Convergence of the series $ sumfrac{1}{a_na_{n+1}} $ if the sequence $(a_n)$ is arithmetic












3












$begingroup$


We know that $(a_n)$ is arithmetic progression and I need to decide if
$$ sum_{n=1}^{infty} frac{1}{a_n cdot a_{n+1}} $$
convergences.
Firstly I check necessary condition:
$ frac{1}{a_n cdot a_{n+1}} rightarrow 0 $
It is true because $$ a_n rightarrow +/- infty $$ because $ a_n = a + (n-1)r $



Ok, now I should check convergence:
Ratio test:
$$ frac{u_{n+1}}{u_n} = frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} rightarrow 1 $$
So it gives me nothing.
Root tests:
$$ sqrt[n]{frac{1}{a_n cdot a_{n+1}}} rightarrow 1 $$
so it fails too...
Have somebody any idea how can I check this?






real-analysis arithmetic-progressions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
    $endgroup$
    – Thomas Andrews
    Dec 13 '18 at 17:51


















3












$begingroup$


We know that $(a_n)$ is arithmetic progression and I need to decide if
$$ sum_{n=1}^{infty} frac{1}{a_n cdot a_{n+1}} $$
convergences.
Firstly I check necessary condition:
$ frac{1}{a_n cdot a_{n+1}} rightarrow 0 $
It is true because $$ a_n rightarrow +/- infty $$ because $ a_n = a + (n-1)r $



Ok, now I should check convergence:
Ratio test:
$$ frac{u_{n+1}}{u_n} = frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} rightarrow 1 $$
So it gives me nothing.
Root tests:
$$ sqrt[n]{frac{1}{a_n cdot a_{n+1}}} rightarrow 1 $$
so it fails too...
Have somebody any idea how can I check this?






real-analysis arithmetic-progressions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
    $endgroup$
    – Thomas Andrews
    Dec 13 '18 at 17:51
















3












3








3


1



$begingroup$


We know that $(a_n)$ is arithmetic progression and I need to decide if
$$ sum_{n=1}^{infty} frac{1}{a_n cdot a_{n+1}} $$
convergences.
Firstly I check necessary condition:
$ frac{1}{a_n cdot a_{n+1}} rightarrow 0 $
It is true because $$ a_n rightarrow +/- infty $$ because $ a_n = a + (n-1)r $



Ok, now I should check convergence:
Ratio test:
$$ frac{u_{n+1}}{u_n} = frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} rightarrow 1 $$
So it gives me nothing.
Root tests:
$$ sqrt[n]{frac{1}{a_n cdot a_{n+1}}} rightarrow 1 $$
so it fails too...
Have somebody any idea how can I check this?






real-analysis arithmetic-progressions







share|cite|improve this question











$endgroup$




We know that $(a_n)$ is arithmetic progression and I need to decide if
$$ sum_{n=1}^{infty} frac{1}{a_n cdot a_{n+1}} $$
convergences.
Firstly I check necessary condition:
$ frac{1}{a_n cdot a_{n+1}} rightarrow 0 $
It is true because $$ a_n rightarrow +/- infty $$ because $ a_n = a + (n-1)r $



Ok, now I should check convergence:
Ratio test:
$$ frac{u_{n+1}}{u_n} = frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} rightarrow 1 $$
So it gives me nothing.
Root tests:
$$ sqrt[n]{frac{1}{a_n cdot a_{n+1}}} rightarrow 1 $$
so it fails too...
Have somebody any idea how can I check this?






real-analysis arithmetic-progressions




real-analysis arithmetic-progressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 17:53









Did

248k23223460




248k23223460










asked Dec 13 '18 at 17:49









VirtualUserVirtualUser

81014




81014








  • 1




    $begingroup$
    Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
    $endgroup$
    – Thomas Andrews
    Dec 13 '18 at 17:51
















  • 1




    $begingroup$
    Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
    $endgroup$
    – Thomas Andrews
    Dec 13 '18 at 17:51










1




1




$begingroup$
Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
$endgroup$
– Thomas Andrews
Dec 13 '18 at 17:51






$begingroup$
Hint: $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$
$endgroup$
– Thomas Andrews
Dec 13 '18 at 17:51












2 Answers
2






active

oldest

votes


















2












$begingroup$

First note that $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$



Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:



$$sum_{n=1}^{N}frac{1}{a_na_{n+1}}=frac{1}{d}left(frac{1}{a_1}-frac{1}{a_{N+1}}right)$$



Which means the series converges to...



We have to assume the $a_ineq 0$ and that $dneq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.




    To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then



    $$frac{1}{a_n cdot a_{n + 1}} = frac{1}{c^2 n^2 + text{ lower order}}$$



    and the comparison is quick from here.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Indeed -- but see @ThomasAndrews' comment on main...
      $endgroup$
      – Did
      Dec 13 '18 at 17:54











    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%2f3038365%2fconvergence-of-the-series-sum-frac1a-na-n1-if-the-sequence-a-n-i%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












    $begingroup$

    First note that $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$



    Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:



    $$sum_{n=1}^{N}frac{1}{a_na_{n+1}}=frac{1}{d}left(frac{1}{a_1}-frac{1}{a_{N+1}}right)$$



    Which means the series converges to...



    We have to assume the $a_ineq 0$ and that $dneq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      First note that $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$



      Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:



      $$sum_{n=1}^{N}frac{1}{a_na_{n+1}}=frac{1}{d}left(frac{1}{a_1}-frac{1}{a_{N+1}}right)$$



      Which means the series converges to...



      We have to assume the $a_ineq 0$ and that $dneq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        First note that $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$



        Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:



        $$sum_{n=1}^{N}frac{1}{a_na_{n+1}}=frac{1}{d}left(frac{1}{a_1}-frac{1}{a_{N+1}}right)$$



        Which means the series converges to...



        We have to assume the $a_ineq 0$ and that $dneq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.






        share|cite|improve this answer









        $endgroup$



        First note that $$frac{1}{a_na_{n+1}}=frac{1}{a_{n+1}-a_n}cdot left(frac{1}{a_{n}}-frac{1}{a_{n+1}}right)$$



        Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:



        $$sum_{n=1}^{N}frac{1}{a_na_{n+1}}=frac{1}{d}left(frac{1}{a_1}-frac{1}{a_{N+1}}right)$$



        Which means the series converges to...



        We have to assume the $a_ineq 0$ and that $dneq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 17:55









        Thomas AndrewsThomas Andrews

        130k11146297




        130k11146297























            1












            $begingroup$

            If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.




            To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then



            $$frac{1}{a_n cdot a_{n + 1}} = frac{1}{c^2 n^2 + text{ lower order}}$$



            and the comparison is quick from here.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Indeed -- but see @ThomasAndrews' comment on main...
              $endgroup$
              – Did
              Dec 13 '18 at 17:54
















            1












            $begingroup$

            If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.




            To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then



            $$frac{1}{a_n cdot a_{n + 1}} = frac{1}{c^2 n^2 + text{ lower order}}$$



            and the comparison is quick from here.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Indeed -- but see @ThomasAndrews' comment on main...
              $endgroup$
              – Did
              Dec 13 '18 at 17:54














            1












            1








            1





            $begingroup$

            If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.




            To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then



            $$frac{1}{a_n cdot a_{n + 1}} = frac{1}{c^2 n^2 + text{ lower order}}$$



            and the comparison is quick from here.






            share|cite|improve this answer









            $endgroup$



            If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.




            To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then



            $$frac{1}{a_n cdot a_{n + 1}} = frac{1}{c^2 n^2 + text{ lower order}}$$



            and the comparison is quick from here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 17:51









            T. BongersT. Bongers

            23.1k54662




            23.1k54662








            • 1




              $begingroup$
              Indeed -- but see @ThomasAndrews' comment on main...
              $endgroup$
              – Did
              Dec 13 '18 at 17:54














            • 1




              $begingroup$
              Indeed -- but see @ThomasAndrews' comment on main...
              $endgroup$
              – Did
              Dec 13 '18 at 17:54








            1




            1




            $begingroup$
            Indeed -- but see @ThomasAndrews' comment on main...
            $endgroup$
            – Did
            Dec 13 '18 at 17:54




            $begingroup$
            Indeed -- but see @ThomasAndrews' comment on main...
            $endgroup$
            – Did
            Dec 13 '18 at 17:54


















            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%2f3038365%2fconvergence-of-the-series-sum-frac1a-na-n1-if-the-sequence-a-n-i%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