Concrete mathematics : a foundation for computer science

62 SUMS
tCj,kiEF a. J, k < A for all finite subsets F s M. Hence xjEG Aj < A, for all
finite subsets G C J.
Finally, let A’ be any real number less than A. Our proof will be complete
if we can find a finite set G C J such that xjeo Aj > A’. We know that
there’s a finite set F C: M such that &j,kIeF oj,k > A’; let G be the set of j’s
in this F, and let Fj = {k 1 (j, k) E F}. Then xjeG A, 3 xjEG tkcF, oj,k =
t(j,k)EF aj,k > A’; QED.
OK, we’re now legitimate! Everything we’ve been doing with infinite
sums is justified, as long a3 there’s a finite bound on all finite sums of the
absolute values of the terms. Since the doubly infinite sum (2.58) gave us
two different answers when we evaluated it in two different ways, its positive s0 whY have f been
hearing a lot lately
terms 1 + i + 5 +. . . must diverge to 03; otherwise we would have gotten the about “harmonic
same answer no matter how we grouped the terms. convergence”?
Exercises
Warmups
1 What does the notation
mean?
0
2 qk
k=4
2 Simplify the expression x . ([x > 01 - [x < 01).
3 Demonstrate your understanding of t-notation by writing out the sums
in full. (Watch out -the second sum is a bit tricky.)
4 Express the triple sum
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