Roger Apéry's proof that zeta(3) is irrational

Transcription

1 Cliff Bott 11 October 2011 Roger Apéry's proof that zeta(3) is irratioal Roger Apéry developed a method for searchig for cotiued fractio represetatios of umbers that have a form such that irratioality criteria ca be applied. At a coferece i 1978 he outlied a proof that (1) is irratioal based o this method. He published a brief ote o his results i the oural Astérisque i 1979 ad a fuller, but still cryptic, accout i Batut ad Olivier i 1980 provided a detailed explaatio of Apéry's method of geeratig rapidly covergig cotiued fractios, but other aspects of the irratioality proof did ot come withi the scope of their paper. Here I attempt to provide a full explaatio of Apéry's method of proof. Apéry's method i outlie We begi with some defiitios ad facts cocerig cotiued fractios. Suppose s 1 is some real umber that we represet i the followig maer a 1 a s 1 ; s ( 2)... (2) b 1 s 2 b s 1 We call this represetatio of s 1 a cotiued fractio. If we defie p 1 a 1 p 2 b 2 p 1 p +1 a +1 p -1 + b +1 p ( 2) q 1 b 1 q 2 a 2 +b 2 q 1 q +1 a +1 q -1 + b +1 q ( 2)... (3) the it is ot hard to show that (Lemma A.1) p 1 s 1 p p s 1 1 ( 2)... (4) q 1 s 2 q q 1 s 1 Coversely, it may be that we commece with a set of real umbers a, b, with p, q defied as i (3). The if the sequece p /q coverges to a limit s 1, we ca represet this limit as i (2), sice whe 1 the s are defied by (2). We call the p /q covergets of the cotiued fractio (whether the sequece coverges or ot) ad the p ad q partial umerators ad partial deomiators respectively. By way of example, the umber x ( 5-1)/2 ca be expressed i the form (2) with s 1 s x ad a b 1. The first of the recurrece relatios (3) is that of the Fiboacci sequece {1, 1, 2,... } {F } ad the secod is that for {F +1 }. Ay umber that ca be expressed as a series ca also be expressed quite easily as a cotiued fractio whose covergets are equal to the partial sums of the series (Lemma A.2). If λ 0, λ 1, λ 2,... is a sequece of o-zero real umbers, λ 0 havig the value 1, the trasformatio a λ -1 λ a ; b λ b ; s λ -1 s applied to (2) yields aother cotiued fractio with the same value s 1 (Lemma A.3) If p,0 ad q,0 are partial umerators ad deomiators respectively of a cotiued fractio ad if σ, is a double sequece of real umbers ( 0) ad if we costruct the ifiite rectagular arrays p, ad q, iteratively by rows as follows

2 2 p,1 p 1, p, 2, q, 1 q 1, q, 2,... (5) the the row, colum ad mai diagoal elemets of these arrays are partial umerators ad partial deomiators respectively of cotiued fractios (Lemma A.4). Clearly p ad q are ratioal if a ad b are ratioal ad itegers if a ad b are itegers. If s 1 ca be represeted as a series with ratioal coefficiets, it is possible usig Lemma A.3 to put the cotiued fractio represetatio of s 1 ito the form (2) with a, b itegers. The coectio betwee cotiued fractios ad tests for ratioality arises as follows. We recall the Theorem (Lemma B) that s 1 is irratioal if there exist itegers h ad such that 0 < s 1 - h < ε for ay give 1>ε> 0 Re-arragig (4) ad usig (3) we obtai q p 1 p q 1 s 1 q s 1 - p 1 1 a a 1... a 1 s 1 T... (6) q q 1 s 1 q q 1 s 1 If lim p q s 1 ad so s 1 p q lim q s 1 p q we also have 1 1 [ p p 1 q q 1 ] [ p p 1 q q 1 ] q 1 a a 1...a 1 T 1 q q... (6a) 1 If we ca fid a represetatio of s 1 such that q ad p are itegers with commo divisor d ad 0 < T /d ε for ay give 0 < ε < 1 whe is sufficietly large, the the theorem applies ad s 1 is irratioal. Suppose that for > 1 s ca be expressed as s σ + ε... (7) where σ is a polyomial i with iteger coefficiets ad ε O(1/) (that is, for all greater tha some specific value, ε <A/ where A is a positive real umber idepedet of ). Equatio (4) suggests that the trasformatio (5) might yield covergets havig the same limit as the origial cotiued fractio ad i fact for ζ(3) ad a variety of more geeral cases this is ideed so. I the case of ζ(3) the partial umerators ad deomiators p, ad q, respectively have large commo factors ad i particular the diagoal elemets p, ad q, yield a versio of (6a) to which the Theorem ca be applied. To apply the method oe must solve two mai problems: evaluatig σ ad fidig represetatios of p, ad q, that eable us to idetify the commo factors. Apéry's method applied to ζ(3) Startig with (1) the by virtue of Lemmas A.2 ad A.3 (with λ 1 1 ad λ 3 ) we obtai for s 1 ζ(3) a cotiued fractio with iteger parameters (Lemma C.1): a 1 b 1 1 ; a 1 6 b ( 2) s q! O O 1 2 ; p! q 3... (8)

3 3 This tells us othig more tha that P 3 1 with (Lemma C.1) Ρ +1 O(1/ 2 ) If L is the smallest positive iteger divisible by all of the itegers 1, 2,..., the (L ) 3 Σ.is a iteger. As a cosequece of Chebyshev's ivestigatios i the mid 19th cetury of the distributio of prime umbers we ow (e.g. Hardy ad Wright Theorem 414) that L O(e ) ad so q s 1 p L 3 3 L 3 icreases with e 3-2 ad the Theorem caot be applied. I the hope of improvig o this situatio, we costruct a hierarchy of cotiued fractios s,1, s,2 ad so o usig the method described by (5). We commece with p,0 p ; q,0 q ad s,0 s as i (8). I Lemma C.2 we show that we obtai i cosequece a set of cotiued fractios where a, s, b, s 1, a 1, p,1! ; b 1, q, 1! a, 1 6 b, ( > 1) s 1, 1, 1, O 1 2 ad partial umerators ad partial deomiators of s 1, are p,! 3 1, 2, ad q,! 3 2,... (9) 1 2 where, ad, are polyomials with iteger coefficiets of degree 2( - 1) (at most) ad 2 respectively i ad cosequetly s 1, lim p, q, lim 3 Table 5.3 of Batut ad Olivier gives the values of p, ad q, for small,. These values are ivariat with respect to exchage of ad ad q, is divisible by (!) 3 (!) 3. The symmetry property ca be show to be geerally true by Lemma A.4(ii), which shows that a, a, ad b, b,. We ca (Lemma C.3) solve the recurreces (5) to fid the 'closed form' expressios (10). I these expressios (r) s - the 'shifted factorial' - is the product of s cosecutive positive itegers, the largest of which is r. (r) s is divisible by s! (see Lemma C.3 - Appedix) ad cosequetly (r) 2s (r) s (r-s) s is divisible by (s!) 2. Thus the expressios (10) ot oly verify the divisibility property of q, but also show that (L ) 3 p, has the same property. 2! 3,! ! 4, 1 1! 3,! [ 0! 4 ] (, Σ 0 defied as 0) (10),

4 4 These results tae us closer to a expressio of the form (6) to which we ca apply the Theorem, sice q, 3 p, 1!6 s 1, q, q 1, s 1, 1!6 s 1, q 1, 1 q 1, O 1 2 1! 3 s 1, 2 1, 1 Sice p, ad q, are divisible by d (!) 3 /(L ) 3 ad s +1, 0 the result is that L 3 2 1, O p, d q, d 3 O 25 The right had side icreases as e (sice must be tae to be costat as icreases). The result suggests however that if we were to use the diagoal elemets p, ad q, (which, by Lemma A.4(iii) are partial umerators ad partial deomiators respectively of a cotiued fractio) it might be possible to obtai a expressio of the form L 3 0 p, q, 3 d d O 22 Oe3 25l which has the desired properties sice it becomes arbitrarily small as icreases. The result we obtai will ot be as strog as this, but still sufficiet to show that ζ(3) is irratioal. By Lemma A.4(iii), the cotiued fractio formed by the diagoal has a 1 6 ; b 1 5 ; a - 3 ( - 1) 9 ; b 3 ( ) ( > 1) ad p, ad q, as defied i (9). If we apply Lemma A.3 with λ 0 1, λ 1/ 6 ( > 0) we obtai a cotiued fractio with a 1 6 ; b 1 5 ; a -( - 1) 3 / 3 ; b ( )/ 3 ( > 1) [ 2 q, 2 0! ] 2 p,, with, 0 1[ 2 2! ] [ 2 ]... (11) This cotiued fractio has the limit ζ(3) (Lemma C.4) ad it is see from (6a) sice that a 1... a L 3 q 3 L 3 p L 3 q q q 1 O e3 3 q It is ot difficult to show (Lemma C.5) that q is a icreasig fuctio of ad that for some positive iteger A idepedet of, q > Ae l 33 > Ae 3.4. This is sufficiet to prove the result.

5 5 Lemmas: A (Theorems about cotiued fractios geerally) Lemma A.1: let s 1 be some real umber that we ca represet i the followig maer: s 1 a 1 a ; s b 1 s 2 b s 1 ( 2)... (a) If we defie p 1 a 1 p 2 b 2 p 1 p +1 a +1 p -1 + b +1 p ( 2) q 1 b 1 q 2 a 2 +b 2 q 1 q +1 a +1 q -1 + b +1 q ( 2)... (b) the s 1 p 1 p p s 1 1 q 1 s 2 q q 1 s 1 ( 2)... (c) Proof: Whe 1 ad 2, the Lemma is clearly correct. Substitutig a 1 s +1 b 1 s 2 i the expressio p p 1 s 1 s 1 q q 1 s 1 we easily see that if (c) is true for the itegers 1, 2,... +1, it is also true for +2. u 1 Lemma A.2: Let s 1 u 1 u 0 1 ; s 1 Ρ 1 ; s 1 P 1 P The s 1 ca be expressed i the form + Ρ +1 be coverget. Defie 1 u ( 1) P s 1 u u 1 u ; s 1s u (2 ) s u 1 1 ad the partial umerator ad partial deomiator of this cotiued fractio are p 1 u ad q 1 Proof: If 1 the s 1 u 1 s 2 ad so s 1 s 1 u 1 1s 2 If 2 the sice Ρ +1 - R -u we have P 1 1 u P P Similarly 1 1 P P 1 u 1 P ad so P u 1 1 u P 1 u u P P 1

6 6 This re-arrages to give P u u 1 P 1 1 u P 1 u 1 P The secod part of the Lemma is easily proved by iductio usig the recurrece formula for p ad q Lemma A.3: Let λ 0 1, λ 1, λ 2,... be a sequece of o-zero real umbers. We defie s 1 as i Lemma 1. The s 1 remais uchaged if we substitute the followig primed quatities for the respective uprimed quatities i equatios (a), (b), (c) of Lemma 1. a' λ -1 λ a b ' λ b s ' λ -1 s ( 2) p ' λ 1 λ 2...λ p q ' λ 1λ 2...λ q Proof: s 1 as defied i (a) is uchaged sice 1 s a 1 ( 1) b s 1 It is true with respect to (c) for 1 ad 2. Suppose it is true for all. The sice (λ +1 λ a +1 )(λ -1...λ 1 p -1 ) + (λ +1 b +1 )(λ...λ 1 p ) (λ +1...λ 1 )(a +1 p -1 + b +1 p ) (λ +1...λ 1 )p +1 it is true for + 1. A similar result applies for q +1. Lemma A.4: Let a ad b ( 1) be sequeces of real umbers ad let u u,0 be a sequece defied as i (3). Let σ, be a double sequece of real umbers ( 0). If we costruct a double sequece iteratively by rows as follows: u, 1 u 1, 2, u, the (i) the elemets i the rows follow the recurrece u, a, u 2, b, u 1, ( 2, 0) (ii) the elemets i the colums follow the recurrece u, d 1, 2 u, 2 2, 1 R 1, 2 u, 1 ( 2) (iii) the diagoal elemets follow the recurrece u, a' u 2, 2 b ' u 1, 1 ( 2) where a, ad b, are defied recursively by a, 1 a, d, d 1, b, 1 R,, d, d 1, ( 2, 0) with R, ad d, defied recursively by R, 2, b 1, d, a 1, 1, R, ad a' ad b' are defied recursively by

7 7 a ' R, 1 R 1, 2 d, 2 a, 2 R, 1 R 1, 2 d 1, 2 a, 1 a 1, 1 b 'R, 1 R d, 2, 2 R 1, 2 d, 1 R, 1 R 1, 1 a, 1 R 1, 2 Proof: (i) The propositio is true for all 2 if 0. If > 0 we have u, 1 u 1, 2, u, a 1, u 1, R, u, a, R, u 2, a 1, b, R, u 1, u 2, 1 u 1,, u 2, u 1, 1 u, 1, u 1, a, u 2, R 1, u 1, a, 1 u 2, 1 b, 1 u 1, 1 a, 1, a, b, 1 u 2, a, 1 b, R 1, u 1, a, R, u 2, [ d, d 1, a,, R 1, R, R 1,] u 1, a, R, u 2, a 1, b, R, u 1, u 1, 1 (ii) The array u, is populated by colums via the expressio u, u 1, 1 2, 1 u, 1 thus: u,-2 u+1,-2 u+2,-2 u,-1 u+1,-1 u, We eed to express u +1,-1 i terms of u,-1 ad u,-2 u 1, 1 u 2, 2 3, 2 u 1, 2 a 2, 2 u, 2 b 2, 2 u 1, 2 2, 2 u, 1 2, 2 u, 2 a 2, 2 3, 2 2, 2 u, 2 3, 2 u, 1 b 2, 2 u, 1 2, 2 u, 2 a 2, 2 2, 2 R 1, 2 u, 2 R 1, 2 u, 1 d 1, 2 u, 2 R 1, 2 u, 1 So u, d 1, 2 u, 2 2, 1 R 1, 2 u, 1 (iii) First we ote that the followig recurreces apply u, u 1, 1 2, 1 u, 1 a 1, 1 u 1, 1 R, 1 u, 1 a 1, 1 u 1, 1 R, 1 u 1,, u 1, 1 d, 1 u 1, 1 R, 1 u 1, We refer to the followig schema

8 8 u-2,-2 u-2,-1 u-1,-1 u,-1 By (i) ad the precedig recurreces, there exist relatioships u, a 1, 1 u 1, 1 R, 1 u, 1 u 1, 1 d 1, 2 u 2, 2 R 1, 2 u 2, 1 u, 1 a, 1 u 2, 1 b, 1 u 1, 1 ad from these we easily fid u, R, 1 R 1, 2 d 1, 2 a, 1 u 2, 2 a 1, 1 R, 1 b, 1 a, 1 R 1, 2 u 1, 1 ad sice R,-1 b,-1 R,-1 R -1,-1 - R,-1 σ +1,-1 R,-1 R -1,-1 +d,-1 - a +1,-1 the first half of (iii) follows. Now referrig to the followig schema u, u-2,-2 u-1,-2 u-1,-1 u-1, we have, usig (ii) ad the recurreces at the head of this sectio u, d,-1 u -1,-1 + R,-1 u -1, u -1,-1 a,-2 u -2,-2 + R -1,-2 u -2,-2 u -1, d,-2 u -1,-2 + (σ +1,-1 + R,-2 )u -1,-1 from which we easily fid u, R, 1 R 1, 2 d, 2 a, 2 u 2, 2 [ u, Lemma B: Theorem: criterio for irratioality R, 1 R d, 2, 2 R 1, 2 a 1, 1] u 1, 1 Theorem: Let x be a real umber. If for ay give real umber 1 > ε > 0, there exist itegers h ad such that, 0 < x - h < ε, the x is irratioal. Proof: sice x - h 0, h/ x. Sice ε < 1 the 0. The rest follows by the couterpositive to the followig propositio: if x is ratioal the there exists a positive iteger such that for every pair of itegers h, where 0 ad h/ x, x - h 1/, with beig idepedet of h,

9 9 The propositio is true if x 0 because the x - h h 1. Otherwise suppose x a/b with a, b 0 ad g.c.d.(a, b) 1. The x - h a - hb / b. The umerator is a iteger ad caot be zero because if it were, we would have a - hb 0 ad h/ a/b x. Therefore the umerator is 1. Idetifyig b (which is uiquely determied by x) with we have the result. Lemmas C: Lemmas specific to the case of ζ(3) Lemma C.1: ζ(3) ca be represeted by the cotiued fractio (8) Proof: By virtue of Lemma A.2 we obtai from the series (1) a cotiued fractio for ζ(3) with a 1 b 1 1 ; a 13 b a ( > 1) Startig with (1) the by virtue of Lemmas A.2 ad A.3 (with λ 1 1 ad λ 3 ) we obtai for s 1 ζ(3) a cotiued fractio with iteger parameters: a 1 b 1 1 ; a 1 6 b ( 2) ad s P with 1 P ad the partial umerators ad partial deomiators are q (!) 3 ; p (!) 3 Σ q Σ From the Euler-Maclauri sum formula (as stated at 7.21 of Whittaer ad Watso) we have 1 where [ 1 2m 1 ] 3 3 P x dx m B m 3 m1 1 2m! [ 1 3] 2m 1 deotes the 2m-1 th derivative of the quatity i the square bracets ad the B m are the Beroullia umbers as defied by W&W with B 1 1/6, B 2 1/30, B 3 1/42 ad so o. Thus P O Usig log divisio to obtai 1/P from P we fid s 1, O 1 4 1,0 O 1 2 Lemma C.2 : Commecig with p,0 ad q,0 as defied i (9) ad s,0 as defied i (11) ad usig (5) we ca costruct by rows a hierarchy of cotiued fractios havig the followig properties: a 1,! ; b 1,! (a) a, 1 6 b, ( > 1)... (b) ad whose partial umerators ad partial deomiators are p,! 3 1, 2, ad

10 10 1 where, ad also q,! 3 2, 2 ad,... (c) are polyomals of degree 2( - 1) (at most) ad 2 respectively i s 1, O (d) Proof: By iductio. Whe 1 we readily obtai: p,1! 3 1 [ 2,1, 1 ] q,1! 3 2,1 where 2, ad 1,1 1 This establishes (c). Sice a 1,1 p 1,1 ad b 1,1 q 1,1 it also establishes (a). (b) is established by applyig Lemma A.5(i). From (4) we have: s 1,1 p q 3,1,1 p 1,1 q 1,1 3 p q 3 p q 3 1,0 1,0 2,0, 0,0 p, 0 q, 0 3 2,0 p 1,0 q 1,0 3 s p q 3 p q 3 2,0 1,0 1,0 2,0, 0,0 p, 0 q,0 3 1,0 p s,0 q, 0 3 1,0 s s 2,0 2,0 1,0 s 1,0 1,0 By (10), the umerator has the form a b 2 O1 3 4 ad usig the biomial series for (1 + 1/) -, the deomiator has the form a b 2a O Thus, carryig our log divisio, we fid s 1,1 s 1,0 1 2 O O Further terms ca be foud through the followig process. By (2) ad Lemma A.5.1 we have If we let s 1,1 a 1,1 6 b 1,1 s 2, s 2,1 s 1, cd e f 2 g 3 O1 4 (where c,..., g are idepedet of ), cross-multiply the equate coefficiets of we fid: 7c -56 ad so c -8 8d + c c -32 ad so d 8 9e + c 2 + 8c + 16d + 2cd 0 so e 0 10f + 8d + 11e + d 2 + cd + 2ec 0 ad so f -32/5 hece s 1, O1 3

11 11 This establishes the four propositios for 1. Suppose they are true for all values of this variable up to some fixed value. We form the partial umerators ad partial deomiators for the row + 1 usig eq' (9). It is ot hard to show that p 1,2 p 2,1 ad similarly for q. Whe > 2 we have from Lemma A.5(ii) p 1, 1 6 p 1, 1 [ ] p, 1 p 1,1 This establishes the truth of (a) for + 1. We establish the truth of (b) sice, with referece to Lemma A.5(i) R, 2, b 1, d, a 1, 1, R, 1 6 a, 1 a, d, d 1, a, 1 6 b, 1 R,, d, d 1, R,, By hypothesis ad usig eq' (9) we obtai Let q,1 1! 3 2 1,! 3 2 2,! 3 2, 1 2, c 2 2 c Maig the appropriate substitutios o the previous equatio we fid 2, 1 42c terms of lesser degree i. I a similar maer we obtai p, 1! 3 [ , 2,,! [, 1, 1 ] 1 2 1, 1 It is ot hard to see from aalysis of its compoets that 1, This proves that (c) is true for +1. As i the case 1, we have from (3b) s 1, 1 s 1, s 2, 2, s 1, 1, O 1 2 O O , 2,, ] has degree at most 2. (We use this step rather tha the ext to obtai the coefficiet of 2 because otherwise we obtai a quadratic diophatie equatio with two iteger solutios.) By (2) ad Lemma A.5 s 1, 1 a 1, 1 6 b, s 1, s 1, 2 As before, we tae s 1, cd e f 2 g 3 O1 4 where c,..., g are idepedet of, ad fid (4+7)c -2( + 2) 2 (4 + 7) ad so c -2( + 2) 2 4( + 2)d + c 2 + 2( + 2)(2 + 5)c + 4( + 2) 2 0 ad so d ( + 2) 3

12 12 (4 + 5)e + 4( + 2)2d + 2cd + 2( + 2) 2 c + c 2 0 so e 0 (4 + 10)f + 2( + 2) 2 d + d 2 + cd + terms with e as a factor 0 so f -(+2) 5 /(4 + 10) 2 1 Lemma C.3: if, ad, are as defied i (9) the they have closed form represetatios as described i (10). 2 Proof: first we deal with,. We write 2,! 3, from which, usig the recurrece (5), we obtai 1 3, , 2,,... (a) with, 0 1. Rather tha usig a polyomial form (liear combiatio of powers of ) we use a shifted factorial form (liear combiatio of shifted factorials i ) to defie μ. Shifted factorials ad some properties are described i the Appedix to the Lemma. The ratioale for usig them is that the recurrece (a) is a form of partial differece equatio. Sice the differece relatio (+1) -() () -1 is the discrete aalogue of the differetial d x /d x x -1, this suggests that it may be fruitful to search for solutios of differece equatios expressed i shifted factorial form. Usig (a), ad maig use of the idetities i the Appedix, we ca calculate, 1, 2, ad these suggest the geeral formula, whe, 0 c 2 with c 2... (b)! 4 We prove this by iductio o. The propositio is true for all whe 0, 1. We suppose it is true for all for each value of the variable less tha or equal to some fixed value. We ca easily trasform (b) to the followig 1 3, ,, ,... (c) O the LHS, we ote that if ay polyomial of degree 2 i ca be writte P, where the c 0 c 2 1 ad d 1, 1 1 d 2 1 are uique. So we ca write 1 c d c 0 for... (d) If we substitute (b) ito the right had side of (c) we obtai a expressio of the form 1 3, 1 c A... (e) 0 It is ot hard to see that A To evaluate for > 0 we ca mae use of the idetities i the Appedix ad fid A [ ] 1 21 A ad [ ]

13 13 1 3, B 1 c 0 A 0 c D 1 c C c 2 E c c [B c 1 C c ] 2 [ D c 1 E c ] c where B 2... (f) C D 2 [ 1 1] 2 1 E 2 4 We see at oce that c 0 1 c 0 1 Now if we compare the coefficiets of 1 22 o the RHS of (d) ad (f) we have 1 c c [2 1]! 1 4 1! 4 If we compare coefficiets of 21 the clearly d For the remaiig c ad d we have, comparig (d) ad (f), the recurreces 1 3 c 1 B c 1 C c 1 3 d 1 D c 1 E c.thus 1 3 c 1 2 1! ! 4 Notig that ad simplifyig the RHS we fid that 1 3 c ! 4 From this we easily obtai 1 3 d ! 4 Thus we have show that for 1 ad ! 4 20 q,! 3 2,! 3! 3,! 3! ! (g) This proves the propositio ad as a cosequece that, is a iteger sice each of ( + ) 2 ad ( + ) 2 is a product of 2 cosecutive itegers ad hece divisible by (!) 2 We ote that the expressio o the RHS is ivariat with respect to iterchage of ad sice whe >, ( + ) 2 0 hece

14 14 mi, 0 0 We have already show elsewhere that q, q, hece (g) is valid for >. Now we deal with p,. We have see that p,! 3 1, q, where 1 1, 0 0 ad if 1, 0 1,1 1 Usig (5) we obtai If we write we obtai 1, 1 We ote that , 1,! 3, is a itegral polyomial of degree 2-2 i with 2, 1 2, 1, 1 3, , 2,, 1,... (h),0 0;,1 1;, ! , ! 4 suggestig that i geeral whe ! 4 3 1, c 2 with c 2 [ 0! 4 ]... (i) where we defie Σ 0 0. We ote that the propositio is true whe 1, 2, 3. We assume it is true for all values of the variable up to some fixed value. (h) ca be trasformed to: 1 3, ,, , 1,... () ad we ca write (c.f. (d)), 1 c 1 2 d c If we substitute (i) ito the RHS of () we obtai a equatio aalogous to (f) 1 3, 1 [ 1 1 4c 1 2!! ] 4 [ B c 1 C c 2! 4 c [ 4c 1 22!! 4... () ] 2 1 ] 2 [ D c 1 E c 2 2! 4 ] (l) Where A, B, C, D are defied as previously. We ca see at oce that 1 c 0 c hece c 0 If we compare coefficiets of (+) 2 i () ad (l) we have

15 c 1 4c 1 2!! ! 4 ad if we compare coefficiets of (+) 2-1 we readily fid that d 1 0 For the remaiig c we have 1 3 c 1 B c 1 C c 2! 4 ad i a maer similar to that used previously we fid 1 c 1 2 [! 4 1 ] For the remaiig d we have 1 3 d 1 D c 1 E c 2 2! 4 ad we easily fid that d 1 0 Appedix - Shifted factorials A shifted factorial () is a product of cosecutive itegers, the largest beig, ad as such ca be see as a geeralisatio of the factorial fuctio! If < the () is zero, otherwise! it is whece we have! 1,!ad ca defie 0 1. Sice!!! where the biomial coefficiet is a iteger, () is divisible by! We ote that whe > 0: 1 1 also 1 3 [1 2 2 ] [ ] ( > 0) [ ] [ ] Lemma C.4: the cotiued fractio defied i (11) has the limit ζ(3) Proof: p q 1 1[ 2 2, 0! ] [ 2 ] 2, [ A A...A B 0! ] 1 B 2...B 2 We show that A B 1 ad this proves the result.

16 16 A [ B ] [ ] [ ][ 3 1 [ ] ] Lemma C.5 Let q be defied by the recurrece (11). The q / q +1 is a icreasig fuctio of ad has limit Proof: From the recurrece relatio (11) we obtai q q 1 q 1 q q 1 3 q (a) So if q /q -1 has a limit it must be a root of x 2-34x+1 0, that is, 17 ± 2. From (10), we see that q > q -1 > 0 ad so if q /q -1 has a limit it must be the larger root sice the smaller is less tha oe. From (a) So q / q -1 < 34 q 1 3 q 1 q 1 q 2 q q 1 q 1 q 2 _[ 1 q 1 q q 2 q 3 ] [ ] If q -1 /q -2 >q -2 /q -3 the both expressios i square bracets are egative ad q /q -1 >q -1 /q -2. Sice q 3 /q /73 >q 2 /q 1 73/6 we have show iductively that q /q -1 is a icreasig fuctio of ad sice it is bouded above it must have a limit. REFERENCES Roger Apéry, Astérisque 61 (1979) p Roger Apéry, Miistère des Uiversités, Comité des travaux historiques et scietifiques i Tome III (Mathématiques) of the Bulleti de la sectio des scieces (1981) p C. Batut ad M. Olivier, Sémiaire de Théorie des Nombres Aée exposé o. 23 (1980) E. T. Whittaer ad G. N. Watso, A Course of Moder Aalysis, 4th Editio G. H. Hardy ad E. M. Wright, 'A Itroductio to the Theory of Numbers', 5th Editio