Evaluate $ limlimits_{n to infty}sumlimits_{k=2}^{n} frac{1}{sqrt[k]{n^k+n+1}+1} $












2












$begingroup$


$$ lim_{n to infty}sum_{k=2}^{n} frac{1}{sqrt[k]{n^k+n+1}+1} $$



I expect the squeeze theorem helps us solving this but I can't find the inequality.



The result should be $1$.






calculus limits







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$endgroup$

















    2












    $begingroup$


    $$ lim_{n to infty}sum_{k=2}^{n} frac{1}{sqrt[k]{n^k+n+1}+1} $$



    I expect the squeeze theorem helps us solving this but I can't find the inequality.



    The result should be $1$.






    calculus limits







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      $$ lim_{n to infty}sum_{k=2}^{n} frac{1}{sqrt[k]{n^k+n+1}+1} $$



      I expect the squeeze theorem helps us solving this but I can't find the inequality.



      The result should be $1$.






      calculus limits







      share|cite|improve this question









      $endgroup$




      $$ lim_{n to infty}sum_{k=2}^{n} frac{1}{sqrt[k]{n^k+n+1}+1} $$



      I expect the squeeze theorem helps us solving this but I can't find the inequality.



      The result should be $1$.






      calculus limits




      calculus limits






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 30 '18 at 17:35









      SADBOYSSADBOYS

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          7












          $begingroup$

          Try $(n+1)^kgt n^k+n+1gt n^k$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Consider
            $s(n)
            =sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}
            $
            where
            $f(n) ge 0$ and
            $f(n)/n^{c} to 0$
            for some $c > 0$.



            $begin{array}\
            s(n)
            &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
            &ltsum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k}}\
            &=sum_{k=2}^{n} dfrac{1}{n}\
            &to ln(n)-1+gamma\
            end{array}
            $



            Similarly,
            since
            $(1+x)^k ge 1+kx,
            (1+x/k)^k ge 1+x$
            so
            $(1+x)^{1/k} le 1+x/k$,



            $begin{array}\
            s(n)
            &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
            &=sum_{k=2}^{n} dfrac{1}{nsqrt[k]{1+f(n)/n^k}+1}\
            &=sum_{k=2}^{n} dfrac1{n}dfrac{1}{sqrt[k]{1+f(n)/n^k}+1/n}\
            &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
            end{array}
            $



            Since $f(n)/n^c to 0$
            for some $c > 0$,
            $f(n) < an^c$
            for some $a > 0$
            so



            $begin{array}\
            f(n)/(kn^k)
            &lt an^c/(kn^k)\
            &=a/(kn^{k-c})\
            &lt a/(kn)
            qquadtext{for } k ge c+1\
            &lt 1/n
            qquadtext{for } k > a\
            end{array}
            $



            Therefore,
            letting
            $p(n) = max(c+1, a)$,
            $f(n)/(kn^k) < 1/n$
            for $k > p(n)$.



            Therefore



            $begin{array}\
            s(n)
            &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
            &gesum_{k=2}^{p(n)} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}+sum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
            &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+2/n}\
            &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{2}\
            &gt frac12ln(n/p(n))\
            &to infty
            qquadtext{since }n/p(n) to infty\
            end{array}
            $






            share|cite|improve this answer









            $endgroup$













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              7












              $begingroup$

              Try $(n+1)^kgt n^k+n+1gt n^k$






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Try $(n+1)^kgt n^k+n+1gt n^k$






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Try $(n+1)^kgt n^k+n+1gt n^k$






                  share|cite|improve this answer









                  $endgroup$



                  Try $(n+1)^kgt n^k+n+1gt n^k$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 17:39









                  Mark BennetMark Bennet

                  80.8k981179




                  80.8k981179























                      1












                      $begingroup$

                      Consider
                      $s(n)
                      =sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}
                      $
                      where
                      $f(n) ge 0$ and
                      $f(n)/n^{c} to 0$
                      for some $c > 0$.



                      $begin{array}\
                      s(n)
                      &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                      &ltsum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k}}\
                      &=sum_{k=2}^{n} dfrac{1}{n}\
                      &to ln(n)-1+gamma\
                      end{array}
                      $



                      Similarly,
                      since
                      $(1+x)^k ge 1+kx,
                      (1+x/k)^k ge 1+x$
                      so
                      $(1+x)^{1/k} le 1+x/k$,



                      $begin{array}\
                      s(n)
                      &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                      &=sum_{k=2}^{n} dfrac{1}{nsqrt[k]{1+f(n)/n^k}+1}\
                      &=sum_{k=2}^{n} dfrac1{n}dfrac{1}{sqrt[k]{1+f(n)/n^k}+1/n}\
                      &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                      end{array}
                      $



                      Since $f(n)/n^c to 0$
                      for some $c > 0$,
                      $f(n) < an^c$
                      for some $a > 0$
                      so



                      $begin{array}\
                      f(n)/(kn^k)
                      &lt an^c/(kn^k)\
                      &=a/(kn^{k-c})\
                      &lt a/(kn)
                      qquadtext{for } k ge c+1\
                      &lt 1/n
                      qquadtext{for } k > a\
                      end{array}
                      $



                      Therefore,
                      letting
                      $p(n) = max(c+1, a)$,
                      $f(n)/(kn^k) < 1/n$
                      for $k > p(n)$.



                      Therefore



                      $begin{array}\
                      s(n)
                      &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                      &gesum_{k=2}^{p(n)} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}+sum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                      &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+2/n}\
                      &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{2}\
                      &gt frac12ln(n/p(n))\
                      &to infty
                      qquadtext{since }n/p(n) to infty\
                      end{array}
                      $






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Consider
                        $s(n)
                        =sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}
                        $
                        where
                        $f(n) ge 0$ and
                        $f(n)/n^{c} to 0$
                        for some $c > 0$.



                        $begin{array}\
                        s(n)
                        &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                        &ltsum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k}}\
                        &=sum_{k=2}^{n} dfrac{1}{n}\
                        &to ln(n)-1+gamma\
                        end{array}
                        $



                        Similarly,
                        since
                        $(1+x)^k ge 1+kx,
                        (1+x/k)^k ge 1+x$
                        so
                        $(1+x)^{1/k} le 1+x/k$,



                        $begin{array}\
                        s(n)
                        &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                        &=sum_{k=2}^{n} dfrac{1}{nsqrt[k]{1+f(n)/n^k}+1}\
                        &=sum_{k=2}^{n} dfrac1{n}dfrac{1}{sqrt[k]{1+f(n)/n^k}+1/n}\
                        &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                        end{array}
                        $



                        Since $f(n)/n^c to 0$
                        for some $c > 0$,
                        $f(n) < an^c$
                        for some $a > 0$
                        so



                        $begin{array}\
                        f(n)/(kn^k)
                        &lt an^c/(kn^k)\
                        &=a/(kn^{k-c})\
                        &lt a/(kn)
                        qquadtext{for } k ge c+1\
                        &lt 1/n
                        qquadtext{for } k > a\
                        end{array}
                        $



                        Therefore,
                        letting
                        $p(n) = max(c+1, a)$,
                        $f(n)/(kn^k) < 1/n$
                        for $k > p(n)$.



                        Therefore



                        $begin{array}\
                        s(n)
                        &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                        &gesum_{k=2}^{p(n)} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}+sum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                        &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+2/n}\
                        &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{2}\
                        &gt frac12ln(n/p(n))\
                        &to infty
                        qquadtext{since }n/p(n) to infty\
                        end{array}
                        $






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Consider
                          $s(n)
                          =sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}
                          $
                          where
                          $f(n) ge 0$ and
                          $f(n)/n^{c} to 0$
                          for some $c > 0$.



                          $begin{array}\
                          s(n)
                          &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                          &ltsum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k}}\
                          &=sum_{k=2}^{n} dfrac{1}{n}\
                          &to ln(n)-1+gamma\
                          end{array}
                          $



                          Similarly,
                          since
                          $(1+x)^k ge 1+kx,
                          (1+x/k)^k ge 1+x$
                          so
                          $(1+x)^{1/k} le 1+x/k$,



                          $begin{array}\
                          s(n)
                          &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                          &=sum_{k=2}^{n} dfrac{1}{nsqrt[k]{1+f(n)/n^k}+1}\
                          &=sum_{k=2}^{n} dfrac1{n}dfrac{1}{sqrt[k]{1+f(n)/n^k}+1/n}\
                          &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          end{array}
                          $



                          Since $f(n)/n^c to 0$
                          for some $c > 0$,
                          $f(n) < an^c$
                          for some $a > 0$
                          so



                          $begin{array}\
                          f(n)/(kn^k)
                          &lt an^c/(kn^k)\
                          &=a/(kn^{k-c})\
                          &lt a/(kn)
                          qquadtext{for } k ge c+1\
                          &lt 1/n
                          qquadtext{for } k > a\
                          end{array}
                          $



                          Therefore,
                          letting
                          $p(n) = max(c+1, a)$,
                          $f(n)/(kn^k) < 1/n$
                          for $k > p(n)$.



                          Therefore



                          $begin{array}\
                          s(n)
                          &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          &gesum_{k=2}^{p(n)} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}+sum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+2/n}\
                          &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{2}\
                          &gt frac12ln(n/p(n))\
                          &to infty
                          qquadtext{since }n/p(n) to infty\
                          end{array}
                          $






                          share|cite|improve this answer









                          $endgroup$



                          Consider
                          $s(n)
                          =sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}
                          $
                          where
                          $f(n) ge 0$ and
                          $f(n)/n^{c} to 0$
                          for some $c > 0$.



                          $begin{array}\
                          s(n)
                          &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                          &ltsum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k}}\
                          &=sum_{k=2}^{n} dfrac{1}{n}\
                          &to ln(n)-1+gamma\
                          end{array}
                          $



                          Similarly,
                          since
                          $(1+x)^k ge 1+kx,
                          (1+x/k)^k ge 1+x$
                          so
                          $(1+x)^{1/k} le 1+x/k$,



                          $begin{array}\
                          s(n)
                          &=sum_{k=2}^{n} dfrac{1}{sqrt[k]{n^k+f(n)}+1}\
                          &=sum_{k=2}^{n} dfrac{1}{nsqrt[k]{1+f(n)/n^k}+1}\
                          &=sum_{k=2}^{n} dfrac1{n}dfrac{1}{sqrt[k]{1+f(n)/n^k}+1/n}\
                          &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          end{array}
                          $



                          Since $f(n)/n^c to 0$
                          for some $c > 0$,
                          $f(n) < an^c$
                          for some $a > 0$
                          so



                          $begin{array}\
                          f(n)/(kn^k)
                          &lt an^c/(kn^k)\
                          &=a/(kn^{k-c})\
                          &lt a/(kn)
                          qquadtext{for } k ge c+1\
                          &lt 1/n
                          qquadtext{for } k > a\
                          end{array}
                          $



                          Therefore,
                          letting
                          $p(n) = max(c+1, a)$,
                          $f(n)/(kn^k) < 1/n$
                          for $k > p(n)$.



                          Therefore



                          $begin{array}\
                          s(n)
                          &gesum_{k=2}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          &gesum_{k=2}^{p(n)} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}+sum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+f(n)/(kn^k)+1/n}\
                          &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{1+2/n}\
                          &gtsum_{k=p(n)}^{n} dfrac1{n}dfrac{1}{2}\
                          &gt frac12ln(n/p(n))\
                          &to infty
                          qquadtext{since }n/p(n) to infty\
                          end{array}
                          $







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 30 '18 at 22:16









                          marty cohenmarty cohen

                          73k549128




                          73k549128






























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                              張江高科駅

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                              張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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