Find the convergence of $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3}$












1












$begingroup$


I want to find what the series $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3}$ converges to exactly or show that it diverges.



By taking the partial sum of the series $S_N$ = $ sum_{n=1}^{N} (n+1)^frac{1}{3} - n^frac{1}{3}$ then $S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... + (N+1)^frac{1}{3} - N^frac{1}{3}$



And at the end I'm left with $S_N = -1 + (N+1)^frac{1}{3}$ and $lim_{N to infty} S_N = -1 + infty= infty $



So $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3} = infty$



Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?



Thank you in prior.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 18:36






  • 1




    $begingroup$
    Check again which terms cancel.
    $endgroup$
    – Martin R
    Jan 18 at 18:37






  • 1




    $begingroup$
    It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
    $endgroup$
    – Jack D'Aurizio
    Jan 18 at 18:37






  • 1




    $begingroup$
    you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
    $endgroup$
    – Will Jagy
    Jan 18 at 18:37










  • $begingroup$
    Thank you guys very much for your comments. I edited it.
    $endgroup$
    – Allorja
    Jan 18 at 18:43
















1












$begingroup$


I want to find what the series $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3}$ converges to exactly or show that it diverges.



By taking the partial sum of the series $S_N$ = $ sum_{n=1}^{N} (n+1)^frac{1}{3} - n^frac{1}{3}$ then $S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... + (N+1)^frac{1}{3} - N^frac{1}{3}$



And at the end I'm left with $S_N = -1 + (N+1)^frac{1}{3}$ and $lim_{N to infty} S_N = -1 + infty= infty $



So $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3} = infty$



Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?



Thank you in prior.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 18:36






  • 1




    $begingroup$
    Check again which terms cancel.
    $endgroup$
    – Martin R
    Jan 18 at 18:37






  • 1




    $begingroup$
    It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
    $endgroup$
    – Jack D'Aurizio
    Jan 18 at 18:37






  • 1




    $begingroup$
    you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
    $endgroup$
    – Will Jagy
    Jan 18 at 18:37










  • $begingroup$
    Thank you guys very much for your comments. I edited it.
    $endgroup$
    – Allorja
    Jan 18 at 18:43














1












1








1





$begingroup$


I want to find what the series $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3}$ converges to exactly or show that it diverges.



By taking the partial sum of the series $S_N$ = $ sum_{n=1}^{N} (n+1)^frac{1}{3} - n^frac{1}{3}$ then $S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... + (N+1)^frac{1}{3} - N^frac{1}{3}$



And at the end I'm left with $S_N = -1 + (N+1)^frac{1}{3}$ and $lim_{N to infty} S_N = -1 + infty= infty $



So $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3} = infty$



Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?



Thank you in prior.






sequences-and-series







share|cite|improve this question











$endgroup$




I want to find what the series $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3}$ converges to exactly or show that it diverges.



By taking the partial sum of the series $S_N$ = $ sum_{n=1}^{N} (n+1)^frac{1}{3} - n^frac{1}{3}$ then $S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... + (N+1)^frac{1}{3} - N^frac{1}{3}$



And at the end I'm left with $S_N = -1 + (N+1)^frac{1}{3}$ and $lim_{N to infty} S_N = -1 + infty= infty $



So $ sum_{n=1}^{infty} (n+1)^frac{1}{3} - n^frac{1}{3} = infty$



Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?



Thank you in prior.






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








edited Jan 18 at 18:42







Allorja

















asked Jan 18 at 18:32









AllorjaAllorja

789




789








  • 1




    $begingroup$
    $-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 18:36






  • 1




    $begingroup$
    Check again which terms cancel.
    $endgroup$
    – Martin R
    Jan 18 at 18:37






  • 1




    $begingroup$
    It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
    $endgroup$
    – Jack D'Aurizio
    Jan 18 at 18:37






  • 1




    $begingroup$
    you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
    $endgroup$
    – Will Jagy
    Jan 18 at 18:37










  • $begingroup$
    Thank you guys very much for your comments. I edited it.
    $endgroup$
    – Allorja
    Jan 18 at 18:43














  • 1




    $begingroup$
    $-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
    $endgroup$
    – Lord Shark the Unknown
    Jan 18 at 18:36






  • 1




    $begingroup$
    Check again which terms cancel.
    $endgroup$
    – Martin R
    Jan 18 at 18:37






  • 1




    $begingroup$
    It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
    $endgroup$
    – Jack D'Aurizio
    Jan 18 at 18:37






  • 1




    $begingroup$
    you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
    $endgroup$
    – Will Jagy
    Jan 18 at 18:37










  • $begingroup$
    Thank you guys very much for your comments. I edited it.
    $endgroup$
    – Allorja
    Jan 18 at 18:43








1




1




$begingroup$
$-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 18:36




$begingroup$
$-1-N^{1/3}$ is negative, yet each $(n+1)^{1/3}-n^{1/3}$ is positive.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 18:36




1




1




$begingroup$
Check again which terms cancel.
$endgroup$
– Martin R
Jan 18 at 18:37




$begingroup$
Check again which terms cancel.
$endgroup$
– Martin R
Jan 18 at 18:37




1




1




$begingroup$
It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
$endgroup$
– Jack D'Aurizio
Jan 18 at 18:37




$begingroup$
It is pretty obvious that such series is divergent, since it is a telescopic series with positive and unbounded partial sums.
$endgroup$
– Jack D'Aurizio
Jan 18 at 18:37




1




1




$begingroup$
you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
$endgroup$
– Will Jagy
Jan 18 at 18:37




$begingroup$
you have not been careful enough. Write out $S_2$ and $S_3$ as separate calculations
$endgroup$
– Will Jagy
Jan 18 at 18:37












$begingroup$
Thank you guys very much for your comments. I edited it.
$endgroup$
– Allorja
Jan 18 at 18:43




$begingroup$
Thank you guys very much for your comments. I edited it.
$endgroup$
– Allorja
Jan 18 at 18:43










4 Answers
4






active

oldest

votes


















2












$begingroup$

Your computation is right, up to this point:
$$S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^frac{1}{3} - N^frac{1}{3}$$



Note that once you cancel all terms you are left with
$$S_N=(N+1)^{frac{1}{3}}-1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
    $endgroup$
    – Will Jagy
    Jan 18 at 18:42






  • 1




    $begingroup$
    @WillJagy good idea, ty
    $endgroup$
    – N. S.
    Jan 18 at 18:43



















1












$begingroup$

we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
$$ small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
Just a visual thing, the pairs that cancel are next to each other this way






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$


    While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.





    METHODOLOGY $1$:



    Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals



    $$(n+1)^{1/3}-n^{1/3}=frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> frac{1}{3(n+1)^{2/3}}>frac{1}{3(n+1)}$$



    Hence, we have



    $$sum_{n=1}^N left((n+1)^{1/3}-n^{1/3} right)>frac13 sum_{n=2}^{N+1}frac1n$$



    Inasmuch as the harmonic series diverges, the series of interest diverges also.




    METHODOLOGY $2$:



    Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}left(1+frac1{3n}+Oleft(frac1{n^2}right)right)-n^{1/3}=frac1{3n^{2/3}}+Oleft(frac1{n^{5/3}}right)$



    Inasmuch as the series $sum_{n=1}^infty frac{1}{n^p}$ diverges for $ple 1$, the series of interest diverges also.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Please let me know how I can improve my answer. I really want to give you the best answer I can.
      $endgroup$
      – Mark Viola
      Jan 19 at 1:21










    • $begingroup$
      You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
      $endgroup$
      – Claude Leibovici
      Jan 19 at 7:03










    • $begingroup$
      Thank you Claude!
      $endgroup$
      – Mark Viola
      Jan 19 at 13:59










    • $begingroup$
      My pleasure, Sir ! Glad to help !!!!
      $endgroup$
      – Claude Leibovici
      Jan 19 at 14:03










    • $begingroup$
      @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
      $endgroup$
      – Mark Viola
      Jan 30 at 5:08



















    0












    $begingroup$

    Another idea:
    $$
    1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
    $$
    Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $sum_n (n+1)^{1/3}-n^{1/3} sim sum_n n^{-2/3}$, which is well-known to be diverging.






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Your computation is right, up to this point:
      $$S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^frac{1}{3} - N^frac{1}{3}$$



      Note that once you cancel all terms you are left with
      $$S_N=(N+1)^{frac{1}{3}}-1$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
        $endgroup$
        – Will Jagy
        Jan 18 at 18:42






      • 1




        $begingroup$
        @WillJagy good idea, ty
        $endgroup$
        – N. S.
        Jan 18 at 18:43
















      2












      $begingroup$

      Your computation is right, up to this point:
      $$S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^frac{1}{3} - N^frac{1}{3}$$



      Note that once you cancel all terms you are left with
      $$S_N=(N+1)^{frac{1}{3}}-1$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
        $endgroup$
        – Will Jagy
        Jan 18 at 18:42






      • 1




        $begingroup$
        @WillJagy good idea, ty
        $endgroup$
        – N. S.
        Jan 18 at 18:43














      2












      2








      2





      $begingroup$

      Your computation is right, up to this point:
      $$S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^frac{1}{3} - N^frac{1}{3}$$



      Note that once you cancel all terms you are left with
      $$S_N=(N+1)^{frac{1}{3}}-1$$






      share|cite|improve this answer











      $endgroup$



      Your computation is right, up to this point:
      $$S_N = 2^frac{1}{3} - 1 + 3^frac{1}{3} - 2^frac{1}{3} +4^frac{1}{3}-3^frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^frac{1}{3} - N^frac{1}{3}$$



      Note that once you cancel all terms you are left with
      $$S_N=(N+1)^{frac{1}{3}}-1$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 18 at 18:43

























      answered Jan 18 at 18:36









      N. S.N. S.

      104k7112208




      104k7112208












      • $begingroup$
        suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
        $endgroup$
        – Will Jagy
        Jan 18 at 18:42






      • 1




        $begingroup$
        @WillJagy good idea, ty
        $endgroup$
        – N. S.
        Jan 18 at 18:43


















      • $begingroup$
        suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
        $endgroup$
        – Will Jagy
        Jan 18 at 18:42






      • 1




        $begingroup$
        @WillJagy good idea, ty
        $endgroup$
        – N. S.
        Jan 18 at 18:43
















      $begingroup$
      suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
      $endgroup$
      – Will Jagy
      Jan 18 at 18:42




      $begingroup$
      suggest editing in penultimate term $+N^{1/3}- (N-1)^{1/3}$
      $endgroup$
      – Will Jagy
      Jan 18 at 18:42




      1




      1




      $begingroup$
      @WillJagy good idea, ty
      $endgroup$
      – N. S.
      Jan 18 at 18:43




      $begingroup$
      @WillJagy good idea, ty
      $endgroup$
      – N. S.
      Jan 18 at 18:43











      1












      $begingroup$

      we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
      $$ small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
      Just a visual thing, the pairs that cancel are next to each other this way






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
        $$ small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
        Just a visual thing, the pairs that cancel are next to each other this way






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
          $$ small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
          Just a visual thing, the pairs that cancel are next to each other this way






          share|cite|improve this answer











          $endgroup$



          we are allowed to write each summand as we wish, so I will try the sum of $-n^{1/3} + (n+1)^{1/3}$ to get, up to $n = N,$
          $$ small -1 + 2^{1/3} - 2^{1/3} + 3^{1/3} - 3^{1/3} + 4^{1/3} cdots -(N-2)^{1/3}+(N-1)^{1/3} - (N-1)^{1/3} + N^{1/3} - N^{1/3} +(N+1)^{1/3} $$
          Just a visual thing, the pairs that cancel are next to each other this way







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 19:02

























          answered Jan 18 at 18:51









          Will JagyWill Jagy

          103k5102200




          103k5102200























              1












              $begingroup$


              While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.





              METHODOLOGY $1$:



              Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals



              $$(n+1)^{1/3}-n^{1/3}=frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> frac{1}{3(n+1)^{2/3}}>frac{1}{3(n+1)}$$



              Hence, we have



              $$sum_{n=1}^N left((n+1)^{1/3}-n^{1/3} right)>frac13 sum_{n=2}^{N+1}frac1n$$



              Inasmuch as the harmonic series diverges, the series of interest diverges also.




              METHODOLOGY $2$:



              Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}left(1+frac1{3n}+Oleft(frac1{n^2}right)right)-n^{1/3}=frac1{3n^{2/3}}+Oleft(frac1{n^{5/3}}right)$



              Inasmuch as the series $sum_{n=1}^infty frac{1}{n^p}$ diverges for $ple 1$, the series of interest diverges also.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Jan 19 at 1:21










              • $begingroup$
                You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
                $endgroup$
                – Claude Leibovici
                Jan 19 at 7:03










              • $begingroup$
                Thank you Claude!
                $endgroup$
                – Mark Viola
                Jan 19 at 13:59










              • $begingroup$
                My pleasure, Sir ! Glad to help !!!!
                $endgroup$
                – Claude Leibovici
                Jan 19 at 14:03










              • $begingroup$
                @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                $endgroup$
                – Mark Viola
                Jan 30 at 5:08
















              1












              $begingroup$


              While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.





              METHODOLOGY $1$:



              Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals



              $$(n+1)^{1/3}-n^{1/3}=frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> frac{1}{3(n+1)^{2/3}}>frac{1}{3(n+1)}$$



              Hence, we have



              $$sum_{n=1}^N left((n+1)^{1/3}-n^{1/3} right)>frac13 sum_{n=2}^{N+1}frac1n$$



              Inasmuch as the harmonic series diverges, the series of interest diverges also.




              METHODOLOGY $2$:



              Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}left(1+frac1{3n}+Oleft(frac1{n^2}right)right)-n^{1/3}=frac1{3n^{2/3}}+Oleft(frac1{n^{5/3}}right)$



              Inasmuch as the series $sum_{n=1}^infty frac{1}{n^p}$ diverges for $ple 1$, the series of interest diverges also.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Jan 19 at 1:21










              • $begingroup$
                You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
                $endgroup$
                – Claude Leibovici
                Jan 19 at 7:03










              • $begingroup$
                Thank you Claude!
                $endgroup$
                – Mark Viola
                Jan 19 at 13:59










              • $begingroup$
                My pleasure, Sir ! Glad to help !!!!
                $endgroup$
                – Claude Leibovici
                Jan 19 at 14:03










              • $begingroup$
                @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                $endgroup$
                – Mark Viola
                Jan 30 at 5:08














              1












              1








              1





              $begingroup$


              While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.





              METHODOLOGY $1$:



              Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals



              $$(n+1)^{1/3}-n^{1/3}=frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> frac{1}{3(n+1)^{2/3}}>frac{1}{3(n+1)}$$



              Hence, we have



              $$sum_{n=1}^N left((n+1)^{1/3}-n^{1/3} right)>frac13 sum_{n=2}^{N+1}frac1n$$



              Inasmuch as the harmonic series diverges, the series of interest diverges also.




              METHODOLOGY $2$:



              Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}left(1+frac1{3n}+Oleft(frac1{n^2}right)right)-n^{1/3}=frac1{3n^{2/3}}+Oleft(frac1{n^{5/3}}right)$



              Inasmuch as the series $sum_{n=1}^infty frac{1}{n^p}$ diverges for $ple 1$, the series of interest diverges also.






              share|cite|improve this answer









              $endgroup$




              While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.





              METHODOLOGY $1$:



              Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals



              $$(n+1)^{1/3}-n^{1/3}=frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> frac{1}{3(n+1)^{2/3}}>frac{1}{3(n+1)}$$



              Hence, we have



              $$sum_{n=1}^N left((n+1)^{1/3}-n^{1/3} right)>frac13 sum_{n=2}^{N+1}frac1n$$



              Inasmuch as the harmonic series diverges, the series of interest diverges also.




              METHODOLOGY $2$:



              Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}left(1+frac1{3n}+Oleft(frac1{n^2}right)right)-n^{1/3}=frac1{3n^{2/3}}+Oleft(frac1{n^{5/3}}right)$



              Inasmuch as the series $sum_{n=1}^infty frac{1}{n^p}$ diverges for $ple 1$, the series of interest diverges also.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 18 at 19:41









              Mark ViolaMark Viola

              132k1276174




              132k1276174












              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Jan 19 at 1:21










              • $begingroup$
                You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
                $endgroup$
                – Claude Leibovici
                Jan 19 at 7:03










              • $begingroup$
                Thank you Claude!
                $endgroup$
                – Mark Viola
                Jan 19 at 13:59










              • $begingroup$
                My pleasure, Sir ! Glad to help !!!!
                $endgroup$
                – Claude Leibovici
                Jan 19 at 14:03










              • $begingroup$
                @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                $endgroup$
                – Mark Viola
                Jan 30 at 5:08


















              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Jan 19 at 1:21










              • $begingroup$
                You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
                $endgroup$
                – Claude Leibovici
                Jan 19 at 7:03










              • $begingroup$
                Thank you Claude!
                $endgroup$
                – Mark Viola
                Jan 19 at 13:59










              • $begingroup$
                My pleasure, Sir ! Glad to help !!!!
                $endgroup$
                – Claude Leibovici
                Jan 19 at 14:03










              • $begingroup$
                @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                $endgroup$
                – Mark Viola
                Jan 30 at 5:08
















              $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              Jan 19 at 1:21




              $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              Jan 19 at 1:21












              $begingroup$
              You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
              $endgroup$
              – Claude Leibovici
              Jan 19 at 7:03




              $begingroup$
              You give a very good answer !(as usual !). $to +1$. Cheers, Mark !
              $endgroup$
              – Claude Leibovici
              Jan 19 at 7:03












              $begingroup$
              Thank you Claude!
              $endgroup$
              – Mark Viola
              Jan 19 at 13:59




              $begingroup$
              Thank you Claude!
              $endgroup$
              – Mark Viola
              Jan 19 at 13:59












              $begingroup$
              My pleasure, Sir ! Glad to help !!!!
              $endgroup$
              – Claude Leibovici
              Jan 19 at 14:03




              $begingroup$
              My pleasure, Sir ! Glad to help !!!!
              $endgroup$
              – Claude Leibovici
              Jan 19 at 14:03












              $begingroup$
              @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:08




              $begingroup$
              @Allorja Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:08











              0












              $begingroup$

              Another idea:
              $$
              1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
              $$
              Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $sum_n (n+1)^{1/3}-n^{1/3} sim sum_n n^{-2/3}$, which is well-known to be diverging.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Another idea:
                $$
                1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
                $$
                Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $sum_n (n+1)^{1/3}-n^{1/3} sim sum_n n^{-2/3}$, which is well-known to be diverging.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Another idea:
                  $$
                  1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
                  $$
                  Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $sum_n (n+1)^{1/3}-n^{1/3} sim sum_n n^{-2/3}$, which is well-known to be diverging.






                  share|cite|improve this answer









                  $endgroup$



                  Another idea:
                  $$
                  1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
                  $$
                  Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $sum_n (n+1)^{1/3}-n^{1/3} sim sum_n n^{-2/3}$, which is well-known to be diverging.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 18:48









                  AaronAaron

                  1,902415




                  1,902415






























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