Concrete mathematics : a foundation for computer science

2.7 INFINITE SUMS 61
in this special case works because tkEF cok = c tkeF ok for all finite Sets F;
the latter fact follows by induction on the size of F.
The associative law (2.16) can be stated as follows: If tkEK ok and
tkeK bk converge absolutely to A and B, respectively, then tkek(ok + bk)
converges absolutely to A + B. This turns out to be a special case of a more
general theorem that we will prove shortly.
The commutative law (2.17) doesn’t really need to be proved, because
we have shown in the discussion following (2.35) how to derive it as a special
case of a general rule for interchanging the order of summation.
The main result we need to prove is the fundamental principle of multiple
sums: Absolutely convergent sums over two or more indices can always be
summed first with respect to any one of those indices. Formally, we shall
Best to skim this prove that if J and the elements of {Ki 1 j E J} are any sets of indices such that
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- Your friendly TA x
iEJ
kEKj
oi,k converges absolutely to A,
then there exist complex numbers Aj for each j E J such that
IL
&K,
oj,k converges absolutely to Aj, and
t
Aj converges absolutely to A.
iEJ
It suffices to prove this assertion when all terms are nonnegative, because we
can prove the general case by breaking everything into real and imaginary,
positive and negative parts as before. Let’s assume therefore that oi,k 3 0 for
all pairs (j, k) E M, where M is the master index set {(j, k) 1 j E J, k E Kj}.
We are given that tCj,k)EM oj,k is finite, namely that
L aj,k 6 A
(j.k)EF
for all finite subsets F C M, and that A is the least such upper bound. If j is
any element of J, each sum of the form xkEFi oj,k where Fj is a finite subset
of Kj is bounded above by A. Hence these finite sums have a least upper
bound Ai 3 0, and tkEKi oj,k = Aj by definition.
We still need to prove that A is the least upper bound of xjEG Aj,
for all finite subsets G G J. Suppose that G is a finite subset of J with
xjEG Aj = A’ > A. We CXI find finite subsets Fi c Kj such that tkeFi oj,k >
(A/A’)Aj for each j E G with Aj > 0. There is at least one such j. But then
~.iEG,kCFi oj,k > (A/A’) xjEG Aj = A, contradicting the fact that we have