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












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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







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    $begingroup$


    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







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      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






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      asked Jan 7 at 18:29









      ZacharyZachary

      1939




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          $begingroup$

          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









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            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









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              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






























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