Find the Limit $lim_{n to infty}sum_{k=1}^{infty}frac{k^{n}}{1+k^{n+2}}$












1












$begingroup$


Find $lim_{n to infty}sum_{k=1}^{infty}frac{k^{n}}{1+k^{n+2}}$



My ideas: let $ n in mathbb N$ be constant, looking at $frac{k^{n}}{1+k^{n+2}}$, we know



$$frac{k^{n}}{1+k^{n+2}}leqfrac{k^{n}}{k^{n+2}}=frac{1}{k^{2}}$$



but this does not help me because $sum_{k=1}^{infty}frac{1}{k^{2}}=pi^{2}/6$



any ideas?






real-analysis sequences-and-series







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  • 2




    $begingroup$
    This question can be solved in the same way as I did here 6 hours ago.
    $endgroup$
    – Masacroso
    Dec 3 '18 at 17:39
















1












$begingroup$


Find $lim_{n to infty}sum_{k=1}^{infty}frac{k^{n}}{1+k^{n+2}}$



My ideas: let $ n in mathbb N$ be constant, looking at $frac{k^{n}}{1+k^{n+2}}$, we know



$$frac{k^{n}}{1+k^{n+2}}leqfrac{k^{n}}{k^{n+2}}=frac{1}{k^{2}}$$



but this does not help me because $sum_{k=1}^{infty}frac{1}{k^{2}}=pi^{2}/6$



any ideas?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This question can be solved in the same way as I did here 6 hours ago.
    $endgroup$
    – Masacroso
    Dec 3 '18 at 17:39














1












1








1





$begingroup$


Find $lim_{n to infty}sum_{k=1}^{infty}frac{k^{n}}{1+k^{n+2}}$



My ideas: let $ n in mathbb N$ be constant, looking at $frac{k^{n}}{1+k^{n+2}}$, we know



$$frac{k^{n}}{1+k^{n+2}}leqfrac{k^{n}}{k^{n+2}}=frac{1}{k^{2}}$$



but this does not help me because $sum_{k=1}^{infty}frac{1}{k^{2}}=pi^{2}/6$



any ideas?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




Find $lim_{n to infty}sum_{k=1}^{infty}frac{k^{n}}{1+k^{n+2}}$



My ideas: let $ n in mathbb N$ be constant, looking at $frac{k^{n}}{1+k^{n+2}}$, we know



$$frac{k^{n}}{1+k^{n+2}}leqfrac{k^{n}}{k^{n+2}}=frac{1}{k^{2}}$$



but this does not help me because $sum_{k=1}^{infty}frac{1}{k^{2}}=pi^{2}/6$



any ideas?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













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edited Dec 3 '18 at 17:39









gt6989b

34.3k22455




34.3k22455










asked Dec 3 '18 at 17:24









SABOYSABOY

649311




649311








  • 2




    $begingroup$
    This question can be solved in the same way as I did here 6 hours ago.
    $endgroup$
    – Masacroso
    Dec 3 '18 at 17:39














  • 2




    $begingroup$
    This question can be solved in the same way as I did here 6 hours ago.
    $endgroup$
    – Masacroso
    Dec 3 '18 at 17:39








2




2




$begingroup$
This question can be solved in the same way as I did here 6 hours ago.
$endgroup$
– Masacroso
Dec 3 '18 at 17:39




$begingroup$
This question can be solved in the same way as I did here 6 hours ago.
$endgroup$
– Masacroso
Dec 3 '18 at 17:39










4 Answers
4






active

oldest

votes


















2












$begingroup$

Note that



$${1over k^2}-{k^nover1+k^{n+2}}={1over k^2(1+k^{n+2})}$$



and, for $kge2$,



$${1over k^2(1+k^{n+2})}le{1over2^nk^2}$$



so that



$$sum_{k=1}^infty{1over k^2(1+k^{n+2})}={1over2}+sum_{k=2}^infty{1over k^2(1+k^{n+2})}le{1over2}+{1over2^n}sum_{k=2}^infty{1over k^2}to{1over2}+0={1over2}$$



as $ntoinfty$. It follows that



$$sum_{k=1}^infty{k^nover1+k^{n+2}}=sum_{k=1}^infty{1over k^2}-sum_{k=1}^infty{1over k^2(1+k^{n+2})}tosum_{k=1}^infty{1over k^2}-{1over2}={pi^2over6}-{1over2}$$



The key step, really, is to realize that the inequality $2^nle1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $kge2$, so the $k=1$ term needs to be split off from the sum.






share|cite|improve this answer









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    1












    $begingroup$

    HINT



    Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
    $$
    frac{k^5}{1+k^7} approx frac{1}{k^2}...
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
      $endgroup$
      – SABOY
      Dec 3 '18 at 17:37










    • $begingroup$
      @SABOY the larger the $n$, the better the approximation in the hint...
      $endgroup$
      – gt6989b
      Dec 3 '18 at 17:38



















    0












    $begingroup$

    I would do in this way
    $$
    eqalign{
    & sumlimits_{1, le ,k} {{{k^{,n} } over {1 + k^{,n + 2} }}}
    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {{{k^{,n + 2} } over {1 + k^{,n + 2} }}} right)}
    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {1 - {1 over {1 + k^{,n + 2} }}} right)} = cr
    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {1 + k^{,n + 2} }}} right)}
    ;mathop approx limits^{n to infty } sumlimits_{1, le ,k} {{1 over {k^{,2} }}}
    - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {k^{,n + 2} }}} right)} = cr
    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - left( {sumlimits_{1, le ,k} {{1 over {k^{,n + 4} }}} - 1} right)
    = zeta (2) + {1 over 2} - zeta (n + 4) to quad zeta (2) - {1 over 2} cr}
    $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The series equals



      $$frac{1}{2}+ sum_{k=2}^{infty}frac{1}{k^{-n} + k^2}.$$



      For each $kge 2,$ the terms in the last series increase to $1/k^2$ as $nto infty.$ By the monotone convergence theorem, the desired limit is



      $$frac{1}{2} +sum_{k=2}^{infty}frac{1}{k^2} = frac{1}{2}+left (frac{pi^2}{6} - 1right) = frac{pi^2}{6} - frac{1}{2}.$$






      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












        $begingroup$

        Note that



        $${1over k^2}-{k^nover1+k^{n+2}}={1over k^2(1+k^{n+2})}$$



        and, for $kge2$,



        $${1over k^2(1+k^{n+2})}le{1over2^nk^2}$$



        so that



        $$sum_{k=1}^infty{1over k^2(1+k^{n+2})}={1over2}+sum_{k=2}^infty{1over k^2(1+k^{n+2})}le{1over2}+{1over2^n}sum_{k=2}^infty{1over k^2}to{1over2}+0={1over2}$$



        as $ntoinfty$. It follows that



        $$sum_{k=1}^infty{k^nover1+k^{n+2}}=sum_{k=1}^infty{1over k^2}-sum_{k=1}^infty{1over k^2(1+k^{n+2})}tosum_{k=1}^infty{1over k^2}-{1over2}={pi^2over6}-{1over2}$$



        The key step, really, is to realize that the inequality $2^nle1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $kge2$, so the $k=1$ term needs to be split off from the sum.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Note that



          $${1over k^2}-{k^nover1+k^{n+2}}={1over k^2(1+k^{n+2})}$$



          and, for $kge2$,



          $${1over k^2(1+k^{n+2})}le{1over2^nk^2}$$



          so that



          $$sum_{k=1}^infty{1over k^2(1+k^{n+2})}={1over2}+sum_{k=2}^infty{1over k^2(1+k^{n+2})}le{1over2}+{1over2^n}sum_{k=2}^infty{1over k^2}to{1over2}+0={1over2}$$



          as $ntoinfty$. It follows that



          $$sum_{k=1}^infty{k^nover1+k^{n+2}}=sum_{k=1}^infty{1over k^2}-sum_{k=1}^infty{1over k^2(1+k^{n+2})}tosum_{k=1}^infty{1over k^2}-{1over2}={pi^2over6}-{1over2}$$



          The key step, really, is to realize that the inequality $2^nle1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $kge2$, so the $k=1$ term needs to be split off from the sum.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Note that



            $${1over k^2}-{k^nover1+k^{n+2}}={1over k^2(1+k^{n+2})}$$



            and, for $kge2$,



            $${1over k^2(1+k^{n+2})}le{1over2^nk^2}$$



            so that



            $$sum_{k=1}^infty{1over k^2(1+k^{n+2})}={1over2}+sum_{k=2}^infty{1over k^2(1+k^{n+2})}le{1over2}+{1over2^n}sum_{k=2}^infty{1over k^2}to{1over2}+0={1over2}$$



            as $ntoinfty$. It follows that



            $$sum_{k=1}^infty{k^nover1+k^{n+2}}=sum_{k=1}^infty{1over k^2}-sum_{k=1}^infty{1over k^2(1+k^{n+2})}tosum_{k=1}^infty{1over k^2}-{1over2}={pi^2over6}-{1over2}$$



            The key step, really, is to realize that the inequality $2^nle1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $kge2$, so the $k=1$ term needs to be split off from the sum.






            share|cite|improve this answer









            $endgroup$



            Note that



            $${1over k^2}-{k^nover1+k^{n+2}}={1over k^2(1+k^{n+2})}$$



            and, for $kge2$,



            $${1over k^2(1+k^{n+2})}le{1over2^nk^2}$$



            so that



            $$sum_{k=1}^infty{1over k^2(1+k^{n+2})}={1over2}+sum_{k=2}^infty{1over k^2(1+k^{n+2})}le{1over2}+{1over2^n}sum_{k=2}^infty{1over k^2}to{1over2}+0={1over2}$$



            as $ntoinfty$. It follows that



            $$sum_{k=1}^infty{k^nover1+k^{n+2}}=sum_{k=1}^infty{1over k^2}-sum_{k=1}^infty{1over k^2(1+k^{n+2})}tosum_{k=1}^infty{1over k^2}-{1over2}={pi^2over6}-{1over2}$$



            The key step, really, is to realize that the inequality $2^nle1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $kge2$, so the $k=1$ term needs to be split off from the sum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 3 '18 at 17:46









            Barry CipraBarry Cipra

            59.7k653126




            59.7k653126























                1












                $begingroup$

                HINT



                Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
                $$
                frac{k^5}{1+k^7} approx frac{1}{k^2}...
                $$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                  $endgroup$
                  – SABOY
                  Dec 3 '18 at 17:37










                • $begingroup$
                  @SABOY the larger the $n$, the better the approximation in the hint...
                  $endgroup$
                  – gt6989b
                  Dec 3 '18 at 17:38
















                1












                $begingroup$

                HINT



                Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
                $$
                frac{k^5}{1+k^7} approx frac{1}{k^2}...
                $$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                  $endgroup$
                  – SABOY
                  Dec 3 '18 at 17:37










                • $begingroup$
                  @SABOY the larger the $n$, the better the approximation in the hint...
                  $endgroup$
                  – gt6989b
                  Dec 3 '18 at 17:38














                1












                1








                1





                $begingroup$

                HINT



                Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
                $$
                frac{k^5}{1+k^7} approx frac{1}{k^2}...
                $$






                share|cite|improve this answer









                $endgroup$



                HINT



                Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
                $$
                frac{k^5}{1+k^7} approx frac{1}{k^2}...
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 17:29









                gt6989bgt6989b

                34.3k22455




                34.3k22455












                • $begingroup$
                  Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                  $endgroup$
                  – SABOY
                  Dec 3 '18 at 17:37










                • $begingroup$
                  @SABOY the larger the $n$, the better the approximation in the hint...
                  $endgroup$
                  – gt6989b
                  Dec 3 '18 at 17:38


















                • $begingroup$
                  Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                  $endgroup$
                  – SABOY
                  Dec 3 '18 at 17:37










                • $begingroup$
                  @SABOY the larger the $n$, the better the approximation in the hint...
                  $endgroup$
                  – gt6989b
                  Dec 3 '18 at 17:38
















                $begingroup$
                Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                $endgroup$
                – SABOY
                Dec 3 '18 at 17:37




                $begingroup$
                Ok then I've got $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ but I cannot say anything about the limit of $lim_{n to infty}sum_{k=1}^{infty}frac{k^n}{1+k^{n+2}}$
                $endgroup$
                – SABOY
                Dec 3 '18 at 17:37












                $begingroup$
                @SABOY the larger the $n$, the better the approximation in the hint...
                $endgroup$
                – gt6989b
                Dec 3 '18 at 17:38




                $begingroup$
                @SABOY the larger the $n$, the better the approximation in the hint...
                $endgroup$
                – gt6989b
                Dec 3 '18 at 17:38











                0












                $begingroup$

                I would do in this way
                $$
                eqalign{
                & sumlimits_{1, le ,k} {{{k^{,n} } over {1 + k^{,n + 2} }}}
                = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {{{k^{,n + 2} } over {1 + k^{,n + 2} }}} right)}
                = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {1 - {1 over {1 + k^{,n + 2} }}} right)} = cr
                & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {1 + k^{,n + 2} }}} right)}
                ;mathop approx limits^{n to infty } sumlimits_{1, le ,k} {{1 over {k^{,2} }}}
                - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {k^{,n + 2} }}} right)} = cr
                & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - left( {sumlimits_{1, le ,k} {{1 over {k^{,n + 4} }}} - 1} right)
                = zeta (2) + {1 over 2} - zeta (n + 4) to quad zeta (2) - {1 over 2} cr}
                $$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  I would do in this way
                  $$
                  eqalign{
                  & sumlimits_{1, le ,k} {{{k^{,n} } over {1 + k^{,n + 2} }}}
                  = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {{{k^{,n + 2} } over {1 + k^{,n + 2} }}} right)}
                  = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {1 - {1 over {1 + k^{,n + 2} }}} right)} = cr
                  & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {1 + k^{,n + 2} }}} right)}
                  ;mathop approx limits^{n to infty } sumlimits_{1, le ,k} {{1 over {k^{,2} }}}
                  - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {k^{,n + 2} }}} right)} = cr
                  & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - left( {sumlimits_{1, le ,k} {{1 over {k^{,n + 4} }}} - 1} right)
                  = zeta (2) + {1 over 2} - zeta (n + 4) to quad zeta (2) - {1 over 2} cr}
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    I would do in this way
                    $$
                    eqalign{
                    & sumlimits_{1, le ,k} {{{k^{,n} } over {1 + k^{,n + 2} }}}
                    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {{{k^{,n + 2} } over {1 + k^{,n + 2} }}} right)}
                    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {1 - {1 over {1 + k^{,n + 2} }}} right)} = cr
                    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {1 + k^{,n + 2} }}} right)}
                    ;mathop approx limits^{n to infty } sumlimits_{1, le ,k} {{1 over {k^{,2} }}}
                    - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {k^{,n + 2} }}} right)} = cr
                    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - left( {sumlimits_{1, le ,k} {{1 over {k^{,n + 4} }}} - 1} right)
                    = zeta (2) + {1 over 2} - zeta (n + 4) to quad zeta (2) - {1 over 2} cr}
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    I would do in this way
                    $$
                    eqalign{
                    & sumlimits_{1, le ,k} {{{k^{,n} } over {1 + k^{,n + 2} }}}
                    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {{{k^{,n + 2} } over {1 + k^{,n + 2} }}} right)}
                    = sumlimits_{1, le ,k} {{1 over {k^{,2} }}left( {1 - {1 over {1 + k^{,n + 2} }}} right)} = cr
                    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {1 + k^{,n + 2} }}} right)}
                    ;mathop approx limits^{n to infty } sumlimits_{1, le ,k} {{1 over {k^{,2} }}}
                    - {1 over 2} - sumlimits_{2, le ,k} {{1 over {k^{,2} }}left( {{1 over {k^{,n + 2} }}} right)} = cr
                    & = sumlimits_{1, le ,k} {{1 over {k^{,2} }}} - {1 over 2} - left( {sumlimits_{1, le ,k} {{1 over {k^{,n + 4} }}} - 1} right)
                    = zeta (2) + {1 over 2} - zeta (n + 4) to quad zeta (2) - {1 over 2} cr}
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 19:06









                    G CabG Cab

                    19.6k31239




                    19.6k31239























                        0












                        $begingroup$

                        The series equals



                        $$frac{1}{2}+ sum_{k=2}^{infty}frac{1}{k^{-n} + k^2}.$$



                        For each $kge 2,$ the terms in the last series increase to $1/k^2$ as $nto infty.$ By the monotone convergence theorem, the desired limit is



                        $$frac{1}{2} +sum_{k=2}^{infty}frac{1}{k^2} = frac{1}{2}+left (frac{pi^2}{6} - 1right) = frac{pi^2}{6} - frac{1}{2}.$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The series equals



                          $$frac{1}{2}+ sum_{k=2}^{infty}frac{1}{k^{-n} + k^2}.$$



                          For each $kge 2,$ the terms in the last series increase to $1/k^2$ as $nto infty.$ By the monotone convergence theorem, the desired limit is



                          $$frac{1}{2} +sum_{k=2}^{infty}frac{1}{k^2} = frac{1}{2}+left (frac{pi^2}{6} - 1right) = frac{pi^2}{6} - frac{1}{2}.$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The series equals



                            $$frac{1}{2}+ sum_{k=2}^{infty}frac{1}{k^{-n} + k^2}.$$



                            For each $kge 2,$ the terms in the last series increase to $1/k^2$ as $nto infty.$ By the monotone convergence theorem, the desired limit is



                            $$frac{1}{2} +sum_{k=2}^{infty}frac{1}{k^2} = frac{1}{2}+left (frac{pi^2}{6} - 1right) = frac{pi^2}{6} - frac{1}{2}.$$






                            share|cite|improve this answer









                            $endgroup$



                            The series equals



                            $$frac{1}{2}+ sum_{k=2}^{infty}frac{1}{k^{-n} + k^2}.$$



                            For each $kge 2,$ the terms in the last series increase to $1/k^2$ as $nto infty.$ By the monotone convergence theorem, the desired limit is



                            $$frac{1}{2} +sum_{k=2}^{infty}frac{1}{k^2} = frac{1}{2}+left (frac{pi^2}{6} - 1right) = frac{pi^2}{6} - frac{1}{2}.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 3 '18 at 22:16









                            zhw.zhw.

                            73.5k43175




                            73.5k43175






























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