 4 2 =0.7048648734... . B B2 C2 A Proof. By Theorem 4.1 we have n B | 2ν | G(n) as n (4.3) 2ν ∼ B2 →∞ ν=1 Y with n 2 n n 4n 2 1 G(n)= n 24 . πe3/2 πe     . We observe that (4.1) and (4.3) are equivalent so that 2 ′F (2n + 1) G(n) as n . B ∼ B2 →∞ We rewrite (4.2) in the suitable form

2 1 n 1 1 1 n +n+ 4 + n + 4πe 2 4 1 1 24 F (2n +1) = 2 2 24 n + . πe3/2 n + 1 2    2  .   Hence, we easily deduce that

2 n 1 1 −n − 2 + 24 1/2 4 1 n 1 e G(n)/F (2n +1) = 1+ e 2 2 24 . 2n 4     It is well known that 2 x n x n x2 lim 1+ = ex and lim e−xn 1+ = e− 2 . n→∞ n n→∞ n Evaluating the asymptotic  terms, we get  

1 1 1 1 1 2 ′/ G(n)/F (2n + 1) e 8 e− 4 2 24 e 8 2− 2 as n , B B2 ∼ ∼ →∞ 1 3 which finally yields ′ =2 24 2− 2 .  B B2 14 BERND C. KELLNER

Theorem 4.4. The Minkowski–Siegel mass formula asymptotically states for positive in- tegers n with 4 n that | n−1 n n2 2 Bn B2ν n 4n 1 M(2n)= | | | | n 24 as n 2n 4ν ∼ B3 πe3/2 πe →∞ ν=1 Y   .   with = √2 . B3 B2 Proof. Let n always be even. By Proposition 2.5 and (1.2) we have B /n ζ(n) n! 4n −n 2−n n =2 πn √2 as n , B /2n ζ(2n) (2n)! ∼ πe →∞ 2n   since ζ(n)/ζ(2n) 1 and ∼ n! 1 log n n log n 2n + log 2 as n . (2n)! ∼ − − 2 →∞     We finally use Theorem 4.1 and (4.3) to obtain B /n n B 4n −n M(2n)=2−n n | 2ν| √2 G(n) as n , B /2n 2ν ∼ B2 πe →∞ 2n ν=1   Y  which gives the result. ′ Result 4.5. The constants , ν (ν =1, 2, 3) mainly depend on the constant 2 and the Glaisher–Kinkelin constant B. B C A Constant Expression Value 1.28242712910062263687... A 2 1.82101745149929239040... C 1 5 1 1 1 2(2π) 2 2 24 e 24 / 2 4.85509664652226751252... B C 5 1 1 A 2 22 24 e 24 / 2 1.93690332773294192068... B C 17 1 A 1 3 22 24 e 24 / 2 2.73919495508550621998... B C 1 5 A1 ′ e 24 /2 4 2 0.70486487346802031057... B C2 A 5. Generalizations In this section we derive a generalization of Theorem 3.1. The results show the structure of the constants k and the generalized constants r,k, which we shall define later, in a wider context. ForF simplification we introduce theF following definitions which arise from the Euler-Maclaurin summation formula. The sum of consecutive integer powers is given by the well known formula

n−1 B (n) B r r nj+1 νr = r+1 − r+1 = B , r 0, r +1 j r−j j +1 ≥ ν=0 j=0 X X   ON ASYMPTOTIC CONSTANTS ... 15

1 where Bm(x) is the mth Bernoulli . Now, the B1 = 2 is responsible for omitting the last power nr in the summation above. Because we further− r need the summation up to n , we change the sign of B1 in the sum as follows: n r r nj+1 S (n)= νr = ( 1)r−jB , r 0. r j − r−j j +1 ≥ ν=1 j=0 X X   This modification also coincides with B ζ( n)=( 1)n+1 n+1 − − n +1 for nonnegative integers n. We define the extended sum r r nj+1f(j + 1) S (n; f( )) = ( 1)r−jB , r 0, r ⋄ j − r−j j +1 ≥ j=0 X   where the symbol is replaced by the index j + 1 in the sum. Note that Sr is linear in the second parameter,⋄ i.e., S (n; α + βf( )) = αS (n)+ βS (n; f( )). r ⋄ r r ⋄ Finally we define B D (x)= ′ B x−(2j−1) where B = m . k 2j,k m,k m(m 1)km−1 j≥1 X − Theorem 5.1. Let r be a nonnegativeb integer. Thenb n r ν ν Q (n) as n , ∼ Ar r →∞ ν=1 Y where is the generalized Glaisher–Kinkelin constant defined by Ar log = ζ( r) H ζ′( r). Ar − − r − − Moreover, log Q Ω with r ∈ r+1 log Q (n)=(S (n) ζ( r)) log n + S (n; H H ). r r − − r r − ⋄ Proof. This formula and the constants easily follow from a more general formula for real r> 1 given in [10, 9.28, p. 595] and after some rearranging of terms.  − Remark 5.2. The case r = 0 reduces to Stirling’s approximation of n! with 0 = √2π. The case r = 1 gives the usual Glaisher–Kinkelin constant = . TheA expression A1 A Sr(n; Hr H⋄) does not depend on the definition of B1, since the term with B1 is cancelled in the sum.− Graham, Knuth, and Patashnik [10, 9.28, p. 595] notice that the constant ζ′( r) has been determined in a book of de Bruijn [7, 3.7] in 1970. The theorem above −has a− long history. In 1894 Alexeiewsky [3] gave the identity§ n r ν ν = exp (ζ′( r, n + 1) ζ′( r)) − − − ν=1 Y 16 BERND C. KELLNER

where ζ′(s, a) is the partial of the with respect to the first variable. Between 1903 and 1913, Ramanujan recorded in his notebooks [5, Entry 27, pp. 273–276] (the first part was published and edited by Berndt [5] in 1985) an for real r > 1 and an analytic expression for the constant C = ζ′( r). − r − − However, Ramanujan only derived closed expressions for C0 and C2r (r 1) in terms of ζ(2r +1); see (5.9) below. In 1933 Bendersky [4] showed that there exist≥ certain con- stants r. Since 1980, several others have investigated the asymptotic formula, including MacLeodA [13], Choudhury [6], and Adamchik [1, 2]. Theorem 5.3. Let k,r be integers with k 1 and r 0. Then n ≥ ≥ 1 r 1 ν 2 k k (kν)! r P (n) Q (n) 2 Q (n) as n . ∼ Fr,k A Ar+1 r,k r r+1 →∞ ν=1 Y The constants r,k and functions Pr,k satisfy that lim r,k =1 and log Pr,k Ωr+2 where F k→∞ F ∈ 1 log P (n)= S (n) log(2πk)+ kS (n) log(k/e) r,k 2 r r+1 r+1 ⌊ 2 ⌋

+ Br+2,k log n + B2j,k Sr+1−2j(n). j=1 X The constants and functions Q bare defined as in Theoremb 5.1. Ar r The determination of exact expressions for the constants r,k seems to be a very com- plicated and extensive task in the case r > 0. The next theoremF gives a partial result for k = 1 and r 0. ≥ Theorem 5.4. Let r be a nonnegative integer. Then 1 log = log log + S (1; B log ). Fr,1 2 Ar − Ar+1 r 1+⋄,1 − A⋄ Case r =0: 1 1 b log = + log 2 log . Fr,1 12 2 A0 − A1 Case r> 0: r+1 log = α + α log Fr,1 r,0 r,j Aj j=1 X where Br+1 2r(r+1) , r j (mod 2), j = 0; r 6≡ r Br−j Bj+2  2 , r j (mod 2), j = 0; j (j+1) (j+2) ≡ αr,j =  j=0  r+1 Br+1−j  δr+1,jP  , r j (mod 2), j> 0;  − − j r+1 6≡  0, r j (mod 2), j> 0  ≡ and δ is Kronecker’s delta. i,j  ON ASYMPTOTIC CONSTANTS ... 17

Proof of Theorem 5.3. Let k and r be fixed. We extend the proof of Theorem 3.1. From (3.1) we have 1 k 1 log(kν)! = log(2πk)+ kν log + kν + log ν + D (ν). (5.1) 2 e 2 k     The summation yields n r ν log(kν)! = F1(n)+ F2(n)+ F3(n) ν=1 X where 1 F (n)= S (n) log(2πk)+ kS (n) log(k/e), 1 2 r r+1 n 1 n F (n)= k νr+1 log ν + νr log ν, 2 2 ν=1 ν=1 nX X r F3(n)= ν Dk(ν). ν=1 X Theorem 5.1 provides 1 F (n)= k (log + log Q (n)) + (log + log Q (n)) + O(n−δ) 2 Ar+1 r+1 2 Ar r with some δ > 0. Let R = r+1 . By definition we have ⌊ 2 ⌋ R r r+1−2j ′ r+1−2j x Dk(x)= B2j,k x + B2j,k x =: E1(x)+ E2(x). j=1 j>R X X Therewith we obtain that b b R n

F3(n)= B2j,k Sr+1−2j(n)+ E2(ν). j=1 ν=1 X X For the second sum above we considerb two cases. We use similar arguments which we have applied to (3.2) and (3.3). If r is odd, then n ′ lim E2(ν)= B2j,k ζ(2j (r + 1)). (5.2) n→∞ − ν=1 X Xj>R b −1 Note that Br+2,k = 0 in that case. If r is even, then we have to take care of the term ν . This gives b n ′ lim E2(ν) Br+2,k log n = γ Br+2,k + B2j,k ζ(2j (r + 1)). (5.3) n→∞ − − ν=1 ! X j>RX+1 b b b 18 BERND C. KELLNER

The right hand side of (5.2), resp. (5.3), defines the constant log . Finally we have to Fr,k collect all results for F1, F2, and F3. This gives the constants and the function Pr,k. It remains to show that lim log = 0. This follows by B 0 as k .  k→∞ Fr,k 2j,k → →∞ The following lemma gives a generalization of Equation (3.7) in Lemma 3.3. After that b we can give a proof of Theorem 5.4. Lemma 5.5. Let n, r be integers with n 1 and r 0. Then ≥ ≥ n n n r r n!Sr(n) ν ν = ν! ν νSr(ν). (5.4) ν=1 ν=1 ν=1 Y Y Y Proof. We regard the following enumeration scheme which can be easily extended to n rows and n columns: 11r 21r 31r 12r 22r 32r 13r 23r 33r The product of all elements, resp. non-framed elements, in the νth row equals n! νr , resp. r ν! ν . The product of the framed elements in the νth column equals νSr(ν−1). Thus n n r r n!Sr(n) = ν! ν νSr(ν)−ν .  ν=1 ν=1 Y Y Proof of Theorem 5.4. Let r 0. We take the of (5.4) to obtain ≥ F1(n)+ F2(n)= F3(n)+ F4(n) (5.5) where n r F1(n)= Sr(n) log n!, F2(n)= ν log ν, ν=1 n Xn r F3(n)= ν log ν!, F4(n)= Sr(ν) log ν. ν=1 ν=1 X X Next we consider the asymptotic expansions F˜j of the functions Fj (j = 1,..., 4) when n . We further reduce the functions F˜ via the maps →∞ j [] ψ C(R+; R) Ω R −→ ∞ −→ to the constant terms which are the asymptotic constants of [F˜j] in Ω∞. Consequently (5.5) turns into

ψ([F˜1]) + ψ([F˜2]) = ψ([F˜3]) + ψ([F˜4]). (5.6) We know from Theorem 5.1 and Theorem 5.3 that 1 ψ([F˜ ]) = log and ψ([F˜ ]) = log + log + log . 2 Ar 3 Fr,1 2 Ar Ar+1 ON ASYMPTOTIC CONSTANTS ... 19

For F˜4 we derive the expression ψ([F˜ ]) = S (1; log ), (5.7) 4 r A⋄ j since each term sjν in Sr(ν) produces the term sj log j. It remains to evaluate F˜1. According to (5.1) we have A 1 1 log n!= log(2π) n + n + log n + D (n) =: E(n)+ D (n). 2 − 2 1 1   Thus

F˜1(x)= Sr(x)E(x)+ Sr(x)D1(x). Since S E Ω has no constant term, we deduce that r ∈ ∞ ψ([F˜1]) = ψ([SrD1]) = Sr(1; B1+⋄,1). The latter equation is similarly derived as (5.7), whereas we regard the constant terms of b the product of the polynomial Sr and the D1. From (5.6) we finally obtain 1 log = log log + S (1; B log ). Fr,1 2 Ar − Ar+1 r 1+⋄,1 − A⋄ Now, we shall evaluate the expression above. For r = 0b we get 1 1 log = + log 2 log , F0,1 12 2 A0 − A1 since 1 S (1; B log )= B log = log . 0 1+⋄,1 − A⋄ 2,1 − A1 12 − A1 For now, let r> 0. We may represent log in terms of log as follows: b Fbr,1 Aj r+1 log = α + α log . Fr,1 r,0 r,j Aj j=1 X The term αr,0 is given by r r B α = S (1; B )= ( 1)r−jB j+2,1 r,0 r 1+⋄,1 j − r−j j +1 j=0 X   b b where the sum runs over even j, since Bj+2,1 = 0 for odd j. If r is odd, then the sum simplifies to the term Br+1/2r(r + 1). Otherwise we derive for even r that r b r B B α = r−j j+2 . r,0 j (j + 1)2(j + 2) j=0 X  1 r r+1 It remains to determine the coefficients αr,j for r+1 j 1. Since 2 x x Sr(x) is an odd, resp. even, polynomial for even, resp. odd, r >≥0, this≥ property transfers− − in a similar 20 BERND C. KELLNER

1 way to 2 log r log r+1 Sr(1; log ⋄), such that αr,j = 0 when 2 r j. Otherwise we get A − A − A | − r B r +1 B α = r−(j−1) δ = r+1−j δ (5.8) r,j − j 1 j − r+1,j − j r +1 − r+1,j  −    for 2 ∤ r j, where the term log is represented by δ .  − − Ar+1 − r+1,j Corollary 5.6. Let r be an odd positive integer. Then

r−1 r! ζ(r + 1) 2 (r + 2)ζ(r + 2) log = + ζ(r +1 2j)ζ(2j + 1) Fr,1 −(2πi)r+1  r − − 2  j=1 X r−1  2  r−1 r! Br+1 Br+1−2j ζ(2j + 1) (r + 2)ζ(r + 2) =( 1) 2 | | + | | . − 2 r(r + 1)! (r +1 2j)!(2π)2j − (2π)r+1  j=1 X −   Proof. As a consequence of the of ζ(s) and its derivative, we have for even positive integers n, cf. [5, p. 276], that 1 n! log = ζ′( n)= ζ(n + 1) (5.9) An − − −2 (2πi)n where the left hand side of (5.9) follows by definition. Theorem 5.4 provides

r+1 B 2 log = r+1 + α log . Fr,1 2r(r + 1) r,2j A2j j=1 X Combining (5.8) and (5.9) gives the second equation above. By Euler’s formula (1.2) we finally derive the first equation.  Remark 5.7. For the sake of completeness, we give an analogue of (5.9) for odd integers. From the logarithmic of Γ(s) and the functional equation of ζ(s), see [5, pp. 183, 276], it follows for even positive integers n, that B B (n 1)! log = n H ζ′(1 n)= n (γ + log(2π))+2 − ζ′(n) An−1 n n−1 − − n (2πi)n where ∞ ζ′(n)= log(ν) ν−n. − ν=2 X However, Mathematica is able to compute values of ζ′ for positive and negative argument values to any given precision. Result 5.8. Exact expressions for in terms of : Fr,1 Aj ON ASYMPTOTIC CONSTANTS ... 21

Constant Expression Value 1 1 12 2 −2 0,1 e 0 1 1.04633506677050318098... F 1A A− 3 24 2 1,1 e 2 0.99600199446870605433... F 7 −A1 − 4 540 6 3 2,1 e 1 3 0.99904614418135586848... F 1 A − 1A − 5 − 720 4 4 3,1 e 2 4 1.00097924030236153773... F 67 A1 A− 1 − 6 − 18900 30 3 5 4,1 e 1 3 5 1.00007169725554110099... F 1 A1 A− 5 A− 7 e 2520 12 12 6 0.99937792615674804266... F5,1 A2 A4 A6 Exact expressions for in terms of ζ(2j + 1): Fr,1 1 3ζ(3) = exp , F1,1 24 − 8π2   1 ζ(3) 15ζ(5) = exp + , F3,1 −720 − 16π2 16π4   1 ζ(3) 5ζ(5) 105ζ(7) = exp + + . F5,1 2520 48π2 16π4 − 16π6   For the first 15 constants (r =0,..., 14) we find that Fr,1 max r,1 1 < 0.05, 0≤r≤14 |F − | but, e.g., 371.61 and 1.16 10−7. F19,1 ≈ F20,1 ≈ · Acknowledgements The author would like to thank Steven Finch for informing about the formula of Milnor– Husemoller and the problem of finding suitable constants; also for giving several references. The author is also grateful to the referee for several remarks and suggestions.

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11. H. Kinkelin, Ueber eine mit der Gammafunction verwandte Transcendente und deren Anwendung auf die Integralrechnung, J. Reine Angew. Math. 57 (1860), 122-158. 12. M. Koecher, A. Krieg, Elliptische Funktionen und Modulformen, Springer-Verlag, 1998. 13. R. A. MacLeod, Fractional part sums and divisor functions, J. 14 (1982), 185–227. 14. J. Milnor, D. Husemoller, Symmetric bilinear forms, Springer-Verlag, 1973. 15. F. W. Sch¨afke, A. Finsterer, On Lindel¨of’s error bound for Stirling’s series, J. Reine Angew. Math. 404 (1990), 135–139. 16. N. J. A. Sloane, Online Encyclopedia of Integer (OEIS), electronically published at: http://www.research.att.com/~njas/sequences. 17. Wolfram Research Inc., Mathematica, Wolfram Research Inc., Champaign, IL.

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