Study the convergence of a series











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I am studying the convergence of the series $$sumlimits_{n<=0} left(sqrt[n]{n} - 1right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.










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  • 2




    Do you really mean $nle 0$?
    – Lord Shark the Unknown
    Nov 17 at 18:26










  • Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
    – user25959
    Nov 27 at 3:12















up vote
1
down vote

favorite












I am studying the convergence of the series $$sumlimits_{n<=0} left(sqrt[n]{n} - 1right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.










share|cite|improve this question




















  • 2




    Do you really mean $nle 0$?
    – Lord Shark the Unknown
    Nov 17 at 18:26










  • Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
    – user25959
    Nov 27 at 3:12













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am studying the convergence of the series $$sumlimits_{n<=0} left(sqrt[n]{n} - 1right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.










share|cite|improve this question















I am studying the convergence of the series $$sumlimits_{n<=0} left(sqrt[n]{n} - 1right)^n$$
I just know that the limit tend to $0$, but I don't know how to prove the convergence.







sequences-and-series






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edited Nov 17 at 21:15









rtybase

10.2k21433




10.2k21433










asked Nov 17 at 18:20









Daniel Guerrero

83




83








  • 2




    Do you really mean $nle 0$?
    – Lord Shark the Unknown
    Nov 17 at 18:26










  • Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
    – user25959
    Nov 27 at 3:12














  • 2




    Do you really mean $nle 0$?
    – Lord Shark the Unknown
    Nov 17 at 18:26










  • Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
    – user25959
    Nov 27 at 3:12








2




2




Do you really mean $nle 0$?
– Lord Shark the Unknown
Nov 17 at 18:26




Do you really mean $nle 0$?
– Lord Shark the Unknown
Nov 17 at 18:26












Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
– user25959
Nov 27 at 3:12




Have you already tried the root test (en.wikipedia.org/wiki/Root_test) ?
– user25959
Nov 27 at 3:12










2 Answers
2






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down vote



accepted










I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have



$$n-1=
left(n^{frac{1}{n}}right)^{n}-1=\
left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$

using AM-GM
$$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$



Or
$$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$





From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means





  • $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
    limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
    limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$


  • or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
    sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
    sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$

    we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
    i.e. converging.






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    $sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
    $$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
    and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.






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      2 Answers
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      2 Answers
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      accepted










      I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have



      $$n-1=
      left(n^{frac{1}{n}}right)^{n}-1=\
      left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$

      using AM-GM
      $$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
      left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
      \
      left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
      left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$



      Or
      $$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$





      From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means





      • $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
        limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
        limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$


      • or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
        sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
        sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$

        we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
        i.e. converging.






      share|cite|improve this answer

























        up vote
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        down vote



        accepted










        I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have



        $$n-1=
        left(n^{frac{1}{n}}right)^{n}-1=\
        left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$

        using AM-GM
        $$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
        left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
        \
        left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
        left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$



        Or
        $$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$





        From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means





        • $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
          limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
          limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$


        • or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
          sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
          sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$

          we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
          i.e. converging.






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          up vote
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          up vote
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          accepted






          I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have



          $$n-1=
          left(n^{frac{1}{n}}right)^{n}-1=\
          left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$

          using AM-GM
          $$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
          \
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$



          Or
          $$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$





          From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means





          • $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
            limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
            limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$


          • or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
            sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
            sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$

            we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
            i.e. converging.






          share|cite|improve this answer












          I assume you meant $n geq 1$. Using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$ we have



          $$n-1=
          left(n^{frac{1}{n}}right)^{n}-1=\
          left(n^{frac{1}{n}}-1right)left(n^{frac{1}{n}(n-1)}+n^{frac{1}{n}(n-2)}+...+n^{frac{1}{n}2}+n^{frac{1}{n}}+1right)geq ...$$

          using AM-GM
          $$... geqleft(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1)}cdot n^{frac{1}{n}(n-2)}cdot ...cdot n^{frac{1}{n}2}cdot n^{frac{1}{n}}}+1right)=\
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}(n-1+n-2+...+1)}}+1right)=
          \
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt[n-1]{n^{frac{1}{n}frac{(n-1)n}{2}}}+1right)=\
          left(n^{frac{1}{n}}-1right)left((n-1)sqrt{n}+1right)$$



          Or
          $$0< n^{frac{1}{n}}-1leq frac{n-1}{(n-1)sqrt{n}+1}<frac{1}{sqrt{n}} tag{1}$$





          From $(1)$ we see that $limlimits_{nrightarrowinfty} left(n^{frac{1}{n}}-1right)=0$, which means





          • $0< left(n^{frac{1}{n}}-1right)^n<frac{1}{n^{frac{n}{2}}}$, thus $0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n leq sumlimits_{ngeq1} frac{1}{n^{frac{n}{2}}}$. The latter is converging by ratio test $$limlimits_{nrightarrowinfty}frac{frac{1}{(n+1)^{frac{n+1}{2}}}}{frac{1}{n^{frac{n}{2}}}}=
            limlimits_{nrightarrowinfty}frac{n^{frac{n}{2}}}{(n+1)^{frac{n+1}{2}}}=
            limlimits_{nrightarrowinfty}frac{1}{left(1+frac{1}{n}right)^{frac{n}{2}}cdot sqrt{n+1}}=frac{1}{sqrt{e}cdot limlimits_{nrightarrowinfty}sqrt{n+1}}=0$$


          • or simpler version from the definition of the limit, for $forall n> N(varepsilon) Rightarrow 0<left(n^{frac{1}{n}}-1right)<varepsilon <1$, thus $0<left(n^{frac{1}{n}}-1right)^n<varepsilon^n$ and $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n <
            sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + sumlimits_{n=N(varepsilon)+1} varepsilon^n=
            sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n + frac{e^{N(varepsilon)+1}}{1-varepsilon}$$

            we just need to find the very first $N(varepsilon)$ for $0<varepsilon<1$ and $sumlimits_{n=1}^{N(varepsilon)} left(n^{frac{1}{n}}-1right)^n$ is a finite number as well as $frac{e^{N(varepsilon)+1}}{1-varepsilon}$. As a result $$0<sumlimits_{ngeq1} left(n^{frac{1}{n}}-1right)^n<infty$$
            i.e. converging.







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          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 20:58









          rtybase

          10.2k21433




          10.2k21433






















              up vote
              0
              down vote













              $sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
              $$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
              and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.






              share|cite|improve this answer

























                up vote
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                down vote













                $sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
                $$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
                and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.






                share|cite|improve this answer























                  up vote
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                  down vote










                  up vote
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                  down vote









                  $sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
                  $$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
                  and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.






                  share|cite|improve this answer












                  $sum_{ngeq 1}left(sqrt[n]{n}-1right)^n $ is convergent, since for any $ngeq 2$ we have
                  $$ sqrt[n]{n} = sqrt[n]{frac{n}{n-1}cdotfrac{n-1}{n-2}cdotldotscdotfrac{2}{1}cdotfrac{1}{1}}leq text{AM}left(1+tfrac{1}{n-1},ldots,1+tfrac{1}{2},2,1right)=1+frac{H_{n-1}}{n} $$
                  and by the Cauchy-Schwarz inequality $H_nleq frac{pi}{sqrt{6}}sqrt{n} $. The series $sum_{ngeq 1}frac{C}{n^{n/2}}$ is trivially convergent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 17 at 20:33









                  Jack D'Aurizio

                  283k33275653




                  283k33275653






























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