When the series $sum_{n=2}^{+infty }(sqrt[n]{n}-1)^a$ is convergent depending on $ainBbb R$?












1












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When the series $sum_{n=2}^{+infty }(sqrt[n]{n}-1)^a$ is convergent depending on $ainBbb R$?



For $a=0 $ the series is divergent.



Else:



$sqrt[n]{n}-1=frac{e^{1/nln n}-1}{1/n ln n}cdot frac{1}{n}ln nsimfrac{ln n}{n}$



So, $sqrt[n]{n}-1$ is asymptotically similar to $frac{ln n}{n}$, and $(sqrt[n]{n}-1)^a$ will converge when $(frac{ln n}{n})^a$ converges. But I don't know how to finish it. Could you help me?






real-analysis sequences-and-series convergence







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    1












    $begingroup$


    When the series $sum_{n=2}^{+infty }(sqrt[n]{n}-1)^a$ is convergent depending on $ainBbb R$?



    For $a=0 $ the series is divergent.



    Else:



    $sqrt[n]{n}-1=frac{e^{1/nln n}-1}{1/n ln n}cdot frac{1}{n}ln nsimfrac{ln n}{n}$



    So, $sqrt[n]{n}-1$ is asymptotically similar to $frac{ln n}{n}$, and $(sqrt[n]{n}-1)^a$ will converge when $(frac{ln n}{n})^a$ converges. But I don't know how to finish it. Could you help me?






    real-analysis sequences-and-series convergence







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      When the series $sum_{n=2}^{+infty }(sqrt[n]{n}-1)^a$ is convergent depending on $ainBbb R$?



      For $a=0 $ the series is divergent.



      Else:



      $sqrt[n]{n}-1=frac{e^{1/nln n}-1}{1/n ln n}cdot frac{1}{n}ln nsimfrac{ln n}{n}$



      So, $sqrt[n]{n}-1$ is asymptotically similar to $frac{ln n}{n}$, and $(sqrt[n]{n}-1)^a$ will converge when $(frac{ln n}{n})^a$ converges. But I don't know how to finish it. Could you help me?






      real-analysis sequences-and-series convergence







      share|cite|improve this question









      $endgroup$




      When the series $sum_{n=2}^{+infty }(sqrt[n]{n}-1)^a$ is convergent depending on $ainBbb R$?



      For $a=0 $ the series is divergent.



      Else:



      $sqrt[n]{n}-1=frac{e^{1/nln n}-1}{1/n ln n}cdot frac{1}{n}ln nsimfrac{ln n}{n}$



      So, $sqrt[n]{n}-1$ is asymptotically similar to $frac{ln n}{n}$, and $(sqrt[n]{n}-1)^a$ will converge when $(frac{ln n}{n})^a$ converges. But I don't know how to finish it. Could you help me?






      real-analysis sequences-and-series convergence




      real-analysis sequences-and-series convergence






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      asked Dec 14 '18 at 2:27









      matematicccmatematiccc

      1275




      1275






















          2 Answers
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          1












          $begingroup$

          For $a>1$, compare to $frac1{n^{1+(a-1)/2}}$:
          $$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n^{1+(a-1)/2}}\
          =lim_{ntoinfty}frac{ln^an}{n^{a/2-1/2}}$$
          Since $a>1$, $frac a2-frac12>0$, hence the limit equals $0$. The series converges.

          For $ale 1$, compare to $frac1n$:$$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n}\
          =lim_{ntoinfty}frac{ln^an}{n^{a-1}}=infty$$
          Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
            $endgroup$
            – Mark Viola
            Dec 14 '18 at 4:01










          • $begingroup$
            @MarkViola Could you please explain it more explicitly
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 4:04



















          1












          $begingroup$


          In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$". To address the OP's request, we now proceed.





          Using $e^x<frac1{1-x}$ for $x<1$ and $minleft(1-frac1nlog(n)right)=1-e^{-1} $ we have



          $$begin{align}
          sqrt[n]{n}-1 &=e^{frac1nlog(n)}-1\\
          &le frac{frac1nlog(n)}{1-frac1nlog(n)}\\
          &le frac{e}{e-1} frac{log(n)}{n}end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 5:54











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          2 Answers
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          2 Answers
          2






          active

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          active

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          1












          $begingroup$

          For $a>1$, compare to $frac1{n^{1+(a-1)/2}}$:
          $$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n^{1+(a-1)/2}}\
          =lim_{ntoinfty}frac{ln^an}{n^{a/2-1/2}}$$
          Since $a>1$, $frac a2-frac12>0$, hence the limit equals $0$. The series converges.

          For $ale 1$, compare to $frac1n$:$$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n}\
          =lim_{ntoinfty}frac{ln^an}{n^{a-1}}=infty$$
          Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
            $endgroup$
            – Mark Viola
            Dec 14 '18 at 4:01










          • $begingroup$
            @MarkViola Could you please explain it more explicitly
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 4:04
















          1












          $begingroup$

          For $a>1$, compare to $frac1{n^{1+(a-1)/2}}$:
          $$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n^{1+(a-1)/2}}\
          =lim_{ntoinfty}frac{ln^an}{n^{a/2-1/2}}$$
          Since $a>1$, $frac a2-frac12>0$, hence the limit equals $0$. The series converges.

          For $ale 1$, compare to $frac1n$:$$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n}\
          =lim_{ntoinfty}frac{ln^an}{n^{a-1}}=infty$$
          Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
            $endgroup$
            – Mark Viola
            Dec 14 '18 at 4:01










          • $begingroup$
            @MarkViola Could you please explain it more explicitly
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 4:04














          1












          1








          1





          $begingroup$

          For $a>1$, compare to $frac1{n^{1+(a-1)/2}}$:
          $$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n^{1+(a-1)/2}}\
          =lim_{ntoinfty}frac{ln^an}{n^{a/2-1/2}}$$
          Since $a>1$, $frac a2-frac12>0$, hence the limit equals $0$. The series converges.

          For $ale 1$, compare to $frac1n$:$$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n}\
          =lim_{ntoinfty}frac{ln^an}{n^{a-1}}=infty$$
          Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.






          share|cite|improve this answer









          $endgroup$



          For $a>1$, compare to $frac1{n^{1+(a-1)/2}}$:
          $$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n^{1+(a-1)/2}}\
          =lim_{ntoinfty}frac{ln^an}{n^{a/2-1/2}}$$
          Since $a>1$, $frac a2-frac12>0$, hence the limit equals $0$. The series converges.

          For $ale 1$, compare to $frac1n$:$$lim_{ntoinfty}left(frac{ln n}{n}right)^aBig/frac1{n}\
          =lim_{ntoinfty}frac{ln^an}{n^{a-1}}=infty$$
          Even if $a<0$, the limit still diverges to infinity. Hence the series diverges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 3:07









          Kemono ChenKemono Chen

          3,0721743




          3,0721743












          • $begingroup$
            $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
            $endgroup$
            – Mark Viola
            Dec 14 '18 at 4:01










          • $begingroup$
            @MarkViola Could you please explain it more explicitly
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 4:04


















          • $begingroup$
            $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
            $endgroup$
            – Mark Viola
            Dec 14 '18 at 4:01










          • $begingroup$
            @MarkViola Could you please explain it more explicitly
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 4:04
















          $begingroup$
          $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
          $endgroup$
          – Mark Viola
          Dec 14 '18 at 4:01




          $begingroup$
          $$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$$
          $endgroup$
          – Mark Viola
          Dec 14 '18 at 4:01












          $begingroup$
          @MarkViola Could you please explain it more explicitly
          $endgroup$
          – Kemono Chen
          Dec 14 '18 at 4:04




          $begingroup$
          @MarkViola Could you please explain it more explicitly
          $endgroup$
          – Kemono Chen
          Dec 14 '18 at 4:04











          1












          $begingroup$


          In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$". To address the OP's request, we now proceed.





          Using $e^x<frac1{1-x}$ for $x<1$ and $minleft(1-frac1nlog(n)right)=1-e^{-1} $ we have



          $$begin{align}
          sqrt[n]{n}-1 &=e^{frac1nlog(n)}-1\\
          &le frac{frac1nlog(n)}{1-frac1nlog(n)}\\
          &le frac{e}{e-1} frac{log(n)}{n}end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 5:54
















          1












          $begingroup$


          In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$". To address the OP's request, we now proceed.





          Using $e^x<frac1{1-x}$ for $x<1$ and $minleft(1-frac1nlog(n)right)=1-e^{-1} $ we have



          $$begin{align}
          sqrt[n]{n}-1 &=e^{frac1nlog(n)}-1\\
          &le frac{frac1nlog(n)}{1-frac1nlog(n)}\\
          &le frac{e}{e-1} frac{log(n)}{n}end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 5:54














          1












          1








          1





          $begingroup$


          In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$". To address the OP's request, we now proceed.





          Using $e^x<frac1{1-x}$ for $x<1$ and $minleft(1-frac1nlog(n)right)=1-e^{-1} $ we have



          $$begin{align}
          sqrt[n]{n}-1 &=e^{frac1nlog(n)}-1\\
          &le frac{frac1nlog(n)}{1-frac1nlog(n)}\\
          &le frac{e}{e-1} frac{log(n)}{n}end{align}$$






          share|cite|improve this answer











          $endgroup$




          In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "$sqrt[n]{n}-1le frac{e}{e-1} frac{log(n)}{n}$". To address the OP's request, we now proceed.





          Using $e^x<frac1{1-x}$ for $x<1$ and $minleft(1-frac1nlog(n)right)=1-e^{-1} $ we have



          $$begin{align}
          sqrt[n]{n}-1 &=e^{frac1nlog(n)}-1\\
          &le frac{frac1nlog(n)}{1-frac1nlog(n)}\\
          &le frac{e}{e-1} frac{log(n)}{n}end{align}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 4:16

























          answered Dec 14 '18 at 4:10









          Mark ViolaMark Viola

          132k1275173




          132k1275173












          • $begingroup$
            I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 5:54


















          • $begingroup$
            I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
            $endgroup$
            – Kemono Chen
            Dec 14 '18 at 5:54
















          $begingroup$
          I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
          $endgroup$
          – Kemono Chen
          Dec 14 '18 at 5:54




          $begingroup$
          I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly.
          $endgroup$
          – Kemono Chen
          Dec 14 '18 at 5:54


















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