aa r X i v : . [ m a t h . G M ] A ug A closed-form expression for ζ (3) Tobias KyrionH¨urth, GermanyAugust 14, 2020
Abstract
In this work we show ζ (3) = 4 π ln( B ) (0.1)with the BenderskyAdamchik constant B = lim n →∞ n Y k =1 k k n n + n + n e − n + n . Contents lim α → P ∞ n =1 1 n − e − nα (1+ e − nα ) M e − snα
34 Applications and Outlook 9
We study the limits e M, s := lim α → ∞ X n =1 n − e − nα (1 + e − nα ) M e − snα (1.1)and give explicit expressions for them for M ≤ s ∈ N . This continues the work in [1], wherethe author showed as a special case of Theorem 3.34 e , = lim α → ∞ X n =1 n sinh( nα )cosh( nα ) = 7 π ζ (3) . (1.2) We derive some limit identities needed for our main results in the next section.
Lemma 2.1.
For fixed N ∈ N we havelim s →∞ s N ∞ X k =1 N +1 Y j =1 k + j + 2 s = 12 N N . roof. On the one hand there holds N Y j =1 k + j + 2 s − N +2 Y j =3 k + j + 2 s = N (4 k + N + 4 s + 1) N +2 Y j =1 k + j + 2 s and on the other hand N +1 Y j =1 k + j + 2 s + N +2 Y j =2 k + j + 2 s = (4 k + N + 4 s + 1) N +2 Y j =1 k + j + 2 s . Therefore we obtain12 N = lim s →∞ s N N Y j =1 s + j + 2= lim s →∞ s N ∞ X k =1 N Y j =1 k + j + 2 s − ∞ X k =1 N +2 Y j =3 k + j + 2 s = N lim s →∞ s N ∞ X k =1 N +1 Y j =1 k + j + 2 s + N lim s →∞ s N ∞ X k =1 N +2 Y j =2 k + j + 2 s . Since N lim s →∞ s N ∞ X k =1 N +1 Y j =1 k + j + 2 s − N lim s →∞ s N ∞ X k =1 N +2 Y j =2 k + j + 2 s = ( N + 1) lim s →∞ s N ∞ X k =1 N +2 Y j =1 k + j + 2 s = 0holds, this proves the claim. (cid:3) Corollary 2.2.
Let d ∈ R an arbitrary but fixed shift value. Then we havelim s →∞ s N ∞ X k =1 N +1 Y j =1 k + j + d + 2 s = 12 N N . (2.1) Proof.
Expand s N = (cid:0) ( s + d ) − d (cid:1) N . (cid:3) Lemma 2.3.
We have lim x →∞ x (cid:16) − x ln (cid:16) x (cid:17)(cid:17) = 12 . (2.2) Proof. l ' Hˆopital’s rule giveslim x →∞ x (cid:16) − x ln (cid:16) x (cid:17)(cid:17) = lim x →∞ − ln (cid:16) x (cid:17) + x +1 − x = lim x →∞ (cid:18) x (cid:16) x ln (cid:16) x (cid:17) − (cid:17) + xx + 1 (cid:19) . (cid:3) Lemma 2.4.
We have lim x →∞ x (cid:16) − (cid:0) x + (cid:1) ln (cid:16) x (cid:17)(cid:17) = − . (2.3)2 roof. l ' Hˆopital’s rule giveslim x →∞ x (cid:16) − (cid:0) x + (cid:1) ln (cid:16) x (cid:17)(cid:17) = lim x →∞ − ln (cid:16) x (cid:17) + (cid:0) x + (cid:1) x ( x +1) − x = lim x →∞ (cid:18) xx + x (cid:16) (cid:0) x + (cid:1) ln (cid:16) x (cid:17) − (cid:17) − x (cid:0) x + (cid:1) ( x + 1) (cid:19) . (cid:3) lim α → P ∞ n =1 1 n − e − nα (1+ e − nα ) M e − snα Our main tool will be the simple recursive formula e M, s + e M, s +2 = e M − , s . (3.1)for e M, s defined in (1.1). With that in mind, it is fairly easy to compute explixit formulae for the e M, s ’s for the first two cases M = 0 and M = 1. Proposition 3.1.
We have the relation e , s = lim α → ∞ X n =1 n (cid:0) − e − nα (cid:1) e − snα = ln (cid:16) s (cid:17) (3.2) Proof.
For x ∈ R + followslim α → ∞ X n =1 n (cid:0) − e − nα (cid:1) e − xnα = lim α → (cid:18) ln (cid:16) − e − xα (cid:17) − ln (cid:16) − e − ( x +2) α (cid:17)(cid:19) = lim α → ln (cid:16) − e − ( x +2) α − e − xα (cid:17) = ln (cid:16) x (cid:17) . Setting x = 2 s gives the result. (cid:3) Proposition 3.2.
For s ∈ N we have the relations e , s +2 = lim α → ∞ X n =1 n − e − nα e − nα e − (4 s +2) nα = ln (cid:16) s Y k =1 (2 k − k + 1)(2 k ) π (cid:17) and e , s +4 = lim α → ∞ X n =1 n − e − nα e − nα e − (4 s +4) nα = ln (cid:16) s Y k =1 k (2 k + 2)(2 k + 1) π (cid:17) . Proof.
First note that lim s →∞ e M, s = 0 for all fixed M . From (3.2) we can deduce by iterativelyeliminating e M, s using (3.1) e , = N X j =1 ( − j +1 e , j + e , N +2 = ln (cid:16) N Y j =1 (2 j ) (2 j − j + 1) (cid:17) + e , N +2 . Sending N → ∞ and Wallis’ product give e , = ln (cid:16) π (cid:17) . Then e , + e , = ln(2) gives e , = ln (cid:16) π (cid:17) .The claim follows by induction over s utilizing (3.1). (cid:3) M = 2 and M = 3 need more effort. We will make use of the generalized hyperfactorials H p ( n ) := n Y k =1 k k p (3.3)and their asymptotic expansions for p = 1 and p = 2. In the following we write H ( n ) = H ( n ). Proposition 3.3.
We have e , s +2 = ln s − Y k =1 (2 k − k − s (2 k + 1) k − s (2 k ) k − s (2 k + 2) k − s s − A π s + e ! and e , s +4 = ln s Y k =1 (2 k ) k − s − (2 k + 2) k − s (2 k − k − s − (2 k + 1) k − s − π s + e s + A ! , with the Glaisher-Kinkelin constant A . Proof.
As before we start with e , = N X j =1 ( − j +1 e , j + e , N +2 = ln N − Y j =0 (cid:16) j Y k =1 (2 k − k + 1)(2 k ) π (cid:17)(cid:16) j Y k =1 (2 k + 1) k (2 k + 2) π (cid:17)! + e , N +2 = ln (cid:16) π (cid:17) N N N − Y j =0 j Y k =1 (2 k − k + 1) (2 k ) (2 k + 2) ! + e , N +2 By induction over N one can show12 N N − Y j =0 j Y k =1 (2 k − k + 1) (2 k ) (2 k + 2) = N Y k =1 (cid:18) k k − (cid:19) k − N − (cid:18) k k + 1 (cid:19) k − N . Then applying Wallis’ product gives e , = ln N Y k =1 (cid:18) k k − (cid:19) k − (cid:18) k k + 1 (cid:19) k ∞ Y k = N +1 (cid:18) (2 k ) (2 k − k + 1) (cid:19) N ! + e , N +2 . We perform an index shift in the second sum inside the last logarithm and replace N by a continuousvariable x . Thus we obtainlim N →∞ ln ∞ Y k = N +1 (cid:18) (2 k ) (2 k − k + 1) (cid:19) N ! = lim x →∞ x ∞ X k =1 (2 ln(2 k + 2 x ) − ln(2 k + 2 x − − ln(2 k + 2 x + 1))= lim x →∞ − x ∞ X k =1 (cid:18) k + 2 x − k + 2 x − − k + 2 x + 1 (cid:19) = lim x →∞ x ∞ X k =1 k + 2 x − k + 2 x )(2 k + 2 x + 1) = 12 , where the latter follows with (2.1). By summarizing the last steps we obtainlim N →∞ ∞ Y k = N +1 (cid:18) (2 k ) (2 k − k + 1) (cid:19) N = e . N Y k =1 (cid:18) k k − (cid:19) k − (cid:18) k k + 1 (cid:19) k = (cid:16) N N Y k =1 k k − (cid:17)(cid:16) N ( N +1) N Y k =1 k k (cid:17) N Y k =1 k k N ( N +1) N Y k =1 k k N +2 Y k =1 k k − ( N +1) N +1 Y k =1 k k − = 2 N +4 N +1 (cid:18) NN (cid:19) (4 N + 2) H ( N ) H ( N + 1) H (2 N ) H (2 N + 2) . We obtain by plugging in (cid:0) NN (cid:1) ∼ N √ πN and H ( N ) ∼ AN N + N + e − N N Y k =1 (cid:18) k k − (cid:19) k − (cid:18) k k + 1 (cid:19) k ∼ A N +4 N +1 N √ π N (cid:18) N (cid:19) N N +3 N + e − N ( N + 1) ( N +1) + N +1+ e − ( N +1) (2 N ) N + N + e − N (2 N + 2) N +1) + N +1+ e − ( N +1) = A √ π (cid:18) N (cid:19) (cid:0) N (cid:1) N +2 N + e N + . This yields ∞ Y k =1 (cid:18) k k − (cid:19) k − (cid:18) k k + 1 (cid:19) k = A √ π e − lim N →∞ e N (cid:0) N (cid:1) N = A √ πe . Here the last step followed by taking logarithms in the limit term and (2.2). Altogether we have shown e , = ln (cid:16) A √ πe e (cid:17) = ln (cid:16) A √ πe (cid:17) . From e , + e , = ln (cid:0) π (cid:1) we get e , = ln (cid:16) π e A (cid:17) . The case for s > s . (cid:3) Proposition 3.4.
There hold e , s +2 = ln s − Y k =1 (2 k − k − sk + s (2 k + 1) k − (6 s − k +3 s − s (2 k ) k − (6 s − k +3 s − s (2 k + 2) k − (2 s − k + s − s π s e s B s − s A s ! and e , s +4 = ln s − Y k =1 (2 k ) k − (6 s +2) k +3 s +2 s (2 k + 2) k − sk + s (2 k − k − (2 s +1) k + s + s (2 k + 1) k − (6 s +1) k +3 s + s s + s − A s +6 π s +2 s + e s + B ! . Proof.
Once more we begin with e , = N X j =1 ( − j +1 e , j + e , N +2 = ln N − Y j =0 j Y k =1 (2 k − k − − j (2 k + 1) k − − j (2 k ) k − − j (2 k + 2) k − j s − s A s π s e s ! + e , N +2 .
5y induction over N we can show N − Y j =0 j Y k =1 (2 k − k − − j (2 k + 1) k − − j (2 k ) k − − j (2 k + 2) k − j = N − Y k =1 (2 k ) k − (6 N − k +3 N − N (2 k + 2) k − (2 N − k + N − N (2 k − k − Nk + N (2 k + 1) k − (6 N − k +3 N − N , hence we obtain e , = ln N − Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k (cid:18) N − Y k =1 (2 k ) (2 k + 2)(2 k − k + 1) (cid:19) N (cid:18) π (cid:19) N × (cid:18) N − Y k =1 (2 k − k (2 k + 1) k +2 (2 k ) k +1 (2 k + 2) k +1 (cid:19) N (cid:18) A e (cid:19) N ! + e , N +2 . We derive the limits of the products which are exponentiated. For convenience we replace N − N . Then we have for the product in the middlelim N →∞ N Y k =1 (2 k ) (2 k + 2)(2 k − k + 1) = lim N →∞ N + 12 N + 2 (cid:18) N Y k =1 k (2 k + 2)(2 k + 1) (cid:19) = π . For the rightmost product we compute N Y k =1 (2 k − k (2 k + 1) k +2 (2 k ) k +1 (2 k + 2) k +1 = N Y k =1 (2 k + 1) k (2 k + 2) N Y k =1 (2 k − k (2 k + 1) k (2 k ) k (2 k + 2) k = N Y k =1 (2 k + 1) k (2 k + 2) N Y k =1 k k +1 N ( N +2) N Y k =1 k k +1 2 N +2 Y k =1 k k − N +6 N +3 N +1 Y k =1 k k − N ( N +1) N Y k =1 k k ( k + 1) k = N Y k =1 (2 k + 1) k (2 k + 2) 12 N +12 N +3 N + 2 1 (cid:0) N +2 N +1 (cid:1) H (2 N ) H (2 N + 2) H ( N ) H ( N + 1) . Applying the asymptotics of H ( N ) and (cid:0) N +2 N +1 (cid:1) gives N Y k =1 (2 k − k (2 k + 1) k +2 (2 k ) k +1 (2 k + 2) k +1 ∼ N Y k =1 (2 k + 1) k (2 k + 2) 12 N +12 N +4 N + 1 π ( N + 1)2 N +4 A (2 N ) N + N + e − N (2 N + 2) N +1) +3 N +3+ e − N +1) N N +4 N + e − N ( N + 1) N +1) +4 N +4+ e − N +1) = N Y k =1 (2 k + 1) k (2 k + 2) 2 πA N + 12 N + 1 (cid:16) N (cid:17) e (cid:16) N (cid:17) N (cid:16) N (cid:17) N e N . Therefore follows with (2.2)lim N →∞ N Y k =1 (2 k − k (2 k + 1) k +2 (2 k ) k +1 (2 k + 2) k +1 = 4 π πA e e e = 2 eA . e , as e , = ln N − Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k (cid:18) ∞ Y k = N (2 k − k + 1) (2 k ) (2 k + 2) (cid:19) N × (cid:18) ∞ Y k = N (2 k ) k +1 (2 k + 2) k +1 (2 k − k (2 k + 1) k +2 (cid:19) N ! + e , N +2 = ln N − Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k (cid:18) ∞ Y k =1 (2 k + 2 N − k + 2 N − (2 k + 2 N − (2 k + 2 N ) (cid:19) N × (cid:18) ∞ Y k =1 (2 k + 2 N − k +6 N − (2 k + 2 N ) k +2 N − (2 k + 2 N − k +2 N − (2 k + 2 N − k +6 N − (cid:19) N ! + e , N +2 We replace N by a continuous variable x and computelim N →∞ ln (cid:18) ∞ Y k =1 (2 k + 2 N − k + 2 N − (2 k + 2 N − (2 k + 2 N ) (cid:19) N ! = lim x →∞ x ∞ X k =1 (ln(2 k + 2 x −
3) + 3 ln(2 k + 2 x − − k + 2 x − − ln(2 k + 2 x ))= lim x →∞ − x ∞ X k =1 (cid:18) k + 2 x − k + 2 x − − k + 2 x − − k + 2 x (cid:19) = lim x →∞ − x ∞ X k =1 k + 2 x − k + 2 x − k + 2 x − k + 2 x ) = − , according to (2.1). Similarly we getlim N →∞ ln (cid:18) ∞ Y k =1 (2 k + 2 N − k +6 N − (2 k + 2 N ) k +2 N − (2 k + 2 N − k +2 N − (2 k + 2 N − k +6 N − (cid:19) N ! = lim x →∞ x ∞ X k =1 (cid:0) (6 k + 6 x −
5) ln(2 k + 2 x −
2) + (2 k + 2 x −
1) ln(2 k + 2 x ) − (2 k + 2 x −
2) ln(2 k + 2 x − − (6 k + 6 x −
4) ln(2 k + 2 x − (cid:1) = lim x →∞ x ∞ X k =1 k + 4 x − k + 2 x − k + 2 x − k + 2 x − k + 2 x ) − x ∞ X k =1 (cid:0) k + 2 x −
2) + 2 ln(2 k + 2 x ) − k + 2 x − − k + 2 x − (cid:1)! = lim x →∞ x ∞ X k =1 k + 2 x − k + 2 x − k + 2 x − − x ∞ X k =1 k + 2 x − k + 2 x − k + 2 x − k + 2 x ) ! = 14 , again according to (2.1). Thus in total we have shownlim N →∞ (cid:18) ∞ Y k = N (2 k − k + 1) (2 k ) (2 k + 2) (cid:19) N (cid:18) ∞ Y k = N (2 k ) k +1 (2 k + 2) k +1 (2 k − k (2 k + 1) k +2 (cid:19) N = e . e , and caluculate N Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k = 2 N +3 N + N N Y k =1 k k + k ( k + 1) k + kN Y k =1 (2 k − (2 k − + (2 k − N Y k =1 (2 k + 1) (2 k +1) − (2 k +1) − = 2 N +3 N + N N Y k =1 k k + k ( k + 1) k + k N Y k =1 k k + k + N + N + N N Y k =1 k k + k + N +2 Y k =1 k k − k − N +4 N + N + N +1 Y k =1 k k − k − = 2 N +8 N + N + N Y k =1 k k +2 k + ( k + 1) k + k N +1 Y k =1 k k − k − N Y k =1 k k + k + N +2 Y k =1 k k − k − = 2 N +8 N + N + (4 N + 2) H ( N ) H (2 N + 2) H ( N ) H ( N + 1) H ( N + 1) H (2 N ) H (2 N ) H (2 N + 2) . Plugging in H ( N ) ∼ AN N + N + e − N and H ( N ) ∼ BN N + N + N e − N + N yields N Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k ∼ N +8 N + N +2 (2 N + 1) N N + N + e − N (2 N + 2) ( N +1) + ( N +1)+ e − ( N +1) ( N + 1) ( N +1) + N +1+ e − ( N +1) (2 N ) N + N + e − N × B N N +2 N + N e − N + N ( N + 1) ( N +1) +2( N +1) + ( N +1) e − ( N +1) + ( N +1) (2 N ) N + N + N e − N + N (2 N + 2) N +1) + ( N +1) + ( N +1) e − ( N +1) + ( N +1) = B (cid:18) N (cid:19) e (cid:0) N (cid:1) N + N e (cid:0) N (cid:1) N ! N e (cid:0) N (cid:1) N + . With (2.2) and (2.3) we obtainlim N →∞ N Y k =1 (2 k ) k + k (2 k + 2) k + k (2 k − k (2 k + 1) k +2 k = B e − e e e = B e − . Everything combined yields e , = ln( B ). Furthermore e , + e , = ln (cid:16) A √ πe (cid:17) gives e , =ln (cid:16) A √ πe B (cid:17) . Again we can proof the general case in the claim by induction over s . (cid:3) Applications and Outlook
Now everything is prepared to proof (0.1).
Theorem 4.1.
There holds ζ (3) = 4 π ln( B ). Proof.
By Theorem 3.3 in [1] we have π ζ (3) = 4 e , and by Proposition 3.4 we have e , = 7 ln( B ). (cid:3) Remark 4.2.
The formula is surprisingly short and elegant.The results in this work can with some effort be generalized for ζ (2 k + 1) using Theorem 3.3 from[1] and asymptotic expansions of the generalized hyperfactorials H p ( n ). References [1] T. Kyrion, “Recurrence Relations for Values of the Riemann Zeta Function in Odd Integers,” preprintpreprint