Proof $sum_{nge1} frac{n!}{x(x+1)cdotdotscdot(x+n-1)}$ ($x>0$) convergence/divergence.












0












$begingroup$


First I noticed that the series can be rewritten to $sum_{nge1}frac1{binom{x}{n}}$. Applying root test: $lim_{ntoinfty} left|frac{binom{x}{n+1}}{binom{x}{n}}right| = lim_{ntoinfty} left|frac{binom{x}{n} cdot frac{x-n}{n-1}}{binom{x}{n}}right| = 1 $. This test is inconclusive. I'm trying hard to find a series to compare to the given one, but I can't seem to find one. Any hints?



Thanks in advance!






calculus sequences-and-series







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    First I noticed that the series can be rewritten to $sum_{nge1}frac1{binom{x}{n}}$. Applying root test: $lim_{ntoinfty} left|frac{binom{x}{n+1}}{binom{x}{n}}right| = lim_{ntoinfty} left|frac{binom{x}{n} cdot frac{x-n}{n-1}}{binom{x}{n}}right| = 1 $. This test is inconclusive. I'm trying hard to find a series to compare to the given one, but I can't seem to find one. Any hints?



    Thanks in advance!






    calculus sequences-and-series







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      First I noticed that the series can be rewritten to $sum_{nge1}frac1{binom{x}{n}}$. Applying root test: $lim_{ntoinfty} left|frac{binom{x}{n+1}}{binom{x}{n}}right| = lim_{ntoinfty} left|frac{binom{x}{n} cdot frac{x-n}{n-1}}{binom{x}{n}}right| = 1 $. This test is inconclusive. I'm trying hard to find a series to compare to the given one, but I can't seem to find one. Any hints?



      Thanks in advance!






      calculus sequences-and-series







      share|cite|improve this question









      $endgroup$




      First I noticed that the series can be rewritten to $sum_{nge1}frac1{binom{x}{n}}$. Applying root test: $lim_{ntoinfty} left|frac{binom{x}{n+1}}{binom{x}{n}}right| = lim_{ntoinfty} left|frac{binom{x}{n} cdot frac{x-n}{n-1}}{binom{x}{n}}right| = 1 $. This test is inconclusive. I'm trying hard to find a series to compare to the given one, but I can't seem to find one. Any hints?



      Thanks in advance!






      calculus sequences-and-series




      calculus sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 18:29









      ZacharyZachary

      1939




      1939






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          It is easy enough to show for integers, but we can rewrite in terms of Gamma functions instead so that we can use related asymptotics
          $$
          frac{n!}{x(x+1)dotsb(x+n-1)}
          = frac{Gamma(n+1)}{Gamma(x+n),/,Gamma(x)}
          = frac{Gamma(x)}{n^{x-1}} frac{Gamma(n)cdot n^x}{Gamma(n+x)}
          $$



          Here:




          Another useful limit for asymptotic approximations is:
          $$lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^{alpha}},quad alphainmathbb{C}$$




          hence
          $$ frac{n!}{x(x+1)dotsb(x+n-1)} / frac{1}{n^{x-1}} to Gamma(x) > 0,quad text{for $x > 0$}$$



          The limit comparison test then says that the series converges or diverges with $sum_{n=1}^infty frac{1}{n^{x-1}}$, a "$p$-series", that converges if and only if $x>2$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Let $u_n$ be the $n$-th term in the series. Then $R_n=frac{u_{n+1}}{u_n}=frac{n+1}{x+n}$.



            So we see that the series diverges if $x leq 1$ because the sequence is nonzero and nondecreasing.



            Now we assume $x=1+y, y > 0$.



            Then $R_n=1-frac{y}{n}frac{1}{1+x/n}= 1-frac{y}{n} + Oleft(frac{1}{n^2}right)$.



            Thus $ln(u_{n+1})-ln(u_n)=ln(R_n)=-yn^{-1}+O(n^{-2})$.



            As a consequence, $ln(u_n)$ goes to $-infty$ and is equal to $-yln(n)+z+o(1)$ for some constant $z$.



            Thus $u_n sim e^zn^{-y}$, hence the series converges iff $y > 1$ ie $x >2$.






            share|cite|improve this answer









            $endgroup$














              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065313%2fproof-sum-n-ge1-fracnxx1-cdot-dots-cdotxn-1-x0-convergence%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              It is easy enough to show for integers, but we can rewrite in terms of Gamma functions instead so that we can use related asymptotics
              $$
              frac{n!}{x(x+1)dotsb(x+n-1)}
              = frac{Gamma(n+1)}{Gamma(x+n),/,Gamma(x)}
              = frac{Gamma(x)}{n^{x-1}} frac{Gamma(n)cdot n^x}{Gamma(n+x)}
              $$



              Here:




              Another useful limit for asymptotic approximations is:
              $$lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^{alpha}},quad alphainmathbb{C}$$




              hence
              $$ frac{n!}{x(x+1)dotsb(x+n-1)} / frac{1}{n^{x-1}} to Gamma(x) > 0,quad text{for $x > 0$}$$



              The limit comparison test then says that the series converges or diverges with $sum_{n=1}^infty frac{1}{n^{x-1}}$, a "$p$-series", that converges if and only if $x>2$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It is easy enough to show for integers, but we can rewrite in terms of Gamma functions instead so that we can use related asymptotics
                $$
                frac{n!}{x(x+1)dotsb(x+n-1)}
                = frac{Gamma(n+1)}{Gamma(x+n),/,Gamma(x)}
                = frac{Gamma(x)}{n^{x-1}} frac{Gamma(n)cdot n^x}{Gamma(n+x)}
                $$



                Here:




                Another useful limit for asymptotic approximations is:
                $$lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^{alpha}},quad alphainmathbb{C}$$




                hence
                $$ frac{n!}{x(x+1)dotsb(x+n-1)} / frac{1}{n^{x-1}} to Gamma(x) > 0,quad text{for $x > 0$}$$



                The limit comparison test then says that the series converges or diverges with $sum_{n=1}^infty frac{1}{n^{x-1}}$, a "$p$-series", that converges if and only if $x>2$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It is easy enough to show for integers, but we can rewrite in terms of Gamma functions instead so that we can use related asymptotics
                  $$
                  frac{n!}{x(x+1)dotsb(x+n-1)}
                  = frac{Gamma(n+1)}{Gamma(x+n),/,Gamma(x)}
                  = frac{Gamma(x)}{n^{x-1}} frac{Gamma(n)cdot n^x}{Gamma(n+x)}
                  $$



                  Here:




                  Another useful limit for asymptotic approximations is:
                  $$lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^{alpha}},quad alphainmathbb{C}$$




                  hence
                  $$ frac{n!}{x(x+1)dotsb(x+n-1)} / frac{1}{n^{x-1}} to Gamma(x) > 0,quad text{for $x > 0$}$$



                  The limit comparison test then says that the series converges or diverges with $sum_{n=1}^infty frac{1}{n^{x-1}}$, a "$p$-series", that converges if and only if $x>2$.






                  share|cite|improve this answer









                  $endgroup$



                  It is easy enough to show for integers, but we can rewrite in terms of Gamma functions instead so that we can use related asymptotics
                  $$
                  frac{n!}{x(x+1)dotsb(x+n-1)}
                  = frac{Gamma(n+1)}{Gamma(x+n),/,Gamma(x)}
                  = frac{Gamma(x)}{n^{x-1}} frac{Gamma(n)cdot n^x}{Gamma(n+x)}
                  $$



                  Here:




                  Another useful limit for asymptotic approximations is:
                  $$lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^{alpha}},quad alphainmathbb{C}$$




                  hence
                  $$ frac{n!}{x(x+1)dotsb(x+n-1)} / frac{1}{n^{x-1}} to Gamma(x) > 0,quad text{for $x > 0$}$$



                  The limit comparison test then says that the series converges or diverges with $sum_{n=1}^infty frac{1}{n^{x-1}}$, a "$p$-series", that converges if and only if $x>2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 21:18









                  adfriedmanadfriedman

                  3,156169




                  3,156169























                      0












                      $begingroup$

                      Let $u_n$ be the $n$-th term in the series. Then $R_n=frac{u_{n+1}}{u_n}=frac{n+1}{x+n}$.



                      So we see that the series diverges if $x leq 1$ because the sequence is nonzero and nondecreasing.



                      Now we assume $x=1+y, y > 0$.



                      Then $R_n=1-frac{y}{n}frac{1}{1+x/n}= 1-frac{y}{n} + Oleft(frac{1}{n^2}right)$.



                      Thus $ln(u_{n+1})-ln(u_n)=ln(R_n)=-yn^{-1}+O(n^{-2})$.



                      As a consequence, $ln(u_n)$ goes to $-infty$ and is equal to $-yln(n)+z+o(1)$ for some constant $z$.



                      Thus $u_n sim e^zn^{-y}$, hence the series converges iff $y > 1$ ie $x >2$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Let $u_n$ be the $n$-th term in the series. Then $R_n=frac{u_{n+1}}{u_n}=frac{n+1}{x+n}$.



                        So we see that the series diverges if $x leq 1$ because the sequence is nonzero and nondecreasing.



                        Now we assume $x=1+y, y > 0$.



                        Then $R_n=1-frac{y}{n}frac{1}{1+x/n}= 1-frac{y}{n} + Oleft(frac{1}{n^2}right)$.



                        Thus $ln(u_{n+1})-ln(u_n)=ln(R_n)=-yn^{-1}+O(n^{-2})$.



                        As a consequence, $ln(u_n)$ goes to $-infty$ and is equal to $-yln(n)+z+o(1)$ for some constant $z$.



                        Thus $u_n sim e^zn^{-y}$, hence the series converges iff $y > 1$ ie $x >2$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Let $u_n$ be the $n$-th term in the series. Then $R_n=frac{u_{n+1}}{u_n}=frac{n+1}{x+n}$.



                          So we see that the series diverges if $x leq 1$ because the sequence is nonzero and nondecreasing.



                          Now we assume $x=1+y, y > 0$.



                          Then $R_n=1-frac{y}{n}frac{1}{1+x/n}= 1-frac{y}{n} + Oleft(frac{1}{n^2}right)$.



                          Thus $ln(u_{n+1})-ln(u_n)=ln(R_n)=-yn^{-1}+O(n^{-2})$.



                          As a consequence, $ln(u_n)$ goes to $-infty$ and is equal to $-yln(n)+z+o(1)$ for some constant $z$.



                          Thus $u_n sim e^zn^{-y}$, hence the series converges iff $y > 1$ ie $x >2$.






                          share|cite|improve this answer









                          $endgroup$



                          Let $u_n$ be the $n$-th term in the series. Then $R_n=frac{u_{n+1}}{u_n}=frac{n+1}{x+n}$.



                          So we see that the series diverges if $x leq 1$ because the sequence is nonzero and nondecreasing.



                          Now we assume $x=1+y, y > 0$.



                          Then $R_n=1-frac{y}{n}frac{1}{1+x/n}= 1-frac{y}{n} + Oleft(frac{1}{n^2}right)$.



                          Thus $ln(u_{n+1})-ln(u_n)=ln(R_n)=-yn^{-1}+O(n^{-2})$.



                          As a consequence, $ln(u_n)$ goes to $-infty$ and is equal to $-yln(n)+z+o(1)$ for some constant $z$.



                          Thus $u_n sim e^zn^{-y}$, hence the series converges iff $y > 1$ ie $x >2$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 7 at 18:38









                          MindlackMindlack

                          4,910211




                          4,910211






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065313%2fproof-sum-n-ge1-fracnxx1-cdot-dots-cdotxn-1-x0-convergence%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Bressuire

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                              Read more

                              Cabo Verde

                              Image
                                Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                              Read more

                              Gyllenstierna

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                              Read more