Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?












1












$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47
















1












$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47














1












1








1


1



$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







share|cite|improve this question









$endgroup$




Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function




calculus integration sequences-and-series gamma-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 31 '13 at 7:33









PandaManPandaMan

1,17911333




1,17911333








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47














  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47








1




1




$begingroup$
Exponential are 'always' quite strong.
$endgroup$
– Felix Marin
Oct 31 '13 at 7:47




$begingroup$
Exponential are 'always' quite strong.
$endgroup$
– Felix Marin
Oct 31 '13 at 7:47




1




1




$begingroup$
Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
$endgroup$
– Lucian
Oct 31 '13 at 7:47




$begingroup$
Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
$endgroup$
– Lucian
Oct 31 '13 at 7:47










2 Answers
2






active

oldest

votes


















7












$begingroup$

We do not need an explicit expression to show that an improper integral converges.



The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



Any increasing sequence which is bounded above converges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07



















0












$begingroup$

Recall the Taylor series for $e^x$ at $x = 0$ is given by:



begin{equation}
e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
end{equation}



Thus,



begin{equation}
e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
end{equation}



Thus,



begin{align}
int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
&= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
end{align}



We now apply the Ratio Test:



begin{align}
R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
&= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
end{align}



Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f546649%2flet-a-n-int-limits-0n-e-x4-dx-does-a-n-n-rightarrow-inf%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07
















    7












    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07














    7












    7








    7





    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$



    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 31 '13 at 7:58

























    answered Oct 31 '13 at 7:40









    André NicolasAndré Nicolas

    453k36428815




    453k36428815












    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07


















    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07
















    $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07




    $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07











    0












    $begingroup$

    Recall the Taylor series for $e^x$ at $x = 0$ is given by:



    begin{equation}
    e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
    end{equation}



    Thus,



    begin{equation}
    e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
    end{equation}



    Thus,



    begin{align}
    int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
    &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
    end{align}



    We now apply the Ratio Test:



    begin{align}
    R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
    &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
    end{align}



    Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Recall the Taylor series for $e^x$ at $x = 0$ is given by:



      begin{equation}
      e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
      end{equation}



      Thus,



      begin{equation}
      e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
      end{equation}



      Thus,



      begin{align}
      int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
      &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
      end{align}



      We now apply the Ratio Test:



      begin{align}
      R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
      &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
      end{align}



      Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Recall the Taylor series for $e^x$ at $x = 0$ is given by:



        begin{equation}
        e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
        end{equation}



        Thus,



        begin{equation}
        e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
        end{equation}



        Thus,



        begin{align}
        int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
        &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
        end{align}



        We now apply the Ratio Test:



        begin{align}
        R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
        &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
        end{align}



        Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






        share|cite|improve this answer









        $endgroup$



        Recall the Taylor series for $e^x$ at $x = 0$ is given by:



        begin{equation}
        e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
        end{equation}



        Thus,



        begin{equation}
        e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
        end{equation}



        Thus,



        begin{align}
        int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
        &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
        end{align}



        We now apply the Ratio Test:



        begin{align}
        R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
        &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
        end{align}



        Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 2:28









        DavidGDavidG

        2,2271724




        2,2271724






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f546649%2flet-a-n-int-limits-0n-e-x4-dx-does-a-n-n-rightarrow-inf%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            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
            Read more

            Marschland

            Image
            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
            Read more

            Dieringhausen

            Image
            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
            Read more