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

    oldest

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    2 Answers
    2






    active

    oldest

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    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

        288k33280660




        288k33280660






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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