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!






sequences-and-series summation







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  • $begingroup$
    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!






sequences-and-series summation







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!






sequences-and-series summation







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




sequences-and-series summation






share|cite|improve this question













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 $-frac{d}{du}left[(1-u)^{x+1}e^{ux}right]$.






share|cite|improve this answer











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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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            Quarter-circle Tiles

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            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
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            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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