How to check the convergence of series?












0












$begingroup$


$displaystyle sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}}), if (a>0,b>0,c>0)$




Edit 1



Sorry, replaced i with n. Obviously, term test is inconclusive
due to limit is equal to zero. Actually, i tried ratio test several
times but didn't succeed and got 1 as a result every time. Also tried
to represent a,b,c as exponents and expand them in Taylor series but
it didn't work too.







sequences-and-series analysis convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Theres no $i$ in the summands
    $endgroup$
    – Calvin Khor
    Jan 9 at 10:37










  • $begingroup$
    you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
    $endgroup$
    – Hayk
    Jan 9 at 10:38










  • $begingroup$
    @user2942898: The question seems fine, but you just need to show your thought.
    $endgroup$
    – Larry
    Jan 9 at 11:03










  • $begingroup$
    Sorry, fixed the question.
    $endgroup$
    – NRJ
    Jan 9 at 12:11
















0












$begingroup$


$displaystyle sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}}), if (a>0,b>0,c>0)$




Edit 1



Sorry, replaced i with n. Obviously, term test is inconclusive
due to limit is equal to zero. Actually, i tried ratio test several
times but didn't succeed and got 1 as a result every time. Also tried
to represent a,b,c as exponents and expand them in Taylor series but
it didn't work too.







sequences-and-series analysis convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Theres no $i$ in the summands
    $endgroup$
    – Calvin Khor
    Jan 9 at 10:37










  • $begingroup$
    you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
    $endgroup$
    – Hayk
    Jan 9 at 10:38










  • $begingroup$
    @user2942898: The question seems fine, but you just need to show your thought.
    $endgroup$
    – Larry
    Jan 9 at 11:03










  • $begingroup$
    Sorry, fixed the question.
    $endgroup$
    – NRJ
    Jan 9 at 12:11














0












0








0





$begingroup$


$displaystyle sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}}), if (a>0,b>0,c>0)$




Edit 1



Sorry, replaced i with n. Obviously, term test is inconclusive
due to limit is equal to zero. Actually, i tried ratio test several
times but didn't succeed and got 1 as a result every time. Also tried
to represent a,b,c as exponents and expand them in Taylor series but
it didn't work too.







sequences-and-series analysis convergence







share|cite|improve this question











$endgroup$




$displaystyle sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}}), if (a>0,b>0,c>0)$




Edit 1



Sorry, replaced i with n. Obviously, term test is inconclusive
due to limit is equal to zero. Actually, i tried ratio test several
times but didn't succeed and got 1 as a result every time. Also tried
to represent a,b,c as exponents and expand them in Taylor series but
it didn't work too.







sequences-and-series analysis convergence




sequences-and-series analysis convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 11:50







NRJ

















asked Jan 9 at 10:33









NRJNRJ

61




61








  • 1




    $begingroup$
    Theres no $i$ in the summands
    $endgroup$
    – Calvin Khor
    Jan 9 at 10:37










  • $begingroup$
    you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
    $endgroup$
    – Hayk
    Jan 9 at 10:38










  • $begingroup$
    @user2942898: The question seems fine, but you just need to show your thought.
    $endgroup$
    – Larry
    Jan 9 at 11:03










  • $begingroup$
    Sorry, fixed the question.
    $endgroup$
    – NRJ
    Jan 9 at 12:11














  • 1




    $begingroup$
    Theres no $i$ in the summands
    $endgroup$
    – Calvin Khor
    Jan 9 at 10:37










  • $begingroup$
    you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
    $endgroup$
    – Hayk
    Jan 9 at 10:38










  • $begingroup$
    @user2942898: The question seems fine, but you just need to show your thought.
    $endgroup$
    – Larry
    Jan 9 at 11:03










  • $begingroup$
    Sorry, fixed the question.
    $endgroup$
    – NRJ
    Jan 9 at 12:11








1




1




$begingroup$
Theres no $i$ in the summands
$endgroup$
– Calvin Khor
Jan 9 at 10:37




$begingroup$
Theres no $i$ in the summands
$endgroup$
– Calvin Khor
Jan 9 at 10:37












$begingroup$
you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
$endgroup$
– Hayk
Jan 9 at 10:38




$begingroup$
you'll get a better chance to receive an answer and will not risk your post to be closed, if you share here what you have tried with the problem.
$endgroup$
– Hayk
Jan 9 at 10:38












$begingroup$
@user2942898: The question seems fine, but you just need to show your thought.
$endgroup$
– Larry
Jan 9 at 11:03




$begingroup$
@user2942898: The question seems fine, but you just need to show your thought.
$endgroup$
– Larry
Jan 9 at 11:03












$begingroup$
Sorry, fixed the question.
$endgroup$
– NRJ
Jan 9 at 12:11




$begingroup$
Sorry, fixed the question.
$endgroup$
– NRJ
Jan 9 at 12:11










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint: $x^{1/n}=1+frac{ln(x)}{n} + O(n^{-2})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
    $endgroup$
    – NRJ
    Jan 10 at 7:27












  • $begingroup$
    $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
    $endgroup$
    – Mindlack
    Jan 10 at 7:39










  • $begingroup$
    I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
    $endgroup$
    – NRJ
    Jan 10 at 12:29












  • $begingroup$
    X obviously should be replaced with n in limit.
    $endgroup$
    – NRJ
    Jan 10 at 12:35










  • $begingroup$
    Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
    $endgroup$
    – Mindlack
    Jan 10 at 12:42



















0












$begingroup$

Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
$sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}})=sum_{n=1}^infty(e^{frac{ln(a)}{n}}-frac{e^{frac{ln(b)}{n}}+e^{frac{ln(c)}{n}}}{2})$

Expanding exponential function as Mclaurin series and using the logarithm properties:
$sum_{n=1}^inftyfrac{(2+2left(frac{ln(a)}{n}right)-(1+left(frac{ln(b)}{n}right)+1+left(frac{ln(c)}{n}right))+O(n^{-2}))}{2}=sum_{n=1}^inftyfrac{ln(frac{a^2}{bc})}{2n}+O(n^{-2})$

The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=sqrt{bc}$.

Answer: diverges everywhere except $a=sqrt{bc}$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint: $x^{1/n}=1+frac{ln(x)}{n} + O(n^{-2})$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
      $endgroup$
      – NRJ
      Jan 10 at 7:27












    • $begingroup$
      $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
      $endgroup$
      – Mindlack
      Jan 10 at 7:39










    • $begingroup$
      I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
      $endgroup$
      – NRJ
      Jan 10 at 12:29












    • $begingroup$
      X obviously should be replaced with n in limit.
      $endgroup$
      – NRJ
      Jan 10 at 12:35










    • $begingroup$
      Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
      $endgroup$
      – Mindlack
      Jan 10 at 12:42
















    0












    $begingroup$

    Hint: $x^{1/n}=1+frac{ln(x)}{n} + O(n^{-2})$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
      $endgroup$
      – NRJ
      Jan 10 at 7:27












    • $begingroup$
      $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
      $endgroup$
      – Mindlack
      Jan 10 at 7:39










    • $begingroup$
      I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
      $endgroup$
      – NRJ
      Jan 10 at 12:29












    • $begingroup$
      X obviously should be replaced with n in limit.
      $endgroup$
      – NRJ
      Jan 10 at 12:35










    • $begingroup$
      Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
      $endgroup$
      – Mindlack
      Jan 10 at 12:42














    0












    0








    0





    $begingroup$

    Hint: $x^{1/n}=1+frac{ln(x)}{n} + O(n^{-2})$.






    share|cite|improve this answer









    $endgroup$



    Hint: $x^{1/n}=1+frac{ln(x)}{n} + O(n^{-2})$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 9 at 11:43









    MindlackMindlack

    4,910211




    4,910211












    • $begingroup$
      Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
      $endgroup$
      – NRJ
      Jan 10 at 7:27












    • $begingroup$
      $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
      $endgroup$
      – Mindlack
      Jan 10 at 7:39










    • $begingroup$
      I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
      $endgroup$
      – NRJ
      Jan 10 at 12:29












    • $begingroup$
      X obviously should be replaced with n in limit.
      $endgroup$
      – NRJ
      Jan 10 at 12:35










    • $begingroup$
      Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
      $endgroup$
      – Mindlack
      Jan 10 at 12:42


















    • $begingroup$
      Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
      $endgroup$
      – NRJ
      Jan 10 at 7:27












    • $begingroup$
      $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
      $endgroup$
      – Mindlack
      Jan 10 at 7:39










    • $begingroup$
      I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
      $endgroup$
      – NRJ
      Jan 10 at 12:29












    • $begingroup$
      X obviously should be replaced with n in limit.
      $endgroup$
      – NRJ
      Jan 10 at 12:35










    • $begingroup$
      Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
      $endgroup$
      – Mindlack
      Jan 10 at 12:42
















    $begingroup$
    Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
    $endgroup$
    – NRJ
    Jan 10 at 7:27






    $begingroup$
    Can you provide a bit more information on how did you expand exponential function into series? As far as i know exp.func can't be expanded in such form due to the fact that each term in Maclaurin series will have $x^{1/0}$
    $endgroup$
    – NRJ
    Jan 10 at 7:27














    $begingroup$
    $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
    $endgroup$
    – Mindlack
    Jan 10 at 7:39




    $begingroup$
    $x^{1/n}=e^{ln(x)/n}=1+ln(x)/n +O(n^{-2})$. In my expansion, $x$ is fixed and the variable term is $n$.
    $endgroup$
    – Mindlack
    Jan 10 at 7:39












    $begingroup$
    I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
    $endgroup$
    – NRJ
    Jan 10 at 12:29






    $begingroup$
    I'm sorry for slowness but still didn't catch it. Using ratio test, ignoring the remainder term get 1 again: $lim_{x to inf}frac{2+2left(frac{ln(a)}{n+1}right)-1-left(frac{ln(b)}{n+1}right)-1-left(frac{ln(c)}{n+1}right)}{2+2left(frac{ln(a)}{n}right)-1-left(frac{ln(b)}{n}right)-1-left(frac{ln(c)}{n}right)}= frac{2ln(a)-ln(b)-ln(c)}{2ln(a)-ln(b)-ln(c)}*frac{n}{n+1}=1$
    $endgroup$
    – NRJ
    Jan 10 at 12:29














    $begingroup$
    X obviously should be replaced with n in limit.
    $endgroup$
    – NRJ
    Jan 10 at 12:35




    $begingroup$
    X obviously should be replaced with n in limit.
    $endgroup$
    – NRJ
    Jan 10 at 12:35












    $begingroup$
    Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
    $endgroup$
    – Mindlack
    Jan 10 at 12:42




    $begingroup$
    Apart from the obvious division by $0$ issue that you seemingly did not notice, why on earth would you bother with a ratio test given an expansion in $alpha/n+O(n^{-2})$ of the term of your series?
    $endgroup$
    – Mindlack
    Jan 10 at 12:42











    0












    $begingroup$

    Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
    $sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}})=sum_{n=1}^infty(e^{frac{ln(a)}{n}}-frac{e^{frac{ln(b)}{n}}+e^{frac{ln(c)}{n}}}{2})$

    Expanding exponential function as Mclaurin series and using the logarithm properties:
    $sum_{n=1}^inftyfrac{(2+2left(frac{ln(a)}{n}right)-(1+left(frac{ln(b)}{n}right)+1+left(frac{ln(c)}{n}right))+O(n^{-2}))}{2}=sum_{n=1}^inftyfrac{ln(frac{a^2}{bc})}{2n}+O(n^{-2})$

    The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=sqrt{bc}$.

    Answer: diverges everywhere except $a=sqrt{bc}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
      $sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}})=sum_{n=1}^infty(e^{frac{ln(a)}{n}}-frac{e^{frac{ln(b)}{n}}+e^{frac{ln(c)}{n}}}{2})$

      Expanding exponential function as Mclaurin series and using the logarithm properties:
      $sum_{n=1}^inftyfrac{(2+2left(frac{ln(a)}{n}right)-(1+left(frac{ln(b)}{n}right)+1+left(frac{ln(c)}{n}right))+O(n^{-2}))}{2}=sum_{n=1}^inftyfrac{ln(frac{a^2}{bc})}{2n}+O(n^{-2})$

      The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=sqrt{bc}$.

      Answer: diverges everywhere except $a=sqrt{bc}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
        $sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}})=sum_{n=1}^infty(e^{frac{ln(a)}{n}}-frac{e^{frac{ln(b)}{n}}+e^{frac{ln(c)}{n}}}{2})$

        Expanding exponential function as Mclaurin series and using the logarithm properties:
        $sum_{n=1}^inftyfrac{(2+2left(frac{ln(a)}{n}right)-(1+left(frac{ln(b)}{n}right)+1+left(frac{ln(c)}{n}right))+O(n^{-2}))}{2}=sum_{n=1}^inftyfrac{ln(frac{a^2}{bc})}{2n}+O(n^{-2})$

        The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=sqrt{bc}$.

        Answer: diverges everywhere except $a=sqrt{bc}$.






        share|cite|improve this answer









        $endgroup$



        Thank's a lot @Mindlack for help. For all who didn't understand above-stated hint here is the solution:
        $sum_{n=1}^infty ({ a^{1/n} - frac{b^{1/n}+c^{1/n}}{2}})=sum_{n=1}^infty(e^{frac{ln(a)}{n}}-frac{e^{frac{ln(b)}{n}}+e^{frac{ln(c)}{n}}}{2})$

        Expanding exponential function as Mclaurin series and using the logarithm properties:
        $sum_{n=1}^inftyfrac{(2+2left(frac{ln(a)}{n}right)-(1+left(frac{ln(b)}{n}right)+1+left(frac{ln(c)}{n}right))+O(n^{-2}))}{2}=sum_{n=1}^inftyfrac{ln(frac{a^2}{bc})}{2n}+O(n^{-2})$

        The second term of sum converges everywhere while the first term is harmonic series which diverges everywhere except $a=sqrt{bc}$.

        Answer: diverges everywhere except $a=sqrt{bc}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 7:08









        NRJNRJ

        61




        61






























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