Exercise 14, § 7
General Topology, IV, 399
Bourbaki marks this one as harder than the rest.
Let be a sequence of finite real numbers such that the series whose general term is is convergent but not absolutely convergent; let . Show that, for each number , there exists a permutation of such that for those indices such that , and such that . [Show by induction on that there is a permutation of such that for all for which and such that, if denotes , there is an integer with the property that for all . Moreover, is such that for all such that and all such that .]
Solution
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