How can I show that the series $sum _{k =1}^{infty} frac{1}{k (1 + (1/k))^k}$ is divergent.












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How can I show that the series $sumlimits_{k =1}^{infty} frac{1}{k left(1 + left(frac{1}{k}right)right)^k}$ is divergent?




I know that I will compare it with the divergent harmonic series, but how is the harmonic series smaller than this?






calculus sequences-and-series convergence







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    yes n is infinity for the other question I do not know @BenW
    $endgroup$
    – hopefully
    Jan 22 at 11:43
















0












$begingroup$



How can I show that the series $sumlimits_{k =1}^{infty} frac{1}{k left(1 + left(frac{1}{k}right)right)^k}$ is divergent?




I know that I will compare it with the divergent harmonic series, but how is the harmonic series smaller than this?






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    yes n is infinity for the other question I do not know @BenW
    $endgroup$
    – hopefully
    Jan 22 at 11:43














0












0








0





$begingroup$



How can I show that the series $sumlimits_{k =1}^{infty} frac{1}{k left(1 + left(frac{1}{k}right)right)^k}$ is divergent?




I know that I will compare it with the divergent harmonic series, but how is the harmonic series smaller than this?






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$





How can I show that the series $sumlimits_{k =1}^{infty} frac{1}{k left(1 + left(frac{1}{k}right)right)^k}$ is divergent?




I know that I will compare it with the divergent harmonic series, but how is the harmonic series smaller than this?






calculus sequences-and-series convergence




calculus sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













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edited Jan 22 at 12:19









Viktor Glombik

1,0121527




1,0121527










asked Jan 22 at 11:35









hopefullyhopefully

250114




250114












  • $begingroup$
    yes n is infinity for the other question I do not know @BenW
    $endgroup$
    – hopefully
    Jan 22 at 11:43


















  • $begingroup$
    yes n is infinity for the other question I do not know @BenW
    $endgroup$
    – hopefully
    Jan 22 at 11:43
















$begingroup$
yes n is infinity for the other question I do not know @BenW
$endgroup$
– hopefully
Jan 22 at 11:43




$begingroup$
yes n is infinity for the other question I do not know @BenW
$endgroup$
– hopefully
Jan 22 at 11:43










2 Answers
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Since $(1+frac{1}{k})^k leq e < 3$ $forall k$ we can infer that your series is bounded below by $sum frac{1}{3n}$ which diverges to $infty$ (the harmonic series).



More generally, one can use the following sometimes called comparison test for series: if $a_n$, $b_n$ are positive sequences such that $lim frac{a_n}{b_n}$ is finite and non-zero, then $sum a_n$ converges iff $sum b_n$ does.






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    $begingroup$

    Hint:



    We have $;Bigl(1+frac 1 kBigr)^k<mathrm e$.






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      2 Answers
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      2 Answers
      2






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      5












      $begingroup$

      Since $(1+frac{1}{k})^k leq e < 3$ $forall k$ we can infer that your series is bounded below by $sum frac{1}{3n}$ which diverges to $infty$ (the harmonic series).



      More generally, one can use the following sometimes called comparison test for series: if $a_n$, $b_n$ are positive sequences such that $lim frac{a_n}{b_n}$ is finite and non-zero, then $sum a_n$ converges iff $sum b_n$ does.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Since $(1+frac{1}{k})^k leq e < 3$ $forall k$ we can infer that your series is bounded below by $sum frac{1}{3n}$ which diverges to $infty$ (the harmonic series).



        More generally, one can use the following sometimes called comparison test for series: if $a_n$, $b_n$ are positive sequences such that $lim frac{a_n}{b_n}$ is finite and non-zero, then $sum a_n$ converges iff $sum b_n$ does.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Since $(1+frac{1}{k})^k leq e < 3$ $forall k$ we can infer that your series is bounded below by $sum frac{1}{3n}$ which diverges to $infty$ (the harmonic series).



          More generally, one can use the following sometimes called comparison test for series: if $a_n$, $b_n$ are positive sequences such that $lim frac{a_n}{b_n}$ is finite and non-zero, then $sum a_n$ converges iff $sum b_n$ does.






          share|cite|improve this answer









          $endgroup$



          Since $(1+frac{1}{k})^k leq e < 3$ $forall k$ we can infer that your series is bounded below by $sum frac{1}{3n}$ which diverges to $infty$ (the harmonic series).



          More generally, one can use the following sometimes called comparison test for series: if $a_n$, $b_n$ are positive sequences such that $lim frac{a_n}{b_n}$ is finite and non-zero, then $sum a_n$ converges iff $sum b_n$ does.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 11:44









          Sorin TircSorin Tirc

          1,855213




          1,855213























              2












              $begingroup$

              Hint:



              We have $;Bigl(1+frac 1 kBigr)^k<mathrm e$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint:



                We have $;Bigl(1+frac 1 kBigr)^k<mathrm e$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint:



                  We have $;Bigl(1+frac 1 kBigr)^k<mathrm e$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  We have $;Bigl(1+frac 1 kBigr)^k<mathrm e$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 22 at 11:43









                  BernardBernard

                  122k741116




                  122k741116






























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