If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty}...












0












$begingroup$


If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$



Using comparison test



$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$



And I am getting nothing.



I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.



How to proceed?






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:36












  • $begingroup$
    I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:42












  • $begingroup$
    no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:43












  • $begingroup$
    Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:56
















0












$begingroup$


If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$



Using comparison test



$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$



And I am getting nothing.



I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.



How to proceed?






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:36












  • $begingroup$
    I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:42












  • $begingroup$
    no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:43












  • $begingroup$
    Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:56














0












0








0





$begingroup$


If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$



Using comparison test



$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$



And I am getting nothing.



I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.



How to proceed?






sequences-and-series







share|cite|improve this question









$endgroup$




If $sum_{n=1}^{infty} a_n$ is a convergent series of positive terms then $sum_{n=1}^{infty} frac{a_n^{1/n}}{n^{4/5}}$ coverges or diverges$?$



Using comparison test



$lim_{n to infty} frac{(a_n)^{1/n}}{n^{4/5} a_n}$



And I am getting nothing.



I know that any such convergent series of positive terms should be of the form $sum_{n=1}^{infty} (1/n^{a})$ where $a>1$. I tried to solve the problem by replacing $a_n$ with $1/n^{a}$, used comparison test, but found no result.



How to proceed?






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 13:31









MathsaddictMathsaddict

3669




3669












  • $begingroup$
    it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:36












  • $begingroup$
    I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:42












  • $begingroup$
    no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:43












  • $begingroup$
    Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:56


















  • $begingroup$
    it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:36












  • $begingroup$
    I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:42












  • $begingroup$
    no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:43












  • $begingroup$
    Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
    $endgroup$
    – Mathsaddict
    Dec 30 '18 at 13:56
















$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36






$begingroup$
it can diverge. $a_n := frac{1}{n^2} implies a_n^{1/n} to 1$
$endgroup$
– mathworker21
Dec 30 '18 at 13:36














$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42






$begingroup$
I tried with this example but since $n^{4/5}$ is in denominator so $n^{th}$ term would tend to zero (in comparison test). So we can't say anything about convergence.
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:42














$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43






$begingroup$
no. the sum $sum_n frac{1}{n^{4/5}}$ diverges
$endgroup$
– mathworker21
Dec 30 '18 at 13:43














$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56




$begingroup$
Can we directly find the behaviour of the series $sum_{n} frac{1}{n^{2/n} n^{4/5}}$ by applying limit on $n^{2/n}$ in all terms$?$
$endgroup$
– Mathsaddict
Dec 30 '18 at 13:56










1 Answer
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$begingroup$

You can have both situations.



1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.



2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    $begingroup$

    You can have both situations.



    1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.



    2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You can have both situations.



      1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.



      2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You can have both situations.



        1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.



        2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.






        share|cite|improve this answer









        $endgroup$



        You can have both situations.



        1) If $a_n = q^n$, with $0<q<1$, then $sum_n a_n$ is convergent but $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{q}{n^{4/5}}$ diverges to $+infty$.



        2) If $a_n = n^{-n}$, then $sum_n a_n$ is convergent and $sum_n frac{a_n^{1/n}}{n^{4/5}} = sum_n frac{1}{n^{1+4/5}}$ is convergent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 13:42









        RigelRigel

        11.4k11320




        11.4k11320






























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