a(n) is an infinite sequence of nonzero real numbers. Prove that it has a subsequence b(n) such that b(n+1)/b(n) tends to 0,1 or infinity as n tends to infinity. I have started by looking at the case a(n) is bounded. Since every sequence has a monotonic subsequence, a(n) too has one. Call it b(n). Therefore b(n) is convergent. If the limit of b(n) is nonzero, then b(n+1)/b(n) tends to 1. What if b(n) tends to 0? Also what if a(n) is unbounded? Can I get a few hints on how to proceed?
Limit of ratio sequence.
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real-analysis
sequences-and-series
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0By "If its limit b(n) is nonzero" do you mean "if the limit of b(n) is nonzero"? – 2017-02-11
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0yeah, i made a mistake. corrected it. – 2017-02-11
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0The following rules might help. $(a_n)\rightarrow +\infty \Rightarrow (1/a_n)\rightarrow 0$. If $(a_n)$ is a null sequence and $(a_n)>0, \forall n \Rightarrow (1/a_n)\rightarrow +\infty$. The proofs of these rules might be instructive as well. – 2017-02-11
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0Start with the case where every a(n) is positive and consider L the limsup of the sequence a=(a(n)). If L=oo, build a sequence b such that b(n+1)/b(n) converges to oo. If L is positive and finite, build b such that b(n+1)/b(n) converges to 1. If L=0, build b such that b(n+1)/b(n) converges to 0. Now slightly modify this argument to deal with a(n) nonzero real valued. – 2017-02-11
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0@shredalert I am afraid that convergent sequences, even including those diverging to oo, are only a small part of the iceberg here. – 2017-02-11
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0@Did New to real analysis here, so was just guessing what might help. – 2017-02-11
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0@Did, i do not understand when you take case L=oo. – 2017-02-11
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0Please be (**much**) more specific. – 2017-02-11