Evaluate the limit of the sequence: $lim_{n_toinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot...












2












$begingroup$


Evaluate the limit of the sequence:



$$lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}$$




My try:



Stolz-cesaro: The limit of the sequence is $frac{infty}{infty}$



$$lim_{ntoinfty}frac{a_n}{b_n}=lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



For our sequence:



$lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{n!}-sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})cdot(1+sqrt{n+1})-(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{(n-1)!}cdot(sqrt{n-1})}{left((1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})right)cdot(sqrt{n}+1)}$



Which got me nowhere.






calculus sequences-and-series limits







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Evaluate the limit of the sequence:



    $$lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}$$




    My try:



    Stolz-cesaro: The limit of the sequence is $frac{infty}{infty}$



    $$lim_{ntoinfty}frac{a_n}{b_n}=lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



    For our sequence:



    $lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{n!}-sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})cdot(1+sqrt{n+1})-(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{(n-1)!}cdot(sqrt{n-1})}{left((1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})right)cdot(sqrt{n}+1)}$



    Which got me nowhere.






    calculus sequences-and-series limits







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Evaluate the limit of the sequence:



      $$lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}$$




      My try:



      Stolz-cesaro: The limit of the sequence is $frac{infty}{infty}$



      $$lim_{ntoinfty}frac{a_n}{b_n}=lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



      For our sequence:



      $lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{n!}-sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})cdot(1+sqrt{n+1})-(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{(n-1)!}cdot(sqrt{n-1})}{left((1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})right)cdot(sqrt{n}+1)}$



      Which got me nowhere.






      calculus sequences-and-series limits







      share|cite|improve this question









      $endgroup$




      Evaluate the limit of the sequence:



      $$lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}$$




      My try:



      Stolz-cesaro: The limit of the sequence is $frac{infty}{infty}$



      $$lim_{ntoinfty}frac{a_n}{b_n}=lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



      For our sequence:



      $lim_{ntoinfty}frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{n!}-sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})cdot(1+sqrt{n+1})-(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})}=lim_{ntoinfty}frac{sqrt{(n-1)!}cdot(sqrt{n-1})}{left((1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})right)cdot(sqrt{n}+1)}$



      Which got me nowhere.






      calculus sequences-and-series limits




      calculus sequences-and-series limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 12 '18 at 10:33









      user605734 MBSuser605734 MBS

      3029




      3029






















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$

          Consider:
          $$
          (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})
          $$



          Take the root from each pair of parentheses and multiply them, then:
          $$
          (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) > sqrt{n!} iff \
          iff frac{1}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} < frac{1}{sqrt{n!}}
          $$
          Going back to original we have that:
          $$
          frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{sqrt{(n-1)!}}{sqrt{n!}} = frac{1}{sqrt n}
          $$



          But the function is greater than $0$ and hence using squeeze theorem we conclude that:
          $$
          0 le lim_{ntoinfty}x_n le lim_{ntoinfty}frac{1}{sqrt n} = 0
          $$



          Hence the limit is $0$.






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            $(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) ge sqrt{1}cdotsqrt{2}cdots sqrt{n}= sqrt{n!}$, hence



            $0 le frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{1}{sqrt{n}}.$



            Can you proceed ?






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              Note that
              $$
              frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac{1}{1+sqrt{n}}
              prod_{k=1}^{n-1}frac{sqrt{k}}{1+sqrt{k}}<frac{1}{1+sqrt{n}}to 0
              $$






              share|cite|improve this answer









              $endgroup$





















                3












                $begingroup$

                We have that



                $$frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac1{sqrt n}prod_{k=1}^nfrac{sqrt{k}}{1+sqrt{k}}le frac1{sqrt n}$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Elements on the denominator are multiplyting
                  $endgroup$
                  – user605734 MBS
                  Dec 12 '18 at 10:39










                • $begingroup$
                  Opssss....thanks!
                  $endgroup$
                  – gimusi
                  Dec 12 '18 at 10:40











                Your Answer





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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                Consider:
                $$
                (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})
                $$



                Take the root from each pair of parentheses and multiply them, then:
                $$
                (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) > sqrt{n!} iff \
                iff frac{1}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} < frac{1}{sqrt{n!}}
                $$
                Going back to original we have that:
                $$
                frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{sqrt{(n-1)!}}{sqrt{n!}} = frac{1}{sqrt n}
                $$



                But the function is greater than $0$ and hence using squeeze theorem we conclude that:
                $$
                0 le lim_{ntoinfty}x_n le lim_{ntoinfty}frac{1}{sqrt n} = 0
                $$



                Hence the limit is $0$.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Consider:
                  $$
                  (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})
                  $$



                  Take the root from each pair of parentheses and multiply them, then:
                  $$
                  (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) > sqrt{n!} iff \
                  iff frac{1}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} < frac{1}{sqrt{n!}}
                  $$
                  Going back to original we have that:
                  $$
                  frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{sqrt{(n-1)!}}{sqrt{n!}} = frac{1}{sqrt n}
                  $$



                  But the function is greater than $0$ and hence using squeeze theorem we conclude that:
                  $$
                  0 le lim_{ntoinfty}x_n le lim_{ntoinfty}frac{1}{sqrt n} = 0
                  $$



                  Hence the limit is $0$.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Consider:
                    $$
                    (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})
                    $$



                    Take the root from each pair of parentheses and multiply them, then:
                    $$
                    (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) > sqrt{n!} iff \
                    iff frac{1}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} < frac{1}{sqrt{n!}}
                    $$
                    Going back to original we have that:
                    $$
                    frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{sqrt{(n-1)!}}{sqrt{n!}} = frac{1}{sqrt n}
                    $$



                    But the function is greater than $0$ and hence using squeeze theorem we conclude that:
                    $$
                    0 le lim_{ntoinfty}x_n le lim_{ntoinfty}frac{1}{sqrt n} = 0
                    $$



                    Hence the limit is $0$.






                    share|cite|improve this answer









                    $endgroup$



                    Consider:
                    $$
                    (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})
                    $$



                    Take the root from each pair of parentheses and multiply them, then:
                    $$
                    (1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) > sqrt{n!} iff \
                    iff frac{1}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} < frac{1}{sqrt{n!}}
                    $$
                    Going back to original we have that:
                    $$
                    frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{sqrt{(n-1)!}}{sqrt{n!}} = frac{1}{sqrt n}
                    $$



                    But the function is greater than $0$ and hence using squeeze theorem we conclude that:
                    $$
                    0 le lim_{ntoinfty}x_n le lim_{ntoinfty}frac{1}{sqrt n} = 0
                    $$



                    Hence the limit is $0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 10:52









                    romanroman

                    2,17321224




                    2,17321224























                        4












                        $begingroup$

                        $(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) ge sqrt{1}cdotsqrt{2}cdots sqrt{n}= sqrt{n!}$, hence



                        $0 le frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{1}{sqrt{n}}.$



                        Can you proceed ?






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          $(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) ge sqrt{1}cdotsqrt{2}cdots sqrt{n}= sqrt{n!}$, hence



                          $0 le frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{1}{sqrt{n}}.$



                          Can you proceed ?






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            $(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) ge sqrt{1}cdotsqrt{2}cdots sqrt{n}= sqrt{n!}$, hence



                            $0 le frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{1}{sqrt{n}}.$



                            Can you proceed ?






                            share|cite|improve this answer









                            $endgroup$



                            $(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n}) ge sqrt{1}cdotsqrt{2}cdots sqrt{n}= sqrt{n!}$, hence



                            $0 le frac{sqrt{(n-1)!}}{(1+sqrt{1})cdot(1+sqrt{2})cdot (1+sqrt{3})cdots (1+sqrt{n})} le frac{1}{sqrt{n}}.$



                            Can you proceed ?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 10:41









                            FredFred

                            45.7k1848




                            45.7k1848























                                3












                                $begingroup$

                                Note that
                                $$
                                frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac{1}{1+sqrt{n}}
                                prod_{k=1}^{n-1}frac{sqrt{k}}{1+sqrt{k}}<frac{1}{1+sqrt{n}}to 0
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  3












                                  $begingroup$

                                  Note that
                                  $$
                                  frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac{1}{1+sqrt{n}}
                                  prod_{k=1}^{n-1}frac{sqrt{k}}{1+sqrt{k}}<frac{1}{1+sqrt{n}}to 0
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$

                                    Note that
                                    $$
                                    frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac{1}{1+sqrt{n}}
                                    prod_{k=1}^{n-1}frac{sqrt{k}}{1+sqrt{k}}<frac{1}{1+sqrt{n}}to 0
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Note that
                                    $$
                                    frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac{1}{1+sqrt{n}}
                                    prod_{k=1}^{n-1}frac{sqrt{k}}{1+sqrt{k}}<frac{1}{1+sqrt{n}}to 0
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 12 '18 at 10:38









                                    Yiorgos S. SmyrlisYiorgos S. Smyrlis

                                    63.3k1384163




                                    63.3k1384163























                                        3












                                        $begingroup$

                                        We have that



                                        $$frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac1{sqrt n}prod_{k=1}^nfrac{sqrt{k}}{1+sqrt{k}}le frac1{sqrt n}$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Elements on the denominator are multiplyting
                                          $endgroup$
                                          – user605734 MBS
                                          Dec 12 '18 at 10:39










                                        • $begingroup$
                                          Opssss....thanks!
                                          $endgroup$
                                          – gimusi
                                          Dec 12 '18 at 10:40
















                                        3












                                        $begingroup$

                                        We have that



                                        $$frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac1{sqrt n}prod_{k=1}^nfrac{sqrt{k}}{1+sqrt{k}}le frac1{sqrt n}$$






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Elements on the denominator are multiplyting
                                          $endgroup$
                                          – user605734 MBS
                                          Dec 12 '18 at 10:39










                                        • $begingroup$
                                          Opssss....thanks!
                                          $endgroup$
                                          – gimusi
                                          Dec 12 '18 at 10:40














                                        3












                                        3








                                        3





                                        $begingroup$

                                        We have that



                                        $$frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac1{sqrt n}prod_{k=1}^nfrac{sqrt{k}}{1+sqrt{k}}le frac1{sqrt n}$$






                                        share|cite|improve this answer











                                        $endgroup$



                                        We have that



                                        $$frac{sqrt{(n-1)!}}{prod_{k=1}^nbig(1+sqrt{k}big)}=frac1{sqrt n}prod_{k=1}^nfrac{sqrt{k}}{1+sqrt{k}}le frac1{sqrt n}$$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 12 '18 at 10:45

























                                        answered Dec 12 '18 at 10:38









                                        gimusigimusi

                                        92.8k84494




                                        92.8k84494












                                        • $begingroup$
                                          Elements on the denominator are multiplyting
                                          $endgroup$
                                          – user605734 MBS
                                          Dec 12 '18 at 10:39










                                        • $begingroup$
                                          Opssss....thanks!
                                          $endgroup$
                                          – gimusi
                                          Dec 12 '18 at 10:40


















                                        • $begingroup$
                                          Elements on the denominator are multiplyting
                                          $endgroup$
                                          – user605734 MBS
                                          Dec 12 '18 at 10:39










                                        • $begingroup$
                                          Opssss....thanks!
                                          $endgroup$
                                          – gimusi
                                          Dec 12 '18 at 10:40
















                                        $begingroup$
                                        Elements on the denominator are multiplyting
                                        $endgroup$
                                        – user605734 MBS
                                        Dec 12 '18 at 10:39




                                        $begingroup$
                                        Elements on the denominator are multiplyting
                                        $endgroup$
                                        – user605734 MBS
                                        Dec 12 '18 at 10:39












                                        $begingroup$
                                        Opssss....thanks!
                                        $endgroup$
                                        – gimusi
                                        Dec 12 '18 at 10:40




                                        $begingroup$
                                        Opssss....thanks!
                                        $endgroup$
                                        – gimusi
                                        Dec 12 '18 at 10:40


















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                                        Marschland

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                                        Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                                        Dieringhausen

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                                        Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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