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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                      The Binding of Isaac: Rebirth/Afterbirth

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                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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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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