Theorem 3.37 in Baby Rudin: Any Counter-Examples In The Other Case?












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$begingroup$


Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:




For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$




I think I fully understand the proof by Rudin.



From Theorem 3.37 in Baby Rudin, we can also conclude that following:




For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.




Am I right?



However, I'm unable to figure out the proof of or come up with any counter-examples to the following:




Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.




What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?






real-analysis sequences-and-series limits analysis limsup-and-liminf







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  • 1




    $begingroup$
    math.stackexchange.com/a/1708611/43949
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    – angryavian
    Dec 3 '18 at 5:34






  • 2




    $begingroup$
    Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
    $endgroup$
    – Idéophage
    Dec 3 '18 at 5:42
















1












$begingroup$


Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:




For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$




I think I fully understand the proof by Rudin.



From Theorem 3.37 in Baby Rudin, we can also conclude that following:




For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.




Am I right?



However, I'm unable to figure out the proof of or come up with any counter-examples to the following:




Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.




What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?






real-analysis sequences-and-series limits analysis limsup-and-liminf







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/a/1708611/43949
    $endgroup$
    – angryavian
    Dec 3 '18 at 5:34






  • 2




    $begingroup$
    Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
    $endgroup$
    – Idéophage
    Dec 3 '18 at 5:42














1












1








1





$begingroup$


Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:




For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$




I think I fully understand the proof by Rudin.



From Theorem 3.37 in Baby Rudin, we can also conclude that following:




For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.




Am I right?



However, I'm unable to figure out the proof of or come up with any counter-examples to the following:




Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.




What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?






real-analysis sequences-and-series limits analysis limsup-and-liminf







share|cite|improve this question









$endgroup$




Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:




For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$




I think I fully understand the proof by Rudin.



From Theorem 3.37 in Baby Rudin, we can also conclude that following:




For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.




Am I right?



However, I'm unable to figure out the proof of or come up with any counter-examples to the following:




Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.




What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?






real-analysis sequences-and-series limits analysis limsup-and-liminf




real-analysis sequences-and-series limits analysis limsup-and-liminf






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asked Dec 3 '18 at 5:27









Saaqib MahmoodSaaqib Mahmood

7,71242477




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  • 1




    $begingroup$
    math.stackexchange.com/a/1708611/43949
    $endgroup$
    – angryavian
    Dec 3 '18 at 5:34






  • 2




    $begingroup$
    Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
    $endgroup$
    – Idéophage
    Dec 3 '18 at 5:42














  • 1




    $begingroup$
    math.stackexchange.com/a/1708611/43949
    $endgroup$
    – angryavian
    Dec 3 '18 at 5:34






  • 2




    $begingroup$
    Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
    $endgroup$
    – Idéophage
    Dec 3 '18 at 5:42








1




1




$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34




$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34




2




2




$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42




$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42










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Concerning the first question: yes, you are right.



On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.






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    $begingroup$

    Concerning the first question: yes, you are right.



    On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Concerning the first question: yes, you are right.



      On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Concerning the first question: yes, you are right.



        On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.






        share|cite|improve this answer









        $endgroup$



        Concerning the first question: yes, you are right.



        On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 6:57









        José Carlos SantosJosé Carlos Santos

        157k22126227




        157k22126227






























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