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Show from the definition of a Cauchy Sequence that Xn=ln(n) is not a Cauchy Series

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  • Yeah it does thanks, I’ll close the question –  Jun 01 '20 at 04:54

2 Answers2

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You need to prove that there is some epsilon > 0 such that for any N > 0 there is some n and m with |ln(n/m)| >= epsilon

for example, is you take epsilon = 1 and n=m+1, then the inequality is satisfied after some index N

ILoveMath
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Assume $\log n$ is Cauchy.

For given $\epsilon >0$ there exists a $n_0$ s.t. for $m\ge n\ge n_0$

$|\log m-\log n| <\epsilon$.

Consider $m=kn$, where $k$ is a positive integer.

Then

$|\log m-\log n|= |\log (m/n)|=|\log k| <\epsilon$.

For $k$ large enough, a contradiction (Why?)

Peter Szilas
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  • becuase it fails eventually for some epsilon, for instance epsilon = 1 will fail after k=e – ILoveMath Jun 01 '20 at 05:12
  • Yes. Or a bit more general, since $\log k$ in not bounded above it will fail for any $\epsilon >0$ choosing $k$ large enough. Cheers. – Peter Szilas Jun 01 '20 at 05:37