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Wednesday, October 15, 2014

Record a problem

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 limnxn=1 and n=11nxn<.Solution.
For every α>1 the series 1/nα converges, thus, for every kN and α=1+1/k, there will be an Nk such that nNk1n1+1/k<12k.  We may assume {Nk} is strictly increasing, then we can define xj=1+1kif Nkj<Nk+1. And of course this choice will do! (with any choices x1,,xN11 whom we don't care)

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