How do i show this equivalence without using integration?












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Without using integration show that $lim_{nrightarrow infty} frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=frac{4}{e}$. Itcould have been easier with integratoon, but i cannot proceed with this one. Can anyone please give me any hint or the solution... i will be very grateful. Please..






real-analysis integration limits







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    $begingroup$


    Without using integration show that $lim_{nrightarrow infty} frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=frac{4}{e}$. Itcould have been easier with integratoon, but i cannot proceed with this one. Can anyone please give me any hint or the solution... i will be very grateful. Please..






    real-analysis integration limits







    share|cite|improve this question







    New contributor




    Kyle walker peters is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Without using integration show that $lim_{nrightarrow infty} frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=frac{4}{e}$. Itcould have been easier with integratoon, but i cannot proceed with this one. Can anyone please give me any hint or the solution... i will be very grateful. Please..






      real-analysis integration limits







      share|cite|improve this question







      New contributor




      Kyle walker peters is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Without using integration show that $lim_{nrightarrow infty} frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=frac{4}{e}$. Itcould have been easier with integratoon, but i cannot proceed with this one. Can anyone please give me any hint or the solution... i will be very grateful. Please..






      real-analysis integration limits




      real-analysis integration limits






      share|cite|improve this question







      New contributor




      Kyle walker peters is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Kyle walker peters is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      asked 1 hour ago









      Kyle walker petersKyle walker peters

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          2 Answers
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          4












          $begingroup$

          If we let $a_n=frac{[(n+1)(n+2)...(2n)]}{n^n}=frac{(2n)!}{n!n^n}$, then $frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=(a_n)^{frac{1}{n}}$. Then $frac{a_{n+1}}{a_n}=frac{(2n+2)!}{(2n)!}×frac{n!}{(n+1)!}×frac{n^n}{(n+1)^{n+1}}=frac{(2n+1)(2n+2)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{2(2n+1)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}rightarrow4×e^{-1}$.



          Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac{1}{n},frac{1}{2n}rightarrow 0$, so we can replace $frac{1}{2n}$ with $frac{1}{n}$. Now hence, $frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}=4×(1+frac{1}{n})^{-n} rightarrow 4× e^{-1}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
            $endgroup$
            – Yves Daoust
            1 hour ago












          • $begingroup$
            I justified....
            $endgroup$
            – Shamim Akhtar
            1 hour ago










          • $begingroup$
            No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
            $endgroup$
            – Yves Daoust
            1 hour ago





















          0












          $begingroup$

          By Stirling,



          $$frac1nsqrt[n]{frac{(2n)!}{n!}}simfrac1nfrac{sqrt[2n]{4pi n}left(dfrac{2n}eright)^2}{sqrt[n]{4pi n}dfrac ne}tofrac 4e.$$






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            If we let $a_n=frac{[(n+1)(n+2)...(2n)]}{n^n}=frac{(2n)!}{n!n^n}$, then $frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=(a_n)^{frac{1}{n}}$. Then $frac{a_{n+1}}{a_n}=frac{(2n+2)!}{(2n)!}×frac{n!}{(n+1)!}×frac{n^n}{(n+1)^{n+1}}=frac{(2n+1)(2n+2)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{2(2n+1)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}rightarrow4×e^{-1}$.



            Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac{1}{n},frac{1}{2n}rightarrow 0$, so we can replace $frac{1}{2n}$ with $frac{1}{n}$. Now hence, $frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}=4×(1+frac{1}{n})^{-n} rightarrow 4× e^{-1}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
              $endgroup$
              – Yves Daoust
              1 hour ago












            • $begingroup$
              I justified....
              $endgroup$
              – Shamim Akhtar
              1 hour ago










            • $begingroup$
              No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
              $endgroup$
              – Yves Daoust
              1 hour ago


















            4












            $begingroup$

            If we let $a_n=frac{[(n+1)(n+2)...(2n)]}{n^n}=frac{(2n)!}{n!n^n}$, then $frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=(a_n)^{frac{1}{n}}$. Then $frac{a_{n+1}}{a_n}=frac{(2n+2)!}{(2n)!}×frac{n!}{(n+1)!}×frac{n^n}{(n+1)^{n+1}}=frac{(2n+1)(2n+2)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{2(2n+1)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}rightarrow4×e^{-1}$.



            Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac{1}{n},frac{1}{2n}rightarrow 0$, so we can replace $frac{1}{2n}$ with $frac{1}{n}$. Now hence, $frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}=4×(1+frac{1}{n})^{-n} rightarrow 4× e^{-1}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
              $endgroup$
              – Yves Daoust
              1 hour ago












            • $begingroup$
              I justified....
              $endgroup$
              – Shamim Akhtar
              1 hour ago










            • $begingroup$
              No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
              $endgroup$
              – Yves Daoust
              1 hour ago
















            4












            4








            4





            $begingroup$

            If we let $a_n=frac{[(n+1)(n+2)...(2n)]}{n^n}=frac{(2n)!}{n!n^n}$, then $frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=(a_n)^{frac{1}{n}}$. Then $frac{a_{n+1}}{a_n}=frac{(2n+2)!}{(2n)!}×frac{n!}{(n+1)!}×frac{n^n}{(n+1)^{n+1}}=frac{(2n+1)(2n+2)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{2(2n+1)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}rightarrow4×e^{-1}$.



            Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac{1}{n},frac{1}{2n}rightarrow 0$, so we can replace $frac{1}{2n}$ with $frac{1}{n}$. Now hence, $frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}=4×(1+frac{1}{n})^{-n} rightarrow 4× e^{-1}$






            share|cite|improve this answer











            $endgroup$



            If we let $a_n=frac{[(n+1)(n+2)...(2n)]}{n^n}=frac{(2n)!}{n!n^n}$, then $frac{[(n+1)(n+2)...(2n)]^{frac{1}{n}}}{n}=(a_n)^{frac{1}{n}}$. Then $frac{a_{n+1}}{a_n}=frac{(2n+2)!}{(2n)!}×frac{n!}{(n+1)!}×frac{n^n}{(n+1)^{n+1}}=frac{(2n+1)(2n+2)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{2(2n+1)}{(n+1)}×frac{n^n}{(n+1)^{n+1}}=frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}rightarrow4×e^{-1}$.



            Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac{1}{n},frac{1}{2n}rightarrow 0$, so we can replace $frac{1}{2n}$ with $frac{1}{n}$. Now hence, $frac{4n(1+frac{1}{2n})}{n(1+frac{1}{n})}×(1+frac{1}{n})^{-n}=4×(1+frac{1}{n})^{-n} rightarrow 4× e^{-1}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Shamim AkhtarShamim Akhtar

            60219




            60219












            • $begingroup$
              The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
              $endgroup$
              – Yves Daoust
              1 hour ago












            • $begingroup$
              I justified....
              $endgroup$
              – Shamim Akhtar
              1 hour ago










            • $begingroup$
              No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
              $endgroup$
              – Yves Daoust
              1 hour ago




















            • $begingroup$
              The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
              $endgroup$
              – Yves Daoust
              1 hour ago












            • $begingroup$
              I justified....
              $endgroup$
              – Shamim Akhtar
              1 hour ago










            • $begingroup$
              No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
              $endgroup$
              – Yves Daoust
              1 hour ago


















            $begingroup$
            The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
            $endgroup$
            – Yves Daoust
            1 hour ago






            $begingroup$
            The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^{-1}$, while you still have an infinite product for all $n$.
            $endgroup$
            – Yves Daoust
            1 hour ago














            $begingroup$
            I justified....
            $endgroup$
            – Shamim Akhtar
            1 hour ago




            $begingroup$
            I justified....
            $endgroup$
            – Shamim Akhtar
            1 hour ago












            $begingroup$
            No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
            $endgroup$
            – Yves Daoust
            1 hour ago






            $begingroup$
            No, you didn't add a word. How is it that $sqrt[n]{prod_{1,n}(1+frac1k)^k}to e$ ?
            $endgroup$
            – Yves Daoust
            1 hour ago













            0












            $begingroup$

            By Stirling,



            $$frac1nsqrt[n]{frac{(2n)!}{n!}}simfrac1nfrac{sqrt[2n]{4pi n}left(dfrac{2n}eright)^2}{sqrt[n]{4pi n}dfrac ne}tofrac 4e.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              By Stirling,



              $$frac1nsqrt[n]{frac{(2n)!}{n!}}simfrac1nfrac{sqrt[2n]{4pi n}left(dfrac{2n}eright)^2}{sqrt[n]{4pi n}dfrac ne}tofrac 4e.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                By Stirling,



                $$frac1nsqrt[n]{frac{(2n)!}{n!}}simfrac1nfrac{sqrt[2n]{4pi n}left(dfrac{2n}eright)^2}{sqrt[n]{4pi n}dfrac ne}tofrac 4e.$$






                share|cite|improve this answer









                $endgroup$



                By Stirling,



                $$frac1nsqrt[n]{frac{(2n)!}{n!}}simfrac1nfrac{sqrt[2n]{4pi n}left(dfrac{2n}eright)^2}{sqrt[n]{4pi n}dfrac ne}tofrac 4e.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Yves DaoustYves Daoust

                134k676233




                134k676233






















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