Bourbaki

Exercise 12, § 7

General Topology, IV, 399

Bourbaki marks this one as harder than the rest.

Let (un)(u_n) be a sequence of finite real numbers. For each permutation σ\sigma of the set N\mathbf{N}, put

r(n)=∣σ(n)−n∣⋅sup⁡m≥n∣um∣. r(n) = |\sigma(n) - n| \cdot \sup_{m \geq n} |u_m|.
a) If the series with general term unu_n is convergent, and if lim⁡n→∞r(n)=0\lim_{n \to \infty} r(n) = 0, show that the series with general term vn=uσ(n)v_n = u_{\sigma(n)} is convergent to the same sum. [Consider the difference ∑k=0nvk−∑k=0nuk\sum_{k=0}^n v_k - \sum_{k=0}^n u_k for large values of nn, and if hh is the smallest integer ≤n\leq n such that σ(h)>n\sigma(h) > n, note that in this difference there are at most σ(h)−h\sigma(h) - h terms in each of the two sums which do not cancel out]. b) Suppose that the series whose general term is unu_n is convergent. Give an example of a permutation σ\sigma such that lim⁡n→∞r(n)=0\lim_{n \to \infty} r(n) = 0 but such that the criterion of Chapter III, § 5, Exercise 6 a) is not satisfied. [Define a family of mutually disjoint intervals Ik=[nk−2k,nk+2k]I_k = [n_k - 2k, n_k + 2k] in N\mathbf{N}, and take σ\sigma such that σ(n)=n\sigma(n) = n whenever nn does not belong to any of the IkI_k, and such that σ(nk−2j)=nk+2j\sigma(n_k - 2j) = n_k + 2j, σ(nk+2j)=nk−2j\sigma(n_k + 2j) = n_k - 2j for all kk and 0≤j≤k0 \leq j \leq k].

Solution

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