Sobre Roger Apéry pode ver em inglês ou francês a biografia escrita por François Apéry, The Mathematical Intelligencer, vol. 18, n° 2, 1996, pp. 54-61.
Adenda de 22.01.10: Foi no livro de Ian Stewart, Os Problemas da Matemática, (Gradiva, 2.ª edição, 1996), que tomei conhecimento da existência da demonstração de
, no qual se lê:
« A função
é agora conhecida como a função zeta de Riemann. Depois de inúmeras dificuldades, Euler conseguiu somar a série para certos valores de
. Em 1734 descobriu que
. Mais tarde provou que, para todo o
par,
é um múlltiplo racional de
. Podemos deduzir daqui que
é irraqcional (de facto, transcendente) para todo o
par. Até muito recentemente, ninguém podia dizer nada deste teor para
ímpar. Devem imaginar a reacção quando, nas Journées Arithmétiques de Marseille-Luminy, em Junho de 1978, R. Apéry, da Universidade de Caen, foi anunciado para falar ‘Sobre a irracionalidade de
‘ . Alf van der Poorten, que estava lá, descreve a conferência nestes termos: ‘O cepticismo era geral. A palestra tendeu a fortalecer esta visão de completa incredulidade. Aqueles que a escutaram sem interesse, ou que estavam limitados por não serem francófonos, pareciam ouvir apenas uma sequência de asserções pouco prováveis’ »

SpringerLink
Estudei o artigo (*) de Alfred van der Poorten (actualmente Professor jubilado de Matemática) A proof that Euler Missed… Apery’s Proof of the Irrationality of
, The Mathematical Intelligencer, Nº 1 (1979) pp. 195-203 (pdf), para o que necessitei de fazer alguns cálculos, dos quais apresento os do parágrafo 3.
– Nota 1 –
Demonstração da identidade

Fazendo

e

vem

donde




Por este motivo,

mas, como





comparando com (3), assim se completa a demonstração de (1).
– Nota 2 –
Dedução de
![\displaystyle\sum_{k=1}^{n-1}\dfrac{\left( -1\right) ^{k-1}\left( k-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left( n^{2}-k^{2}\right) }=\dfrac{1}{n^{2}}-\dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }.\quad (5) \displaystyle\sum_{k=1}^{n-1}\dfrac{\left( -1\right) ^{k-1}\left( k-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left( n^{2}-k^{2}\right) }=\dfrac{1}{n^{2}}-\dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }.\quad (5)](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7Bn-1%7D%5Cdfrac%7B%5Cleft%28+-1%5Cright%29+%5E%7Bk-1%7D%5Cleft%28+k-1%5Cright%29+%21%5E%7B2%7D%7D%7B%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%28+n%5E%7B2%7D-k%5E%7B2%7D%5Cright%29+%7D%3D%5Cdfrac%7B1%7D%7Bn%5E%7B2%7D%7D-%5Cdfrac%7B%5Cleft%28+-1%5Cright%29+%5E%7Bn-1%7D%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7Bn%5E%7B2%7D%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D.%5Cquad+%285%29&bg=ffffff&fg=000000&s=0)
Para

e

na identidade (1), vem, do lado esquerdo:

![=\displaystyle\sum_{k=1}^{K}\dfrac{\left( -1^{2}\right) \left( -2^{2}\right) ...\left[ -\left( k-1\right) ^{2}\right] }{(n^{2}-1^{2})(n^{2}-2^{2})...(n^{2}-k^{2})} =\displaystyle\sum_{k=1}^{K}\dfrac{\left( -1^{2}\right) \left( -2^{2}\right) ...\left[ -\left( k-1\right) ^{2}\right] }{(n^{2}-1^{2})(n^{2}-2^{2})...(n^{2}-k^{2})}](http://l.wordpress.com/latex.php?latex=%3D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7BK%7D%5Cdfrac%7B%5Cleft%28+-1%5E%7B2%7D%5Cright%29+%5Cleft%28+-2%5E%7B2%7D%5Cright%29+...%5Cleft%5B+-%5Cleft%28+k-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D%7B%28n%5E%7B2%7D-1%5E%7B2%7D%29%28n%5E%7B2%7D-2%5E%7B2%7D%29...%28n%5E%7B2%7D-k%5E%7B2%7D%29%7D&bg=ffffff&fg=000000&s=0)

![=\displaystyle\sum_{k=1}^{K}\dfrac{\left( -1\right) ^{k-1}\left[ 1\cdot 2\cdot ...\cdot\left( k-1\right) \right] ^{2}}{(n^{2}-1^{2})...(n^{2}-k^{2})} =\displaystyle\sum_{k=1}^{K}\dfrac{\left( -1\right) ^{k-1}\left[ 1\cdot 2\cdot ...\cdot\left( k-1\right) \right] ^{2}}{(n^{2}-1^{2})...(n^{2}-k^{2})}](http://l.wordpress.com/latex.php?latex=%3D%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5E%7BK%7D%5Cdfrac%7B%5Cleft%28+-1%5Cright%29+%5E%7Bk-1%7D%5Cleft%5B+1%5Ccdot+2%5Ccdot+...%5Ccdot%5Cleft%28+k-1%5Cright%29+%5Cright%5D+%5E%7B2%7D%7D%7B%28n%5E%7B2%7D-1%5E%7B2%7D%29...%28n%5E%7B2%7D-k%5E%7B2%7D%29%7D&bg=ffffff&fg=000000&s=0)

e, do lado direito:





como se queria deduzir, para
.
Para demonstrar

falta, portanto, deduzir
![\dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2\left( -1\right) ^{n-1}}{n^{2}\dbinom{2n}{n}}; \dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2\left( -1\right) ^{n-1}}{n^{2}\dbinom{2n}{n}};](http://l.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Cleft%28+-1%5Cright%29+%5E%7Bn-1%7D%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7Bn%5E%7B2%7D%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D%3D%5Cdfrac%7B2%5Cleft%28+-1%5Cright%29+%5E%7Bn-1%7D%7D%7Bn%5E%7B2%7D%5Cdbinom%7B2n%7D%7Bn%7D%7D%3B&bg=ffffff&fg=000000&s=0)
ou seja, simplificando
![\dfrac{\left( n-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2}{\dbinom{2n}{n}}.\qquad (11) \dfrac{\left( n-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2}{\dbinom{2n}{n}}.\qquad (11)](http://l.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7B%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D%3D%5Cdfrac%7B2%7D%7B%5Cdbinom%7B2n%7D%7Bn%7D%7D.%5Cqquad+%2811%29&bg=ffffff&fg=000000&s=0)
Para o denominador do membro esquerdo vem, sucessivamente:
![\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] \left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right]](http://l.wordpress.com/latex.php?latex=%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+&bg=ffffff&fg=000000&s=0)

![=\left[ (n-1)(n-2)...2\cdot 1\right] \left[ (n+1)(n+2)...(2n-1)\right] =\left[ (n-1)(n-2)...2\cdot 1\right] \left[ (n+1)(n+2)...(2n-1)\right]](http://l.wordpress.com/latex.php?latex=%3D%5Cleft%5B+%28n-1%29%28n-2%29...2%5Ccdot+1%5Cright%5D+%5Cleft%5B+%28n%2B1%29%28n%2B2%29...%282n-1%29%5Cright%5D+&bg=ffffff&fg=000000&s=0)
![=\left( n-1\right) !\dfrac{1}{n!}\left[ n!(n+1)(n+2)...(2n-1)\right] =\left( n-1\right) !\dfrac{1}{n!}\left[ n!(n+1)(n+2)...(2n-1)\right]](http://l.wordpress.com/latex.php?latex=%3D%5Cleft%28+n-1%5Cright%29+%21%5Cdfrac%7B1%7D%7Bn%21%7D%5Cleft%5B+n%21%28n%2B1%29%28n%2B2%29...%282n-1%29%5Cright%5D+&bg=ffffff&fg=000000&s=0)


Em resumo:
![\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] =\dfrac{\left( 2n\right) !}{2n^{2}}. \left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] =\dfrac{\left( 2n\right) !}{2n^{2}}.](http://l.wordpress.com/latex.php?latex=%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%3D%5Cdfrac%7B%5Cleft%28+2n%5Cright%29+%21%7D%7B2n%5E%7B2%7D%7D.&bg=ffffff&fg=000000&s=0)
e
![\dfrac{\left( n-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{\left( n-1\right) !^{2}}{\dfrac{\left( 2n\right) !}{2n^{2}}}=\dfrac{2n^{2}\left( n-1\right) !^{2}}{\left( 2n\right) !} \dfrac{\left( n-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{\left( n-1\right) !^{2}}{\dfrac{\left( 2n\right) !}{2n^{2}}}=\dfrac{2n^{2}\left( n-1\right) !^{2}}{\left( 2n\right) !}](http://l.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7B%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D%3D%5Cdfrac%7B%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7B%5Cdfrac%7B%5Cleft%28+2n%5Cright%29+%21%7D%7B2n%5E%7B2%7D%7D%7D%3D%5Cdfrac%7B2n%5E%7B2%7D%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7B%5Cleft%28+2n%5Cright%29+%21%7D&bg=ffffff&fg=000000&s=0)
![=\dfrac{2n^{2}\left( n-1\right) !^{2}}{n!^{2}\dbinom{2n}{n}}=\dfrac{2\left[ n\left( n-1\right) !\right] ^{2}}{n!^{2}\dbinom{2n}{n}} =\dfrac{2n^{2}\left( n-1\right) !^{2}}{n!^{2}\dbinom{2n}{n}}=\dfrac{2\left[ n\left( n-1\right) !\right] ^{2}}{n!^{2}\dbinom{2n}{n}}](http://l.wordpress.com/latex.php?latex=%3D%5Cdfrac%7B2n%5E%7B2%7D%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7Bn%21%5E%7B2%7D%5Cdbinom%7B2n%7D%7Bn%7D%7D%3D%5Cdfrac%7B2%5Cleft%5B+n%5Cleft%28+n-1%5Cright%29+%21%5Cright%5D+%5E%7B2%7D%7D%7Bn%21%5E%7B2%7D%5Cdbinom%7B2n%7D%7Bn%7D%7D&bg=ffffff&fg=000000&s=0)

e, portanto
![\dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2\left( -1\right) ^{n-1}}{n^{2}\dbinom{2n}{n}}; \dfrac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }=\dfrac{2\left( -1\right) ^{n-1}}{n^{2}\dbinom{2n}{n}};](http://l.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Cleft%28+-1%5Cright%29+%5E%7Bn-1%7D%5Cleft%28+n-1%5Cright%29+%21%5E%7B2%7D%7D%7Bn%5E%7B2%7D%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%5B+n%5E%7B2%7D-%5Cleft%28+n-1%5Cright%29+%5E%7B2%7D%5Cright%5D+%7D%3D%5Cdfrac%7B2%5Cleft%28+-1%5Cright%29+%5E%7Bn-1%7D%7D%7Bn%5E%7B2%7D%5Cdbinom%7B2n%7D%7Bn%7D%7D%3B&bg=ffffff&fg=000000&s=0)
donde, se obtem a identidade atrás, que se repete:

– Nota 3 –
Dedução de [no original falta o factor 2 do segundo membro. No entanto, se
fosse definido com este factor no denominador, a fórmula seria a que aparece no original.]

em que
.
Partindo desta definição temos

e, sucessivamente

![=\dfrac{k!^{2}}{k^{3}}\left[ \dfrac{\left( n-k\right) !}{\left( n+k\right) !}-\dfrac{\left( n-1-k\right) !}{\left( n-1+k\right) !}\right] =\dfrac{k!^{2}}{k^{3}}\left[ \dfrac{\left( n-k\right) !}{\left( n+k\right) !}-\dfrac{\left( n-1-k\right) !}{\left( n-1+k\right) !}\right]](http://l.wordpress.com/latex.php?latex=%3D%5Cdfrac%7Bk%21%5E%7B2%7D%7D%7Bk%5E%7B3%7D%7D%5Cleft%5B+%5Cdfrac%7B%5Cleft%28+n-k%5Cright%29+%21%7D%7B%5Cleft%28+n%2Bk%5Cright%29+%21%7D-%5Cdfrac%7B%5Cleft%28+n-1-k%5Cright%29+%21%7D%7B%5Cleft%28+n-1%2Bk%5Cright%29+%21%7D%5Cright%5D+&bg=ffffff&fg=000000&s=0)
![=\dfrac{k!^{2}}{k^{3}}\left[ \dfrac{\left( n-1\right) \left( n-1-k\right) !}{\left( n+k\right) \left( n-1+k\right) !}-\dfrac{\left( n-1-k\right) !}{\left( n-1+k\right) !}\right] =\dfrac{k!^{2}}{k^{3}}\left[ \dfrac{\left( n-1\right) \left( n-1-k\right) !}{\left( n+k\right) \left( n-1+k\right) !}-\dfrac{\left( n-1-k\right) !}{\left( n-1+k\right) !}\right]](http://l.wordpress.com/latex.php?latex=%3D%5Cdfrac%7Bk%21%5E%7B2%7D%7D%7Bk%5E%7B3%7D%7D%5Cleft%5B+%5Cdfrac%7B%5Cleft%28+n-1%5Cright%29+%5Cleft%28+n-1-k%5Cright%29+%21%7D%7B%5Cleft%28+n%2Bk%5Cright%29+%5Cleft%28+n-1%2Bk%5Cright%29+%21%7D-%5Cdfrac%7B%5Cleft%28+n-1-k%5Cright%29+%21%7D%7B%5Cleft%28+n-1%2Bk%5Cright%29+%21%7D%5Cright%5D+&bg=ffffff&fg=000000&s=0)




mas como

![=n\left[ \left( n-1\right) ...\left( n-k\right) \right] \left[ \left( n+1\right) ...\left( n+k\right) \right] =n\left[ \left( n-1\right) ...\left( n-k\right) \right] \left[ \left( n+1\right) ...\left( n+k\right) \right]](http://l.wordpress.com/latex.php?latex=%3Dn%5Cleft%5B+%5Cleft%28+n-1%5Cright%29+...%5Cleft%28+n-k%5Cright%29+%5Cright%5D+%5Cleft%5B+%5Cleft%28+n%2B1%5Cright%29+...%5Cleft%28+n%2Bk%5Cright%29+%5Cright%5D+&bg=ffffff&fg=000000&s=0)



resulta


![=-2\left[ \dfrac{k\left( k-1\right) !}{k}\right] ^{2}\dfrac{1}{n\left( n^{2}-1^{2}\right) ...\left( n^{2}-k^{2}\right) } =-2\left[ \dfrac{k\left( k-1\right) !}{k}\right] ^{2}\dfrac{1}{n\left( n^{2}-1^{2}\right) ...\left( n^{2}-k^{2}\right) }](http://l.wordpress.com/latex.php?latex=%3D-2%5Cleft%5B+%5Cdfrac%7Bk%5Cleft%28+k-1%5Cright%29+%21%7D%7Bk%7D%5Cright%5D+%5E%7B2%7D%5Cdfrac%7B1%7D%7Bn%5Cleft%28+n%5E%7B2%7D-1%5E%7B2%7D%5Cright%29+...%5Cleft%28+n%5E%7B2%7D-k%5E%7B2%7D%5Cright%29+%7D&bg=ffffff&fg=000000&s=0)

donde



somando ambos os membros, vem


ou

e, finalmente

como se pretendia deduzir.
– Nota 4 –
Dedução de

Usando a notação de Iverson, o lado esquerdo pode escrever-se


Esta notação significa neste caso

Agora já podemos trocar a ordem dos dois somatórios

O somatório do lado direito, no qual se utiliza a notação de Iverson, atendendo a que
significa
, escreve-se na notação habitual

Provou-se que

Mas como

conclui-se que

– Nota 5 –
Dedução de

Atendendo à definição de 

obtém-se

Mas

logo

Agrupando agora os dois somatórios da mesma quantidade que figuram nos dois membros desta identidade, obtemos

isto é

– Nota 6 –
Dedução de

Vai-se mostrar que

Em virtude de no intervalo
se ter

e
,
então
;
donde resulta que

e, portanto,
,
quando 
Se a soma dos termos em valores absolutos converge para
então também o faz a própria soma. Fazendo agora em

tender para
obtém-se, como pretendíamos

A este respeito R. Apéry escreveu em Irrationalité de ζ(2) et ζ(3), Astérisque 61 (1979), 11–13:
« Pour étudier
, nous posons:

L’ utilisation de la diagonale
donne la série

qui à defaut de prouver immédiatement l’ irrationalité de
converge mieux que
. »
(*) A. van der Poorten, A proof that Euler missed… Apéry’s proof of the irrationality of ζ(3) (An informal report), Math. Intelligencer 1:4 (1978/79), 195–203
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