How to show that $exists N in mathbbN$ s.t $|a_n|^1/n < frac1R <frac1r$, where $limsup |a_n|^1/n = frac1R $

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In the book of Functions of One Complex Variables by Conway, at page 31, it is claimed that



enter image description here



However, I cannot understand the existence of such an $N$ that makes $|a_n|^1/n < 1/r$. I mean as far as I see, the only assumption that we are making is $|z| < R$, and we also do know that $sup |a_n|^1/n$ is a decreasing sequence of $n$ converging to $frac1R $ in this case, but the claim is not about $sup |a_n|^1/n$, it is just for $|a_n|^1/n$.



tl:dr;



How to show that the existence of such an $N$ ?




sequences-and-series complex-analysis complex-numbers




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  • 6




    Write down the definition of $limsup$, and you will find out.
    – Crostul
    Aug 18 at 7:53














up vote
0
down vote

favorite












In the book of Functions of One Complex Variables by Conway, at page 31, it is claimed that



enter image description here



However, I cannot understand the existence of such an $N$ that makes $|a_n|^1/n < 1/r$. I mean as far as I see, the only assumption that we are making is $|z| < R$, and we also do know that $sup |a_n|^1/n$ is a decreasing sequence of $n$ converging to $frac1R $ in this case, but the claim is not about $sup |a_n|^1/n$, it is just for $|a_n|^1/n$.



tl:dr;



How to show that the existence of such an $N$ ?




sequences-and-series complex-analysis complex-numbers




share|cite|improve this question


















  • 6




    Write down the definition of $limsup$, and you will find out.
    – Crostul
    Aug 18 at 7:53












up vote
0
down vote

favorite









up vote
0
down vote

favorite











In the book of Functions of One Complex Variables by Conway, at page 31, it is claimed that



enter image description here



However, I cannot understand the existence of such an $N$ that makes $|a_n|^1/n < 1/r$. I mean as far as I see, the only assumption that we are making is $|z| < R$, and we also do know that $sup |a_n|^1/n$ is a decreasing sequence of $n$ converging to $frac1R $ in this case, but the claim is not about $sup |a_n|^1/n$, it is just for $|a_n|^1/n$.



tl:dr;



How to show that the existence of such an $N$ ?




sequences-and-series complex-analysis complex-numbers




share|cite|improve this question














In the book of Functions of One Complex Variables by Conway, at page 31, it is claimed that



enter image description here



However, I cannot understand the existence of such an $N$ that makes $|a_n|^1/n < 1/r$. I mean as far as I see, the only assumption that we are making is $|z| < R$, and we also do know that $sup |a_n|^1/n$ is a decreasing sequence of $n$ converging to $frac1R $ in this case, but the claim is not about $sup |a_n|^1/n$, it is just for $|a_n|^1/n$.



tl:dr;



How to show that the existence of such an $N$ ?





sequences-and-series complex-analysis complex-numbers





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 8:53









Bernard

111k635103




111k635103










asked Aug 18 at 7:46









onurcanbektas

3,1221834




3,1221834







  • 6




    Write down the definition of $limsup$, and you will find out.
    – Crostul
    Aug 18 at 7:53












  • 6




    Write down the definition of $limsup$, and you will find out.
    – Crostul
    Aug 18 at 7:53







6




6




Write down the definition of $limsup$, and you will find out.
– Crostul
Aug 18 at 7:53




Write down the definition of $limsup$, and you will find out.
– Crostul
Aug 18 at 7:53















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