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I have been told, that Cauchy second limit theorem doesn't always work both ways - if the limit of $n$th root of sequence exist it doesn't immediately mean the limit of $a_{n+1}/a_n$ is equal to nth root limit. Can you explain why? Show an example? Thank you a lot.

Martin Sleziak
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mmm
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1 Answers1

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Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n \to 0.$ But depending on parity of $n$, the limit as $n \to \infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.

When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} \to 0.$

When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1} \to \infty.$

Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$

Martin Sleziak
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coffeemath
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  • Just added explanation of different limits $0,\infty.$ – coffeemath Jan 23 '19 at 01:30
  • mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd. – coffeemath Jan 23 '19 at 01:52
  • i made some thinking, and you are right!! – mmm Jan 23 '19 at 01:53
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    Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet. – mmm Jan 23 '19 at 02:00
  • @mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,\infty$ as before. – coffeemath Jan 23 '19 at 04:26
  • thanks, i see it – mmm Jan 23 '19 at 14:34