Convergency of $sum_{n=0}^{infty}frac{n!}{(kn)!}$, where $k > 1$












0












$begingroup$


I am confident that $$sum_{n=1}^{infty}frac{n!}{(2n)!}approx1.5923$$
converges. Other series such as
$$sum_{n=1}^{infty}frac{n!}{(1.1n)!}approx5.5690$$
appear to converge as well, and
my hypothesis is that the series
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges.



I know that the $p$-series converges for $p>1$ due to the following relationship:
$$int_{1}^{infty}frac{1}{x^p}dx < sum_{1}^{infty}frac{1}{x^p}<1+int_{1}^{infty}frac{1}{x^p}dx$$
However, I am not sure how to prove that
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges with the same method that is used to prove the convergency of the $p$-series.
Could you provide me some hints?






calculus sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Have you tried ratio test?
    $endgroup$
    – Anurag A
    Oct 2 '18 at 21:47










  • $begingroup$
    What is $(1.1n)!$?
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 21:59






  • 1




    $begingroup$
    Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 22:00












  • $begingroup$
    @Hagen It is a coefficient. 11/10
    $endgroup$
    – Larry
    Oct 2 '18 at 22:01










  • $begingroup$
    @Larry: then what is $1.1! $ ?
    $endgroup$
    – Yves Daoust
    Oct 2 '18 at 22:06


















0












$begingroup$


I am confident that $$sum_{n=1}^{infty}frac{n!}{(2n)!}approx1.5923$$
converges. Other series such as
$$sum_{n=1}^{infty}frac{n!}{(1.1n)!}approx5.5690$$
appear to converge as well, and
my hypothesis is that the series
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges.



I know that the $p$-series converges for $p>1$ due to the following relationship:
$$int_{1}^{infty}frac{1}{x^p}dx < sum_{1}^{infty}frac{1}{x^p}<1+int_{1}^{infty}frac{1}{x^p}dx$$
However, I am not sure how to prove that
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges with the same method that is used to prove the convergency of the $p$-series.
Could you provide me some hints?






calculus sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Have you tried ratio test?
    $endgroup$
    – Anurag A
    Oct 2 '18 at 21:47










  • $begingroup$
    What is $(1.1n)!$?
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 21:59






  • 1




    $begingroup$
    Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 22:00












  • $begingroup$
    @Hagen It is a coefficient. 11/10
    $endgroup$
    – Larry
    Oct 2 '18 at 22:01










  • $begingroup$
    @Larry: then what is $1.1! $ ?
    $endgroup$
    – Yves Daoust
    Oct 2 '18 at 22:06
















0












0








0





$begingroup$


I am confident that $$sum_{n=1}^{infty}frac{n!}{(2n)!}approx1.5923$$
converges. Other series such as
$$sum_{n=1}^{infty}frac{n!}{(1.1n)!}approx5.5690$$
appear to converge as well, and
my hypothesis is that the series
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges.



I know that the $p$-series converges for $p>1$ due to the following relationship:
$$int_{1}^{infty}frac{1}{x^p}dx < sum_{1}^{infty}frac{1}{x^p}<1+int_{1}^{infty}frac{1}{x^p}dx$$
However, I am not sure how to prove that
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges with the same method that is used to prove the convergency of the $p$-series.
Could you provide me some hints?






calculus sequences-and-series







share|cite|improve this question









$endgroup$




I am confident that $$sum_{n=1}^{infty}frac{n!}{(2n)!}approx1.5923$$
converges. Other series such as
$$sum_{n=1}^{infty}frac{n!}{(1.1n)!}approx5.5690$$
appear to converge as well, and
my hypothesis is that the series
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges.



I know that the $p$-series converges for $p>1$ due to the following relationship:
$$int_{1}^{infty}frac{1}{x^p}dx < sum_{1}^{infty}frac{1}{x^p}<1+int_{1}^{infty}frac{1}{x^p}dx$$
However, I am not sure how to prove that
$$sum_{n=0}^{infty}frac{n!}{(kn)!}, k > 1$$
converges with the same method that is used to prove the convergency of the $p$-series.
Could you provide me some hints?






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 Oct 2 '18 at 21:45









LarryLarry

2,40831129




2,40831129








  • 1




    $begingroup$
    Have you tried ratio test?
    $endgroup$
    – Anurag A
    Oct 2 '18 at 21:47










  • $begingroup$
    What is $(1.1n)!$?
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 21:59






  • 1




    $begingroup$
    Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 22:00












  • $begingroup$
    @Hagen It is a coefficient. 11/10
    $endgroup$
    – Larry
    Oct 2 '18 at 22:01










  • $begingroup$
    @Larry: then what is $1.1! $ ?
    $endgroup$
    – Yves Daoust
    Oct 2 '18 at 22:06
















  • 1




    $begingroup$
    Have you tried ratio test?
    $endgroup$
    – Anurag A
    Oct 2 '18 at 21:47










  • $begingroup$
    What is $(1.1n)!$?
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 21:59






  • 1




    $begingroup$
    Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
    $endgroup$
    – Hagen von Eitzen
    Oct 2 '18 at 22:00












  • $begingroup$
    @Hagen It is a coefficient. 11/10
    $endgroup$
    – Larry
    Oct 2 '18 at 22:01










  • $begingroup$
    @Larry: then what is $1.1! $ ?
    $endgroup$
    – Yves Daoust
    Oct 2 '18 at 22:06










1




1




$begingroup$
Have you tried ratio test?
$endgroup$
– Anurag A
Oct 2 '18 at 21:47




$begingroup$
Have you tried ratio test?
$endgroup$
– Anurag A
Oct 2 '18 at 21:47












$begingroup$
What is $(1.1n)!$?
$endgroup$
– Hagen von Eitzen
Oct 2 '18 at 21:59




$begingroup$
What is $(1.1n)!$?
$endgroup$
– Hagen von Eitzen
Oct 2 '18 at 21:59




1




1




$begingroup$
Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
$endgroup$
– Hagen von Eitzen
Oct 2 '18 at 22:00






$begingroup$
Note that $frac{n!}{(kn)!}=frac{1}{{knchoose n}(kn-n)!}<frac1{n!}$ for $kge 2$
$endgroup$
– Hagen von Eitzen
Oct 2 '18 at 22:00














$begingroup$
@Hagen It is a coefficient. 11/10
$endgroup$
– Larry
Oct 2 '18 at 22:01




$begingroup$
@Hagen It is a coefficient. 11/10
$endgroup$
– Larry
Oct 2 '18 at 22:01












$begingroup$
@Larry: then what is $1.1! $ ?
$endgroup$
– Yves Daoust
Oct 2 '18 at 22:06






$begingroup$
@Larry: then what is $1.1! $ ?
$endgroup$
– Yves Daoust
Oct 2 '18 at 22:06












3 Answers
3






active

oldest

votes


















2












$begingroup$

$logGamma(s+1)$ is a convex function on $mathbb{R}^+$ and
$$ frac{n!}{(kn)!} = frac{Gamma(n+1)}{Gamma(kn+1)}leq frac{1}{n^{(k-1)n}}.$$
Something similar but more accurate can be deduced from Stirling's inequality
$$ left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m+1}}leqGamma(m+1) leq left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m}}.$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    By elementary means:



    For any natural $m$,



    $$kn-nge m$$ for sufficiently large $n$, so that $$dfrac{(kn)!}{n!}gelfloor knrfloorlfloor kn-1rfloorcdots(n+1)>n^m.$$



    Hence the series converges.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
      $endgroup$
      – AOrtiz
      Oct 2 '18 at 22:48






    • 1




      $begingroup$
      @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
      $endgroup$
      – Yves Daoust
      Oct 2 '18 at 22:49












    • $begingroup$
      Nice, thanks for the clarification! +1
      $endgroup$
      – AOrtiz
      Oct 2 '18 at 22:50



















    2












    $begingroup$

    Since $k > 1$, write $k > 1 + frac{1}{N}$ for some $N$ sufficiently large. Then if $n > 2N$, we have $$kn > big(1+frac{1}{N}big)n = n + frac{n}{N} > n+2.$$
    Thus for such $N$, we have
    $$sum_{n> 2N} frac{n!}{(kn)!} le sum_{n>2N} frac{n!}{(n+2)!} =sum_{n>2N}frac{1}{(n+2)(n+1)} le sum_{n>2N}frac{1}{n^2},$$ which converges by the $p$-test.






    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









      2












      $begingroup$

      $logGamma(s+1)$ is a convex function on $mathbb{R}^+$ and
      $$ frac{n!}{(kn)!} = frac{Gamma(n+1)}{Gamma(kn+1)}leq frac{1}{n^{(k-1)n}}.$$
      Something similar but more accurate can be deduced from Stirling's inequality
      $$ left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m+1}}leqGamma(m+1) leq left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m}}.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $logGamma(s+1)$ is a convex function on $mathbb{R}^+$ and
        $$ frac{n!}{(kn)!} = frac{Gamma(n+1)}{Gamma(kn+1)}leq frac{1}{n^{(k-1)n}}.$$
        Something similar but more accurate can be deduced from Stirling's inequality
        $$ left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m+1}}leqGamma(m+1) leq left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m}}.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $logGamma(s+1)$ is a convex function on $mathbb{R}^+$ and
          $$ frac{n!}{(kn)!} = frac{Gamma(n+1)}{Gamma(kn+1)}leq frac{1}{n^{(k-1)n}}.$$
          Something similar but more accurate can be deduced from Stirling's inequality
          $$ left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m+1}}leqGamma(m+1) leq left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m}}.$$






          share|cite|improve this answer











          $endgroup$



          $logGamma(s+1)$ is a convex function on $mathbb{R}^+$ and
          $$ frac{n!}{(kn)!} = frac{Gamma(n+1)}{Gamma(kn+1)}leq frac{1}{n^{(k-1)n}}.$$
          Something similar but more accurate can be deduced from Stirling's inequality
          $$ left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m+1}}leqGamma(m+1) leq left(frac{m}{e}right)^msqrt{2pi m}, e^{frac{1}{12m}}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 2 '18 at 22:31

























          answered Oct 2 '18 at 22:14









          Jack D'AurizioJack D'Aurizio

          289k33281661




          289k33281661























              2












              $begingroup$

              By elementary means:



              For any natural $m$,



              $$kn-nge m$$ for sufficiently large $n$, so that $$dfrac{(kn)!}{n!}gelfloor knrfloorlfloor kn-1rfloorcdots(n+1)>n^m.$$



              Hence the series converges.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:48






              • 1




                $begingroup$
                @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
                $endgroup$
                – Yves Daoust
                Oct 2 '18 at 22:49












              • $begingroup$
                Nice, thanks for the clarification! +1
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:50
















              2












              $begingroup$

              By elementary means:



              For any natural $m$,



              $$kn-nge m$$ for sufficiently large $n$, so that $$dfrac{(kn)!}{n!}gelfloor knrfloorlfloor kn-1rfloorcdots(n+1)>n^m.$$



              Hence the series converges.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:48






              • 1




                $begingroup$
                @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
                $endgroup$
                – Yves Daoust
                Oct 2 '18 at 22:49












              • $begingroup$
                Nice, thanks for the clarification! +1
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:50














              2












              2








              2





              $begingroup$

              By elementary means:



              For any natural $m$,



              $$kn-nge m$$ for sufficiently large $n$, so that $$dfrac{(kn)!}{n!}gelfloor knrfloorlfloor kn-1rfloorcdots(n+1)>n^m.$$



              Hence the series converges.






              share|cite|improve this answer











              $endgroup$



              By elementary means:



              For any natural $m$,



              $$kn-nge m$$ for sufficiently large $n$, so that $$dfrac{(kn)!}{n!}gelfloor knrfloorlfloor kn-1rfloorcdots(n+1)>n^m.$$



              Hence the series converges.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Oct 2 '18 at 22:43

























              answered Oct 2 '18 at 22:25









              Yves DaoustYves Daoust

              127k673226




              127k673226












              • $begingroup$
                I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:48






              • 1




                $begingroup$
                @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
                $endgroup$
                – Yves Daoust
                Oct 2 '18 at 22:49












              • $begingroup$
                Nice, thanks for the clarification! +1
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:50


















              • $begingroup$
                I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:48






              • 1




                $begingroup$
                @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
                $endgroup$
                – Yves Daoust
                Oct 2 '18 at 22:49












              • $begingroup$
                Nice, thanks for the clarification! +1
                $endgroup$
                – AOrtiz
                Oct 2 '18 at 22:50
















              $begingroup$
              I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
              $endgroup$
              – AOrtiz
              Oct 2 '18 at 22:48




              $begingroup$
              I like the direction your answer is taking, but could you clarify how $m$ appears in the product you have written out? Maybe with underbrace?
              $endgroup$
              – AOrtiz
              Oct 2 '18 at 22:48




              1




              1




              $begingroup$
              @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
              $endgroup$
              – Yves Daoust
              Oct 2 '18 at 22:49






              $begingroup$
              @AOrtiz: the number of factors in the product is the difference between the largest and the smallest, $lfloor knrfloor-n$.
              $endgroup$
              – Yves Daoust
              Oct 2 '18 at 22:49














              $begingroup$
              Nice, thanks for the clarification! +1
              $endgroup$
              – AOrtiz
              Oct 2 '18 at 22:50




              $begingroup$
              Nice, thanks for the clarification! +1
              $endgroup$
              – AOrtiz
              Oct 2 '18 at 22:50











              2












              $begingroup$

              Since $k > 1$, write $k > 1 + frac{1}{N}$ for some $N$ sufficiently large. Then if $n > 2N$, we have $$kn > big(1+frac{1}{N}big)n = n + frac{n}{N} > n+2.$$
              Thus for such $N$, we have
              $$sum_{n> 2N} frac{n!}{(kn)!} le sum_{n>2N} frac{n!}{(n+2)!} =sum_{n>2N}frac{1}{(n+2)(n+1)} le sum_{n>2N}frac{1}{n^2},$$ which converges by the $p$-test.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Since $k > 1$, write $k > 1 + frac{1}{N}$ for some $N$ sufficiently large. Then if $n > 2N$, we have $$kn > big(1+frac{1}{N}big)n = n + frac{n}{N} > n+2.$$
                Thus for such $N$, we have
                $$sum_{n> 2N} frac{n!}{(kn)!} le sum_{n>2N} frac{n!}{(n+2)!} =sum_{n>2N}frac{1}{(n+2)(n+1)} le sum_{n>2N}frac{1}{n^2},$$ which converges by the $p$-test.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Since $k > 1$, write $k > 1 + frac{1}{N}$ for some $N$ sufficiently large. Then if $n > 2N$, we have $$kn > big(1+frac{1}{N}big)n = n + frac{n}{N} > n+2.$$
                  Thus for such $N$, we have
                  $$sum_{n> 2N} frac{n!}{(kn)!} le sum_{n>2N} frac{n!}{(n+2)!} =sum_{n>2N}frac{1}{(n+2)(n+1)} le sum_{n>2N}frac{1}{n^2},$$ which converges by the $p$-test.






                  share|cite|improve this answer











                  $endgroup$



                  Since $k > 1$, write $k > 1 + frac{1}{N}$ for some $N$ sufficiently large. Then if $n > 2N$, we have $$kn > big(1+frac{1}{N}big)n = n + frac{n}{N} > n+2.$$
                  Thus for such $N$, we have
                  $$sum_{n> 2N} frac{n!}{(kn)!} le sum_{n>2N} frac{n!}{(n+2)!} =sum_{n>2N}frac{1}{(n+2)(n+1)} le sum_{n>2N}frac{1}{n^2},$$ which converges by the $p$-test.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 20:45

























                  answered Oct 2 '18 at 22:26









                  AOrtizAOrtiz

                  10.6k21441




                  10.6k21441






























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                      Mario Kart Wii

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                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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                      What does “Dominus providebit” mean?

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                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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                      Antonio Litta Visconti Arese

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                      Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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