§3. Infinite products
395
Theorem 13. Every infinite product p = fl(1 + un) for which the series Un
is absolutely convergent is itself convergent, with p =I 0 if none of factors of
the product is zero.
One says that such an infinite product is absolutely convergent. This, for
example, is the case of the infinite product
00
sin x = x II (1 - x 2 /n 2 7r 2 )
n=l
for the function sin x which we have mentioned in Chap. II, nO 6, valid for
all x E C. On replacing x by ix one finds an analogous product for the
function sinhx. The proof - we have already mentioned Euler's first, with his
"algebraic equation of infinite degree", at the end of nO 21 of Chap. II - will
be the object of the following nO.
The condition L: lunl < +00, sufficient to ensure convergence of the product in the general case, is also necessary if the Un are real and all of the same
sign for n large. If Un 2: 0 for all n, it is clear that Ul + ... + Un < Pn ::; P
for all n, whence the result. If Un ::; 0 for n large, then also 0 < 1 + Un ::; 1
since Un ---+ OJ we may therefore assume these inequalities valid for all n, so
that 1 2: Pn 2: Pn-l > 0 for all nj the relation Un = (Pn - Pn-t}/Pn-l now
shows that Un rv (Pn - Pn-l)/Pj since the sequence (Pn) is decreasing and
converges, the right hand side is the general term of a convergent series of
negative terms, whence the convergence of L: Un in this case.
Example 1 {infinite product for the Riemann (( s) function). Let P be a prime
number and consider
Clearly this eliminates all the terms whose index n is divisible by P from the
zeta series. On multiplying the result by 1 - 1/ qS, where q is another prime
number, one removes from the remaining series all the terms whose indices
are divisible by q and so all the terms of the initial series whose indices are
divisible by P or q (or by both). If one denotes the sequence
2,3,5,7, 11, 13, 17, ...
of prime numbers by (Pn), one sees that
(17.1 )
(1 - 1/pV ... (1 - pk) ((s) =
n not divisible
by PI or ... or Pk
The integers n figuring in the remaining series are, leaving n = 1 aside, all
2: Pkj any integer n is a product of prime factors, which must be ::; n and
so ::; Pk if n ::; Pk· Since the sequence (Pk) is unending if one believes Euclid
395
Theorem 13. Every infinite product p = fl(1 + un) for which the series Un
is absolutely convergent is itself convergent, with p =I 0 if none of factors of
the product is zero.
One says that such an infinite product is absolutely convergent. This, for
example, is the case of the infinite product
00
sin x = x II (1 - x 2 /n 2 7r 2 )
n=l
for the function sin x which we have mentioned in Chap. II, nO 6, valid for
all x E C. On replacing x by ix one finds an analogous product for the
function sinhx. The proof - we have already mentioned Euler's first, with his
"algebraic equation of infinite degree", at the end of nO 21 of Chap. II - will
be the object of the following nO.
The condition L: lunl < +00, sufficient to ensure convergence of the product in the general case, is also necessary if the Un are real and all of the same
sign for n large. If Un 2: 0 for all n, it is clear that Ul + ... + Un < Pn ::; P
for all n, whence the result. If Un ::; 0 for n large, then also 0 < 1 + Un ::; 1
since Un ---+ OJ we may therefore assume these inequalities valid for all n, so
that 1 2: Pn 2: Pn-l > 0 for all nj the relation Un = (Pn - Pn-t}/Pn-l now
shows that Un rv (Pn - Pn-l)/Pj since the sequence (Pn) is decreasing and
converges, the right hand side is the general term of a convergent series of
negative terms, whence the convergence of L: Un in this case.
Example 1 {infinite product for the Riemann (( s) function). Let P be a prime
number and consider
Clearly this eliminates all the terms whose index n is divisible by P from the
zeta series. On multiplying the result by 1 - 1/ qS, where q is another prime
number, one removes from the remaining series all the terms whose indices
are divisible by q and so all the terms of the initial series whose indices are
divisible by P or q (or by both). If one denotes the sequence
2,3,5,7, 11, 13, 17, ...
of prime numbers by (Pn), one sees that
(17.1 )
(1 - 1/pV ... (1 - pk) ((s) =
n not divisible
by PI or ... or Pk
The integers n figuring in the remaining series are, leaving n = 1 aside, all
2: Pkj any integer n is a product of prime factors, which must be ::; n and
so ::; Pk if n ::; Pk· Since the sequence (Pk) is unending if one believes Euclid