§3. Infinite products
405
converges unconditionally for all n 2: 1, the series L An converges absolutely,
and
(19.7)
The proof in fact shows a little more. Suppose that the Un are functions
un(x) defined on a set X and that the series un(x) is normally convergent
on X: lun(x)1 ~ Vn with LVn < +00. It is then clear that all the series
featuring in the proof are dominated by the analogous series relative to the
product Il(1 + vn). In consequence, the sums An(x) converge normally as
well as the series L An(x). If for example X is an open subset G of C, if the
Un (z) are analytic on C and if the series L Un (z) converges normally on all
compact subsets KeG, one can conclude that all the functions appearing in
the proof are analytic, on condition, as always, that we know that a normally
convergent series of analytic functions is again analytic (Chap. VII) .
. We can even, in such cases, sometimes obtain an expansion of the infinite
product in power series if we assume that the Un are themselves power series
without constant term
Un(Z) = L an (p)zP
p~l
converging on a disc Izi < R and that the series LUn(z) converges normally
on every disc Izi ~ r < R. The general term of the series (6) is then a product
of absolutely convergent series
(19.8) Vj(Z)=Uil(Z) ... Uin(Z)=
L ail(PI) ... ain(Pn)zPl+ ... +Pn
Pl,···,Pn~1
by the formula for multiplying absolutely convergent series, where we put
j = (il , ... , in) E I n as above. Calculating formally, the sum of An(z) can
be written
(19.9)
for this to converge unconditionally it is necessary and sufficient that the
analogous sum obtained on replacing the an(p) and Z by their absolute values
should converge. This amounts to substituting the product Il[l + wn(z)] for
the given infinite product Il[l + Un (z)J, where we have put
(19.10)
Wn(Z) = L lan(p)zPI,
P
a convergent series for Izi < R. We are allowed to regroup the terms, in order
to test the convergence of the new sum (9), which has positive terms (Chap. II,