Asymptotic formula for $sum_{nle x}frac 1n$












2












$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56


















2












$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56
















2












2








2


1



$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







share|cite|improve this question











$endgroup$





Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory




complex-analysis number-theory analytic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 19:29







Topo

















asked Dec 25 '18 at 4:59









TopoTopo

330314




330314












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56




















  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56


















$begingroup$
I don't understand the thing in yellow. I've never seen all big Os in an equation
$endgroup$
– mathworker21
Dec 25 '18 at 5:35




$begingroup$
I don't understand the thing in yellow. I've never seen all big Os in an equation
$endgroup$
– mathworker21
Dec 25 '18 at 5:35












$begingroup$
@mathworker21 Fixed.
$endgroup$
– Topo
Dec 25 '18 at 5:50




$begingroup$
@mathworker21 Fixed.
$endgroup$
– Topo
Dec 25 '18 at 5:50












$begingroup$
Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
$endgroup$
– mathworker21
Dec 25 '18 at 5:56






$begingroup$
Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
$endgroup$
– mathworker21
Dec 25 '18 at 5:56












1 Answer
1






active

oldest

votes


















4












$begingroup$

All you need is
$log(1+z) < z$
for $z > 0$.



Then
$log(x+2)-log(x)
=log(1+frac{2}{x})
lt frac{2}{x}
=O(frac1{x})
$.



To show
$log(1+z) < z$
for $z > 0$:



$log(1+z)
=int_0^z frac{dt}{1+t}
lt int_0^z dt
=z
$.



You can also use the integral
to get a lower bound:



$begin{array}\
log(1+z)
&=int_0^z frac{dt}{1+t}\
&gt int_0^z frac{dt}{1+z}\
&=frac{z}{1+z}\
&=frac{z+z^2-z^2}{1+z}\
&=z-frac{z^2}{1+z}\
&gt z-z^2\
end{array}
$






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3051851%2fasymptotic-formula-for-sum-n-le-x-frac-1n%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    All you need is
    $log(1+z) < z$
    for $z > 0$.



    Then
    $log(x+2)-log(x)
    =log(1+frac{2}{x})
    lt frac{2}{x}
    =O(frac1{x})
    $    .



    To show
    $log(1+z) < z$
    for $z > 0$:



    $log(1+z)
    =int_0^z frac{dt}{1+t}
    lt int_0^z dt
    =z
    $    .



    You can also use the integral
    to get a lower bound:



    $begin{array}\
    log(1+z)
    &=int_0^z frac{dt}{1+t}\
    &gt int_0^z frac{dt}{1+z}\
    &=frac{z}{1+z}\
    &=frac{z+z^2-z^2}{1+z}\
    &=z-frac{z^2}{1+z}\
    &gt z-z^2\
    end{array}
    $






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      All you need is
      $log(1+z) < z$
      for $z > 0$.



      Then
      $log(x+2)-log(x)
      =log(1+frac{2}{x})
      lt frac{2}{x}
      =O(frac1{x})
      $      .



      To show
      $log(1+z) < z$
      for $z > 0$:



      $log(1+z)
      =int_0^z frac{dt}{1+t}
      lt int_0^z dt
      =z
      $      .



      You can also use the integral
      to get a lower bound:



      $begin{array}\
      log(1+z)
      &=int_0^z frac{dt}{1+t}\
      &gt int_0^z frac{dt}{1+z}\
      &=frac{z}{1+z}\
      &=frac{z+z^2-z^2}{1+z}\
      &=z-frac{z^2}{1+z}\
      &gt z-z^2\
      end{array}
      $






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        All you need is
        $log(1+z) < z$
        for $z > 0$.



        Then
        $log(x+2)-log(x)
        =log(1+frac{2}{x})
        lt frac{2}{x}
        =O(frac1{x})
        $        .



        To show
        $log(1+z) < z$
        for $z > 0$:



        $log(1+z)
        =int_0^z frac{dt}{1+t}
        lt int_0^z dt
        =z
        $        .



        You can also use the integral
        to get a lower bound:



        $begin{array}\
        log(1+z)
        &=int_0^z frac{dt}{1+t}\
        &gt int_0^z frac{dt}{1+z}\
        &=frac{z}{1+z}\
        &=frac{z+z^2-z^2}{1+z}\
        &=z-frac{z^2}{1+z}\
        &gt z-z^2\
        end{array}
        $






        share|cite|improve this answer









        $endgroup$



        All you need is
        $log(1+z) < z$
        for $z > 0$.



        Then
        $log(x+2)-log(x)
        =log(1+frac{2}{x})
        lt frac{2}{x}
        =O(frac1{x})
        $        .



        To show
        $log(1+z) < z$
        for $z > 0$:



        $log(1+z)
        =int_0^z frac{dt}{1+t}
        lt int_0^z dt
        =z
        $        .



        You can also use the integral
        to get a lower bound:



        $begin{array}\
        log(1+z)
        &=int_0^z frac{dt}{1+t}\
        &gt int_0^z frac{dt}{1+z}\
        &=frac{z}{1+z}\
        &=frac{z+z^2-z^2}{1+z}\
        &=z-frac{z^2}{1+z}\
        &gt z-z^2\
        end{array}
        $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 5:09









        marty cohenmarty cohen

        76.1k549130




        76.1k549130






























            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%2f3051851%2fasymptotic-formula-for-sum-n-le-x-frac-1n%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more