Is it true that $xgeq sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}$ for $0<x<1$












1












$begingroup$


I would like to prove or disprove $xgeq sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}$. From numerical examples it looks like the inequality holds. However, I have no idea how to prove it.



I have found a similar question
sum of an infinite series $sum_{k=1}^infty left( prod_{m=1}^kfrac{1}{1+mgamma}right) $, where the function looks very similar to mine, but not the same.



Looking forward to any hints. Thanks in advance!










share|cite|improve this question









$endgroup$












  • $begingroup$
    It may help to know where this expression came from. Can you provide context?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:39
















1












$begingroup$


I would like to prove or disprove $xgeq sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}$. From numerical examples it looks like the inequality holds. However, I have no idea how to prove it.



I have found a similar question
sum of an infinite series $sum_{k=1}^infty left( prod_{m=1}^kfrac{1}{1+mgamma}right) $, where the function looks very similar to mine, but not the same.



Looking forward to any hints. Thanks in advance!










share|cite|improve this question









$endgroup$












  • $begingroup$
    It may help to know where this expression came from. Can you provide context?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:39














1












1








1





$begingroup$


I would like to prove or disprove $xgeq sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}$. From numerical examples it looks like the inequality holds. However, I have no idea how to prove it.



I have found a similar question
sum of an infinite series $sum_{k=1}^infty left( prod_{m=1}^kfrac{1}{1+mgamma}right) $, where the function looks very similar to mine, but not the same.



Looking forward to any hints. Thanks in advance!










share|cite|improve this question









$endgroup$




I would like to prove or disprove $xgeq sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}$. From numerical examples it looks like the inequality holds. However, I have no idea how to prove it.



I have found a similar question
sum of an infinite series $sum_{k=1}^infty left( prod_{m=1}^kfrac{1}{1+mgamma}right) $, where the function looks very similar to mine, but not the same.



Looking forward to any hints. Thanks in advance!







sequences-and-series summation






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 3:29









hdiuwhciuhdiuwhciu

629




629












  • $begingroup$
    It may help to know where this expression came from. Can you provide context?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:39


















  • $begingroup$
    It may help to know where this expression came from. Can you provide context?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:39
















$begingroup$
It may help to know where this expression came from. Can you provide context?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:39




$begingroup$
It may help to know where this expression came from. Can you provide context?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

As you noticed, the equality hold and it hold for any $x >0$.



$$prod_{n=1}^kfrac{x}{x+n}=frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and
$$sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}=sum_{k=1}^inftyfrac{k,x^k}{(x+1)_k}=frac{x^2,Gamma (x)}{Gamma (x+1)}=x$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just a little nitpick: any $x>0$. (+1) anyway.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:48










  • $begingroup$
    Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
    $endgroup$
    – hdiuwhciu
    Dec 7 '18 at 3:50












  • $begingroup$
    @JackD'Aurizio. Thanks for pointing it !
    $endgroup$
    – Claude Leibovici
    Dec 7 '18 at 3:57






  • 1




    $begingroup$
    I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:59










  • $begingroup$
    Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 4:00



















2












$begingroup$

For a proof of $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.



$$frac{1}{(x+1)_k}=frac{Gamma(x+1)}{Gamma(x+k+1)}=frac{1}{Gamma(k)}B(k,x+1)=frac{1}{(k-1)!}int_{0}^{1}u^{k-1}(1-u)^x,du.$$
Multiplying both sides by $kx^k$ and summing over $kgeq 1$ we get



$$ sum_{kgeq 1}frac{k x^k}{(x+1)_k} = int_{0}^{1}(1-u)^x sum_{kgeq 1}frac{k x^k u^k}{(k-1)!}cdotfrac{du}{u}=xint_{0}^{1}(1-u)^x (1+ux)e^{ux},du$$
where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.






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    2 Answers
    2






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    2 Answers
    2






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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    As you noticed, the equality hold and it hold for any $x >0$.



    $$prod_{n=1}^kfrac{x}{x+n}=frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and
    $$sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}=sum_{k=1}^inftyfrac{k,x^k}{(x+1)_k}=frac{x^2,Gamma (x)}{Gamma (x+1)}=x$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just a little nitpick: any $x>0$. (+1) anyway.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:48










    • $begingroup$
      Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
      $endgroup$
      – hdiuwhciu
      Dec 7 '18 at 3:50












    • $begingroup$
      @JackD'Aurizio. Thanks for pointing it !
      $endgroup$
      – Claude Leibovici
      Dec 7 '18 at 3:57






    • 1




      $begingroup$
      I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:59










    • $begingroup$
      Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 4:00
















    2












    $begingroup$

    As you noticed, the equality hold and it hold for any $x >0$.



    $$prod_{n=1}^kfrac{x}{x+n}=frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and
    $$sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}=sum_{k=1}^inftyfrac{k,x^k}{(x+1)_k}=frac{x^2,Gamma (x)}{Gamma (x+1)}=x$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just a little nitpick: any $x>0$. (+1) anyway.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:48










    • $begingroup$
      Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
      $endgroup$
      – hdiuwhciu
      Dec 7 '18 at 3:50












    • $begingroup$
      @JackD'Aurizio. Thanks for pointing it !
      $endgroup$
      – Claude Leibovici
      Dec 7 '18 at 3:57






    • 1




      $begingroup$
      I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:59










    • $begingroup$
      Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 4:00














    2












    2








    2





    $begingroup$

    As you noticed, the equality hold and it hold for any $x >0$.



    $$prod_{n=1}^kfrac{x}{x+n}=frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and
    $$sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}=sum_{k=1}^inftyfrac{k,x^k}{(x+1)_k}=frac{x^2,Gamma (x)}{Gamma (x+1)}=x$$






    share|cite|improve this answer











    $endgroup$



    As you noticed, the equality hold and it hold for any $x >0$.



    $$prod_{n=1}^kfrac{x}{x+n}=frac{x^k}{(x+1)_k}$$ where appear Pochhammer symbols and
    $$sum_{k=1}^infty k prod_{n=1}^k frac{x}{x+n}=sum_{k=1}^inftyfrac{k,x^k}{(x+1)_k}=frac{x^2,Gamma (x)}{Gamma (x+1)}=x$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 3:57

























    answered Dec 7 '18 at 3:45









    Claude LeiboviciClaude Leibovici

    120k1157132




    120k1157132








    • 1




      $begingroup$
      Just a little nitpick: any $x>0$. (+1) anyway.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:48










    • $begingroup$
      Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
      $endgroup$
      – hdiuwhciu
      Dec 7 '18 at 3:50












    • $begingroup$
      @JackD'Aurizio. Thanks for pointing it !
      $endgroup$
      – Claude Leibovici
      Dec 7 '18 at 3:57






    • 1




      $begingroup$
      I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:59










    • $begingroup$
      Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 4:00














    • 1




      $begingroup$
      Just a little nitpick: any $x>0$. (+1) anyway.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:48










    • $begingroup$
      Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
      $endgroup$
      – hdiuwhciu
      Dec 7 '18 at 3:50












    • $begingroup$
      @JackD'Aurizio. Thanks for pointing it !
      $endgroup$
      – Claude Leibovici
      Dec 7 '18 at 3:57






    • 1




      $begingroup$
      I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 3:59










    • $begingroup$
      Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
      $endgroup$
      – Jack D'Aurizio
      Dec 7 '18 at 4:00








    1




    1




    $begingroup$
    Just a little nitpick: any $x>0$. (+1) anyway.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:48




    $begingroup$
    Just a little nitpick: any $x>0$. (+1) anyway.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:48












    $begingroup$
    Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
    $endgroup$
    – hdiuwhciu
    Dec 7 '18 at 3:50






    $begingroup$
    Sorry for this stupid comment. How does the second last equality follow? Thanks! @JackD'Aurizio
    $endgroup$
    – hdiuwhciu
    Dec 7 '18 at 3:50














    $begingroup$
    @JackD'Aurizio. Thanks for pointing it !
    $endgroup$
    – Claude Leibovici
    Dec 7 '18 at 3:57




    $begingroup$
    @JackD'Aurizio. Thanks for pointing it !
    $endgroup$
    – Claude Leibovici
    Dec 7 '18 at 3:57




    1




    1




    $begingroup$
    I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:59




    $begingroup$
    I guess such identity is just part of the arsenal of Monsieur Leibovici, but for a proof from scratch, one might start by noticing that $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is a meromorphic function with potential simple poles at $-1,-2,-3,-4,ldots$. By computing the residues (they're all zero) we get that those potential poles are regularity points, hence $sum_{kgeq 1}frac{k x^k}{(x+1)_k}$ is an entire function fixed by its derivatives at the origin. This leads to $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:59












    $begingroup$
    Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 4:00




    $begingroup$
    Then $frac{k x^k}{(x+1)_k}$ is a positive function on any subset of $mathbb{R}^+$, hence $sum_{k=1}^{N}frac{k x^k}{(x+1)_k}<x$ for any $x>0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 4:00











    2












    $begingroup$

    For a proof of $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.



    $$frac{1}{(x+1)_k}=frac{Gamma(x+1)}{Gamma(x+k+1)}=frac{1}{Gamma(k)}B(k,x+1)=frac{1}{(k-1)!}int_{0}^{1}u^{k-1}(1-u)^x,du.$$
    Multiplying both sides by $kx^k$ and summing over $kgeq 1$ we get



    $$ sum_{kgeq 1}frac{k x^k}{(x+1)_k} = int_{0}^{1}(1-u)^x sum_{kgeq 1}frac{k x^k u^k}{(k-1)!}cdotfrac{du}{u}=xint_{0}^{1}(1-u)^x (1+ux)e^{ux},du$$
    where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      For a proof of $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.



      $$frac{1}{(x+1)_k}=frac{Gamma(x+1)}{Gamma(x+k+1)}=frac{1}{Gamma(k)}B(k,x+1)=frac{1}{(k-1)!}int_{0}^{1}u^{k-1}(1-u)^x,du.$$
      Multiplying both sides by $kx^k$ and summing over $kgeq 1$ we get



      $$ sum_{kgeq 1}frac{k x^k}{(x+1)_k} = int_{0}^{1}(1-u)^x sum_{kgeq 1}frac{k x^k u^k}{(k-1)!}cdotfrac{du}{u}=xint_{0}^{1}(1-u)^x (1+ux)e^{ux},du$$
      where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        For a proof of $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.



        $$frac{1}{(x+1)_k}=frac{Gamma(x+1)}{Gamma(x+k+1)}=frac{1}{Gamma(k)}B(k,x+1)=frac{1}{(k-1)!}int_{0}^{1}u^{k-1}(1-u)^x,du.$$
        Multiplying both sides by $kx^k$ and summing over $kgeq 1$ we get



        $$ sum_{kgeq 1}frac{k x^k}{(x+1)_k} = int_{0}^{1}(1-u)^x sum_{kgeq 1}frac{k x^k u^k}{(k-1)!}cdotfrac{du}{u}=xint_{0}^{1}(1-u)^x (1+ux)e^{ux},du$$
        where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.






        share|cite|improve this answer











        $endgroup$



        For a proof of $sum_{kgeq 1}frac{k x^k}{(x+1)_k}=x$ avoiding Complex Analysis, one might invoke Euler's Beta function.



        $$frac{1}{(x+1)_k}=frac{Gamma(x+1)}{Gamma(x+k+1)}=frac{1}{Gamma(k)}B(k,x+1)=frac{1}{(k-1)!}int_{0}^{1}u^{k-1}(1-u)^x,du.$$
        Multiplying both sides by $kx^k$ and summing over $kgeq 1$ we get



        $$ sum_{kgeq 1}frac{k x^k}{(x+1)_k} = int_{0}^{1}(1-u)^x sum_{kgeq 1}frac{k x^k u^k}{(k-1)!}cdotfrac{du}{u}=xint_{0}^{1}(1-u)^x (1+ux)e^{ux},du$$
        where the last integral does not really depend on $x$: it equals $1$ for any $x>0$, since the integrand function is $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 4:19

























        answered Dec 7 '18 at 4:10









        Jack D'AurizioJack D'Aurizio

        1




        1






























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