How find this $lim_{ntoinfty}left(frac{1}{a_{n}+1}+frac{1}{a_{n}+2}+cdots+frac{1}{a_{n}+b_{n}}right)=x$












3












$begingroup$


Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
$$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$



I only know this
$$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$



But for my problem I can't. Thank you






calculus sequences-and-series analysis limits integers







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    3












    $begingroup$


    Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
    $$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$



    I only know this
    $$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$



    But for my problem I can't. Thank you






    calculus sequences-and-series analysis limits integers







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
      $$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$



      I only know this
      $$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$



      But for my problem I can't. Thank you






      calculus sequences-and-series analysis limits integers







      share|cite|improve this question











      $endgroup$




      Prove that for any $xin[0,infty)$ there exist sequences of positive integers ${a_{n}}_{ninmathbb N}$ and ${b_{n}}_{ninmathbb N}$, such that
      $$lim_{ntoinfty}left(dfrac{1}{a_{n}+1}+dfrac{1}{a_{n}+2}+cdots+dfrac{1}{a_{n}+b_{n}}right)=x.$$



      I only know this
      $$lim_{ntoinfty}dfrac{1}{n+1}+dfrac{1}{n+2}+cdots+dfrac{1}{n+n}=ln{2}$$



      But for my problem I can't. Thank you






      calculus sequences-and-series analysis limits integers




      calculus sequences-and-series analysis limits integers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 28 '14 at 15:05









      Yiorgos S. Smyrlis

      63.2k1384163




      63.2k1384163










      asked Jan 19 '14 at 14:41









      china mathchina math

      10.2k631117




      10.2k631117






















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          Note that (integral comparison test)
          $$
          ln big(tfrac{ell-1/n}{k-1/n}big)
          =int_{kn-1}^{ell n-1}frac{dx}{x}<
          frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
          =ln (tfrac{ell}{k})
          $$
          Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
          $$
          frac{ell_n}{k_n}tomathrm{e}^x.
          $$
          Then
          $$
          lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
          frac{1}{k_nn+j}to x.
          $$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Hint: Change the limit in a Riemann sum :-)



            like here






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
              $endgroup$
              – china math
              Jan 19 '14 at 14:45





















            0












            $begingroup$

            Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
            $$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
            (Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7












              $begingroup$

              Note that (integral comparison test)
              $$
              ln big(tfrac{ell-1/n}{k-1/n}big)
              =int_{kn-1}^{ell n-1}frac{dx}{x}<
              frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
              =ln (tfrac{ell}{k})
              $$
              Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
              $$
              frac{ell_n}{k_n}tomathrm{e}^x.
              $$
              Then
              $$
              lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
              frac{1}{k_nn+j}to x.
              $$






              share|cite|improve this answer











              $endgroup$


















                7












                $begingroup$

                Note that (integral comparison test)
                $$
                ln big(tfrac{ell-1/n}{k-1/n}big)
                =int_{kn-1}^{ell n-1}frac{dx}{x}<
                frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
                =ln (tfrac{ell}{k})
                $$
                Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
                $$
                frac{ell_n}{k_n}tomathrm{e}^x.
                $$
                Then
                $$
                lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
                frac{1}{k_nn+j}to x.
                $$






                share|cite|improve this answer











                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Note that (integral comparison test)
                  $$
                  ln big(tfrac{ell-1/n}{k-1/n}big)
                  =int_{kn-1}^{ell n-1}frac{dx}{x}<
                  frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
                  =ln (tfrac{ell}{k})
                  $$
                  Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
                  $$
                  frac{ell_n}{k_n}tomathrm{e}^x.
                  $$
                  Then
                  $$
                  lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
                  frac{1}{k_nn+j}to x.
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  Note that (integral comparison test)
                  $$
                  ln big(tfrac{ell-1/n}{k-1/n}big)
                  =int_{kn-1}^{ell n-1}frac{dx}{x}<
                  frac{1}{kn+1}+frac{1}{kn+2}+cdots+frac{1}{ell n}<int_{kn}^{ell n}frac{dx}{x}
                  =ln (tfrac{ell}{k})
                  $$
                  Let strictly increasing sequences of integers $k_n$, $ell_n$, such that
                  $$
                  frac{ell_n}{k_n}tomathrm{e}^x.
                  $$
                  Then
                  $$
                  lim_{ntoinfty}sum_{j=1}^{(ell_m-k_m)n}
                  frac{1}{k_nn+j}to x.
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 9 '18 at 17:34

























                  answered Jan 19 '14 at 14:58









                  Yiorgos S. SmyrlisYiorgos S. Smyrlis

                  63.2k1384163




                  63.2k1384163























                      1












                      $begingroup$

                      Hint: Change the limit in a Riemann sum :-)



                      like here






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                        $endgroup$
                        – china math
                        Jan 19 '14 at 14:45


















                      1












                      $begingroup$

                      Hint: Change the limit in a Riemann sum :-)



                      like here






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                        $endgroup$
                        – china math
                        Jan 19 '14 at 14:45
















                      1












                      1








                      1





                      $begingroup$

                      Hint: Change the limit in a Riemann sum :-)



                      like here






                      share|cite|improve this answer











                      $endgroup$



                      Hint: Change the limit in a Riemann sum :-)



                      like here







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 13 '17 at 12:21









                      Community♦

                      1




                      1










                      answered Jan 19 '14 at 14:43









                      Bman72Bman72

                      1,9221922




                      1,9221922












                      • $begingroup$
                        can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                        $endgroup$
                        – china math
                        Jan 19 '14 at 14:45




















                      • $begingroup$
                        can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                        $endgroup$
                        – china math
                        Jan 19 '14 at 14:45


















                      $begingroup$
                      can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                      $endgroup$
                      – china math
                      Jan 19 '14 at 14:45






                      $begingroup$
                      can you post your solution? why xsit sequences of nonnegative integers ${a_{n}},{b_{n}}$?Thank you
                      $endgroup$
                      – china math
                      Jan 19 '14 at 14:45













                      0












                      $begingroup$

                      Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
                      $$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
                      (Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
                        $$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
                        (Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
                          $$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
                          (Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)






                          share|cite|improve this answer









                          $endgroup$



                          Use the fact that any real number can be approximated arbitrarily closely by rationals. The numbers that you are looking for are any sequences such that:
                          $$e^x-1=lim_{ntoinfty}frac{b_n}{a_n}$$
                          (Can you see why? Hint: $H_n=dfrac11+dfrac12+dfrac13+dotsb+dfrac1napproxln n+gamma$, and what you have above is $H_{a_n+b_n}-H_{a_n}$.)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 28 '14 at 15:39









                          Akiva WeinbergerAkiva Weinberger

                          13.8k12167




                          13.8k12167






























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