how to write the general formula for a telescoping series












1












$begingroup$


Im trying to check if the series



$sum_{k=2}^{infty} frac{1}{sqrt{k-1}} - frac{1}{sqrt{k+1}} $



is converging or not.



I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.



For the series with even k >=2 I have started writing down the terms:



$S_k = (1-frac{1}{sqrt{3}})+ (frac{1}{sqrt{3}}-frac{1}{sqrt{5}})+(frac{1}{sqrt{5}}-frac{1}{sqrt{7}})+...$



But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.






sequences-and-series convergence summation telescopic-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
    $endgroup$
    – Matti P.
    14 hours ago
















1












$begingroup$


Im trying to check if the series



$sum_{k=2}^{infty} frac{1}{sqrt{k-1}} - frac{1}{sqrt{k+1}} $



is converging or not.



I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.



For the series with even k >=2 I have started writing down the terms:



$S_k = (1-frac{1}{sqrt{3}})+ (frac{1}{sqrt{3}}-frac{1}{sqrt{5}})+(frac{1}{sqrt{5}}-frac{1}{sqrt{7}})+...$



But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.






sequences-and-series convergence summation telescopic-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
    $endgroup$
    – Matti P.
    14 hours ago














1












1








1





$begingroup$


Im trying to check if the series



$sum_{k=2}^{infty} frac{1}{sqrt{k-1}} - frac{1}{sqrt{k+1}} $



is converging or not.



I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.



For the series with even k >=2 I have started writing down the terms:



$S_k = (1-frac{1}{sqrt{3}})+ (frac{1}{sqrt{3}}-frac{1}{sqrt{5}})+(frac{1}{sqrt{5}}-frac{1}{sqrt{7}})+...$



But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.






sequences-and-series convergence summation telescopic-series







share|cite|improve this question











$endgroup$




Im trying to check if the series



$sum_{k=2}^{infty} frac{1}{sqrt{k-1}} - frac{1}{sqrt{k+1}} $



is converging or not.



I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.



For the series with even k >=2 I have started writing down the terms:



$S_k = (1-frac{1}{sqrt{3}})+ (frac{1}{sqrt{3}}-frac{1}{sqrt{5}})+(frac{1}{sqrt{5}}-frac{1}{sqrt{7}})+...$



But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.






sequences-and-series convergence summation telescopic-series




sequences-and-series convergence summation telescopic-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 13 hours ago









José Carlos Santos

172k23132240




172k23132240










asked 14 hours ago









F WiF Wi

454




454








  • 1




    $begingroup$
    Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
    $endgroup$
    – Matti P.
    14 hours ago














  • 1




    $begingroup$
    Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
    $endgroup$
    – Matti P.
    14 hours ago








1




1




$begingroup$
Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
$endgroup$
– Matti P.
14 hours ago




$begingroup$
Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
$endgroup$
– Matti P.
14 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Since$$sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt{k+1}}right)=sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt k}right)+sum_{k=2}^inftyleft(frac1{sqrt k}-frac1{sqrt{k+1}}right),$$the sum of your series is $1+dfrac1{sqrt2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
    $endgroup$
    – Evargalo
    9 hours ago








  • 1




    $begingroup$
    @Evargalo I agree that it is a good example for that discussion.
    $endgroup$
    – José Carlos Santos
    8 hours ago



















4












$begingroup$

To remain on the safe side of strictness consider the partial sum



$$s(n) = sum_{k=2}^n left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k+1}}right)$$



and then look for the limit $ntoinfty$.



We have



$$s(n) = left( frac{1}{sqrt{1}} +frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+ ...+ frac{1}{sqrt{n-1}}right)\
-left( frac{1}{sqrt{3}} +frac{1}{sqrt{4}}+ ...+ frac{1}{sqrt{n-1}}+frac{1}{sqrt{n}}+frac{1}{sqrt{n+1}}right)\
= 1+frac{1}{sqrt{2}} -frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} $$



and we get finally



$$lim_{nto infty } , s(n) =1+frac{1}{sqrt{2}} $$



EDIT



The generalization to an arbitrary sequence ${a(k)}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by



$$s(d,n) = sum_{k=1}^n left( a_{n}-a_{k+d} right)= sum_{j=1}^d a_j-sum_{j=1}^d a_{n+j}$$



And if $lim_{nto infty } , a_n =0$ we find



$$s(d) = lim_{nto infty } , s(d,n) =sum_{j=1}^d a_j$$






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%2f3172963%2fhow-to-write-the-general-formula-for-a-telescoping-series%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









    2












    $begingroup$

    Since$$sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt{k+1}}right)=sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt k}right)+sum_{k=2}^inftyleft(frac1{sqrt k}-frac1{sqrt{k+1}}right),$$the sum of your series is $1+dfrac1{sqrt2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
      $endgroup$
      – Evargalo
      9 hours ago








    • 1




      $begingroup$
      @Evargalo I agree that it is a good example for that discussion.
      $endgroup$
      – José Carlos Santos
      8 hours ago
















    2












    $begingroup$

    Since$$sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt{k+1}}right)=sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt k}right)+sum_{k=2}^inftyleft(frac1{sqrt k}-frac1{sqrt{k+1}}right),$$the sum of your series is $1+dfrac1{sqrt2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
      $endgroup$
      – Evargalo
      9 hours ago








    • 1




      $begingroup$
      @Evargalo I agree that it is a good example for that discussion.
      $endgroup$
      – José Carlos Santos
      8 hours ago














    2












    2








    2





    $begingroup$

    Since$$sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt{k+1}}right)=sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt k}right)+sum_{k=2}^inftyleft(frac1{sqrt k}-frac1{sqrt{k+1}}right),$$the sum of your series is $1+dfrac1{sqrt2}$.






    share|cite|improve this answer









    $endgroup$



    Since$$sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt{k+1}}right)=sum_{k=2}^inftyleft(frac1{sqrt{k-1}}-frac1{sqrt k}right)+sum_{k=2}^inftyleft(frac1{sqrt k}-frac1{sqrt{k+1}}right),$$the sum of your series is $1+dfrac1{sqrt2}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 14 hours ago









    José Carlos SantosJosé Carlos Santos

    172k23132240




    172k23132240












    • $begingroup$
      Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
      $endgroup$
      – Evargalo
      9 hours ago








    • 1




      $begingroup$
      @Evargalo I agree that it is a good example for that discussion.
      $endgroup$
      – José Carlos Santos
      8 hours ago


















    • $begingroup$
      Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
      $endgroup$
      – Evargalo
      9 hours ago








    • 1




      $begingroup$
      @Evargalo I agree that it is a good example for that discussion.
      $endgroup$
      – José Carlos Santos
      8 hours ago
















    $begingroup$
    Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
    $endgroup$
    – Evargalo
    9 hours ago






    $begingroup$
    Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
    $endgroup$
    – Evargalo
    9 hours ago






    1




    1




    $begingroup$
    @Evargalo I agree that it is a good example for that discussion.
    $endgroup$
    – José Carlos Santos
    8 hours ago




    $begingroup$
    @Evargalo I agree that it is a good example for that discussion.
    $endgroup$
    – José Carlos Santos
    8 hours ago











    4












    $begingroup$

    To remain on the safe side of strictness consider the partial sum



    $$s(n) = sum_{k=2}^n left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k+1}}right)$$



    and then look for the limit $ntoinfty$.



    We have



    $$s(n) = left( frac{1}{sqrt{1}} +frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+ ...+ frac{1}{sqrt{n-1}}right)\
    -left( frac{1}{sqrt{3}} +frac{1}{sqrt{4}}+ ...+ frac{1}{sqrt{n-1}}+frac{1}{sqrt{n}}+frac{1}{sqrt{n+1}}right)\
    = 1+frac{1}{sqrt{2}} -frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} $$



    and we get finally



    $$lim_{nto infty } , s(n) =1+frac{1}{sqrt{2}} $$



    EDIT



    The generalization to an arbitrary sequence ${a(k)}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by



    $$s(d,n) = sum_{k=1}^n left( a_{n}-a_{k+d} right)= sum_{j=1}^d a_j-sum_{j=1}^d a_{n+j}$$



    And if $lim_{nto infty } , a_n =0$ we find



    $$s(d) = lim_{nto infty } , s(d,n) =sum_{j=1}^d a_j$$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      To remain on the safe side of strictness consider the partial sum



      $$s(n) = sum_{k=2}^n left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k+1}}right)$$



      and then look for the limit $ntoinfty$.



      We have



      $$s(n) = left( frac{1}{sqrt{1}} +frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+ ...+ frac{1}{sqrt{n-1}}right)\
      -left( frac{1}{sqrt{3}} +frac{1}{sqrt{4}}+ ...+ frac{1}{sqrt{n-1}}+frac{1}{sqrt{n}}+frac{1}{sqrt{n+1}}right)\
      = 1+frac{1}{sqrt{2}} -frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} $$



      and we get finally



      $$lim_{nto infty } , s(n) =1+frac{1}{sqrt{2}} $$



      EDIT



      The generalization to an arbitrary sequence ${a(k)}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by



      $$s(d,n) = sum_{k=1}^n left( a_{n}-a_{k+d} right)= sum_{j=1}^d a_j-sum_{j=1}^d a_{n+j}$$



      And if $lim_{nto infty } , a_n =0$ we find



      $$s(d) = lim_{nto infty } , s(d,n) =sum_{j=1}^d a_j$$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        To remain on the safe side of strictness consider the partial sum



        $$s(n) = sum_{k=2}^n left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k+1}}right)$$



        and then look for the limit $ntoinfty$.



        We have



        $$s(n) = left( frac{1}{sqrt{1}} +frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+ ...+ frac{1}{sqrt{n-1}}right)\
        -left( frac{1}{sqrt{3}} +frac{1}{sqrt{4}}+ ...+ frac{1}{sqrt{n-1}}+frac{1}{sqrt{n}}+frac{1}{sqrt{n+1}}right)\
        = 1+frac{1}{sqrt{2}} -frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} $$



        and we get finally



        $$lim_{nto infty } , s(n) =1+frac{1}{sqrt{2}} $$



        EDIT



        The generalization to an arbitrary sequence ${a(k)}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by



        $$s(d,n) = sum_{k=1}^n left( a_{n}-a_{k+d} right)= sum_{j=1}^d a_j-sum_{j=1}^d a_{n+j}$$



        And if $lim_{nto infty } , a_n =0$ we find



        $$s(d) = lim_{nto infty } , s(d,n) =sum_{j=1}^d a_j$$






        share|cite|improve this answer











        $endgroup$



        To remain on the safe side of strictness consider the partial sum



        $$s(n) = sum_{k=2}^n left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k+1}}right)$$



        and then look for the limit $ntoinfty$.



        We have



        $$s(n) = left( frac{1}{sqrt{1}} +frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+ ...+ frac{1}{sqrt{n-1}}right)\
        -left( frac{1}{sqrt{3}} +frac{1}{sqrt{4}}+ ...+ frac{1}{sqrt{n-1}}+frac{1}{sqrt{n}}+frac{1}{sqrt{n+1}}right)\
        = 1+frac{1}{sqrt{2}} -frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} $$



        and we get finally



        $$lim_{nto infty } , s(n) =1+frac{1}{sqrt{2}} $$



        EDIT



        The generalization to an arbitrary sequence ${a(k)}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by



        $$s(d,n) = sum_{k=1}^n left( a_{n}-a_{k+d} right)= sum_{j=1}^d a_j-sum_{j=1}^d a_{n+j}$$



        And if $lim_{nto infty } , a_n =0$ we find



        $$s(d) = lim_{nto infty } , s(d,n) =sum_{j=1}^d a_j$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 8 hours ago

























        answered 14 hours ago









        Dr. Wolfgang HintzeDr. Wolfgang Hintze

        3,880620




        3,880620






























            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%2f3172963%2fhow-to-write-the-general-formula-for-a-telescoping-series%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

            Lallio

            Image
            Coordenadas: 45° 40' N 9° 38' E Lallio      Comuna    Lallio Localização de Lallio na Itália Coordenadas 45° 40' N 9° 38' E Região Lombardia Província Bérgamo Área - Total 2 km² Altitude 216 m População  - Total 3 826     • Densidade 1 913 hab./km² Outros dados Comunas limítrofes Bergamo, Dalmine, Stezzano, Treviolo Código ISTAT 016123 Código cadastral E422 Código postal 24040 Prefixo telefônico 035 Website www.comune.lallio.bg.it Lallio é uma comuna italiana da região da Lombardia, província de Bérgamo, com cerca de 3.826 habitantes. Estende-se por uma área de 2 km², tendo uma densidade populacional de 1913 hab/km². Faz fronteira com Bergamo, Dalmine, Stezzano, Treviolo. [ 1 ] [ 2 ] [ 3 ] Demografia | Variação demográfica do município entre 1861 e 2011 [ 3 ] Fonte : Istituto Nazionale di Statistica (ISTAT) - Elabor...
            Read more

            Futebolista

            Image
            Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
            Read more

            Jornalista

            Image
            Uma repórter de televisão segurando um microfone na frente de um cinegrafista. Fotojornalistas no Campeonato Mundial de Atletismo 2013 Uma repórter entrevistando Boris Johnson, Prefeito de Londres. Jornalista é o profissional formado em Jornalismo. É a pessoa responsável pela apuração, investigação e apresentação de notícias, reportagens, entrevistas ou distribuição de notícias ou outra informação de interesse coletivo. O trabalho do jornalista é chamado jornalismo. Um jornalista pode trabalhar com questões gerais ou especializar-se em determinadas áreas. No entanto, a maioria dos jornalistas tendem a se especializar, e cooperando com outros jornalistas, produzir publicações que abrangem muitos tópicos. [ 1 ] Por exemplo, um jornalista esportivo cobre notícias dentro do mundo dos esportes, mas este jornalista pode ser uma parte de um jornal que cobre diversos temas. O exercício do Jornalismo é privativo de jornalista. Entre as áreas em que o jornalista trabalha estão o...
            Read more