Problem related to convergence of a series












1












$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04
















1












$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04














1












1








1





$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







share|cite|improve this question











$endgroup$




Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 6:14









Bungo

13.7k22148




13.7k22148










asked Dec 17 '18 at 5:13









blue boyblue boy

1,241613




1,241613








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04














  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04








1




1




$begingroup$
You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 6:02




$begingroup$
You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 6:02












$begingroup$
Oh my bad . Sorry.
$endgroup$
– blue boy
Dec 17 '18 at 6:04




$begingroup$
Oh my bad . Sorry.
$endgroup$
– blue boy
Dec 17 '18 at 6:04










2 Answers
2






active

oldest

votes


















1












$begingroup$

The idea is correct but we don't need to use absolute convergence, we have indeed that



$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



and both series converge.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Errors:





    1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





      1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




    It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



    It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You had the right idea but the devil is in the details.
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:14










    • $begingroup$
      I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
      $endgroup$
      – blue boy
      Dec 17 '18 at 6:15











    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%2f3043556%2fproblem-related-to-convergence-of-a-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









    1












    $begingroup$

    The idea is correct but we don't need to use absolute convergence, we have indeed that



    $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



    and both series converge.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The idea is correct but we don't need to use absolute convergence, we have indeed that



      $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



      and both series converge.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The idea is correct but we don't need to use absolute convergence, we have indeed that



        $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



        and both series converge.






        share|cite|improve this answer









        $endgroup$



        The idea is correct but we don't need to use absolute convergence, we have indeed that



        $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



        and both series converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 7:34









        gimusigimusi

        92.9k84494




        92.9k84494























            2












            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15
















            2












            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15














            2












            2








            2





            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$



            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 6:11









            DanielWainfleetDanielWainfleet

            35.2k31648




            35.2k31648












            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15


















            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15
















            $begingroup$
            You had the right idea but the devil is in the details.
            $endgroup$
            – DanielWainfleet
            Dec 17 '18 at 6:14




            $begingroup$
            You had the right idea but the devil is in the details.
            $endgroup$
            – DanielWainfleet
            Dec 17 '18 at 6:14












            $begingroup$
            I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
            $endgroup$
            – blue boy
            Dec 17 '18 at 6:15




            $begingroup$
            I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
            $endgroup$
            – blue boy
            Dec 17 '18 at 6:15


















            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%2f3043556%2fproblem-related-to-convergence-of-a-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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more