In PG office PhD students are preparing their "Qualifying Exam" in Advanced Calculus, one of them discusses with me the following interesting question:
Problem. Find a sequence {xn} of real numbers such that limn→∞xn=1 and ∞∑n=11nxn<∞.Solution.
For every α>1 the series ∑1/nα converges, thus, for every k∈N and α=1+1/k, there will be an Nk such that ∑n≥Nk1n1+1/k<12k. We may assume {Nk} is strictly increasing, then we can define xj=1+1kif Nk≤j<Nk+1. And of course this choice will do! (with any choices x1,…,xN1−1 whom we don't care)◼
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