Convergence test $sum_{n=1}^{+infty}(-1)^{leftlfloor{sqrt{n}}rightrfloor}frac{1}{ln(n)}$












1












$begingroup$


How can I try to check the convergence of series
$$
sum_{n=2}^{+infty}(-1)^{leftlfloor{sqrt{n}}rightrfloor}frac{1}{ln(n)}
$$






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure the summation starts from $ n = 1$ ?
    $endgroup$
    – Ahmad Bazzi
    Dec 19 '18 at 19:20










  • $begingroup$
    No, from $n=2$, thanks
    $endgroup$
    – avan1235
    Dec 19 '18 at 19:21
















1












$begingroup$


How can I try to check the convergence of series
$$
sum_{n=2}^{+infty}(-1)^{leftlfloor{sqrt{n}}rightrfloor}frac{1}{ln(n)}
$$






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure the summation starts from $ n = 1$ ?
    $endgroup$
    – Ahmad Bazzi
    Dec 19 '18 at 19:20










  • $begingroup$
    No, from $n=2$, thanks
    $endgroup$
    – avan1235
    Dec 19 '18 at 19:21














1












1








1





$begingroup$


How can I try to check the convergence of series
$$
sum_{n=2}^{+infty}(-1)^{leftlfloor{sqrt{n}}rightrfloor}frac{1}{ln(n)}
$$






sequences-and-series convergence







share|cite|improve this question











$endgroup$




How can I try to check the convergence of series
$$
sum_{n=2}^{+infty}(-1)^{leftlfloor{sqrt{n}}rightrfloor}frac{1}{ln(n)}
$$






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 19:21







avan1235

















asked Dec 19 '18 at 19:16









avan1235avan1235

3297




3297












  • $begingroup$
    Are you sure the summation starts from $ n = 1$ ?
    $endgroup$
    – Ahmad Bazzi
    Dec 19 '18 at 19:20










  • $begingroup$
    No, from $n=2$, thanks
    $endgroup$
    – avan1235
    Dec 19 '18 at 19:21


















  • $begingroup$
    Are you sure the summation starts from $ n = 1$ ?
    $endgroup$
    – Ahmad Bazzi
    Dec 19 '18 at 19:20










  • $begingroup$
    No, from $n=2$, thanks
    $endgroup$
    – avan1235
    Dec 19 '18 at 19:21
















$begingroup$
Are you sure the summation starts from $ n = 1$ ?
$endgroup$
– Ahmad Bazzi
Dec 19 '18 at 19:20




$begingroup$
Are you sure the summation starts from $ n = 1$ ?
$endgroup$
– Ahmad Bazzi
Dec 19 '18 at 19:20












$begingroup$
No, from $n=2$, thanks
$endgroup$
– avan1235
Dec 19 '18 at 19:21




$begingroup$
No, from $n=2$, thanks
$endgroup$
– avan1235
Dec 19 '18 at 19:21










2 Answers
2






active

oldest

votes


















2












$begingroup$

Use Cauchy's convergence test and consider the following subsums
$$sum_{n=k^2}^{k^2+2k} (-1)^{lfloor sqrt n rfloor} frac{1}{ln(n)}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We can write the partial sums of the series of interest as



    $$begin{align}
    sum_{n=4}^{(N+1)^2-1} frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}&=sum_{m=2}^N sum_{n=m^2}^{(m+1)^2-1}frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}\\
    &=sum_{m=2}^N (-1)^m sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}tag1
    end{align}$$



    It is sufficient to show that the sequence $S_m=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $lim_{mtoinfty}S_mne0$). Proceeding we have the crude estimates



    $$begin{align}
    S_m-S_{m-1}&=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}-sum_{n=(m-1)^2}^{m^2-1} frac{1}{log(n)}\\
    &ge frac{2m+1}{2log(m+1)}-frac{2m-1}{2log(m-1)}\\
    &=frac{(m+1/2)log(m-1)-(m-1/2)log(m+1)}{log(m+1)log(m-1)}tag2
    end{align}$$



    We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$, $T_m=(m+1/2)log(m-1)-(m-1/2)log(m+1)$ as



    $$begin{align}
    T_m&=(m+1/2)log(m-1)-(m-1/2)log(m+1)&\\
    &=log(m)+mlogleft(1-frac2{m+1}right)+frac12logleft(1-frac1{m^2}right)\\
    &ge log(m)-frac{2m}{m-1}-frac1{m^2}tag3
    end{align}$$



    From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.



    And we are done!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Just curious ... was there a reason that you revoked accepting this answer?
      $endgroup$
      – Mark Viola
      Dec 23 '18 at 17:00











    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%2f3046773%2fconvergence-test-sum-n-1-infty-1-left-lfloor-sqrtn-right-rfloor%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












    $begingroup$

    Use Cauchy's convergence test and consider the following subsums
    $$sum_{n=k^2}^{k^2+2k} (-1)^{lfloor sqrt n rfloor} frac{1}{ln(n)}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Use Cauchy's convergence test and consider the following subsums
      $$sum_{n=k^2}^{k^2+2k} (-1)^{lfloor sqrt n rfloor} frac{1}{ln(n)}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Use Cauchy's convergence test and consider the following subsums
        $$sum_{n=k^2}^{k^2+2k} (-1)^{lfloor sqrt n rfloor} frac{1}{ln(n)}$$






        share|cite|improve this answer









        $endgroup$



        Use Cauchy's convergence test and consider the following subsums
        $$sum_{n=k^2}^{k^2+2k} (-1)^{lfloor sqrt n rfloor} frac{1}{ln(n)}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 20:08









        Мокин ВасилийМокин Василий

        361




        361























            1












            $begingroup$

            We can write the partial sums of the series of interest as



            $$begin{align}
            sum_{n=4}^{(N+1)^2-1} frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}&=sum_{m=2}^N sum_{n=m^2}^{(m+1)^2-1}frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}\\
            &=sum_{m=2}^N (-1)^m sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}tag1
            end{align}$$



            It is sufficient to show that the sequence $S_m=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $lim_{mtoinfty}S_mne0$). Proceeding we have the crude estimates



            $$begin{align}
            S_m-S_{m-1}&=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}-sum_{n=(m-1)^2}^{m^2-1} frac{1}{log(n)}\\
            &ge frac{2m+1}{2log(m+1)}-frac{2m-1}{2log(m-1)}\\
            &=frac{(m+1/2)log(m-1)-(m-1/2)log(m+1)}{log(m+1)log(m-1)}tag2
            end{align}$$



            We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$, $T_m=(m+1/2)log(m-1)-(m-1/2)log(m+1)$ as



            $$begin{align}
            T_m&=(m+1/2)log(m-1)-(m-1/2)log(m+1)&\\
            &=log(m)+mlogleft(1-frac2{m+1}right)+frac12logleft(1-frac1{m^2}right)\\
            &ge log(m)-frac{2m}{m-1}-frac1{m^2}tag3
            end{align}$$



            From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.



            And we are done!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Just curious ... was there a reason that you revoked accepting this answer?
              $endgroup$
              – Mark Viola
              Dec 23 '18 at 17:00
















            1












            $begingroup$

            We can write the partial sums of the series of interest as



            $$begin{align}
            sum_{n=4}^{(N+1)^2-1} frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}&=sum_{m=2}^N sum_{n=m^2}^{(m+1)^2-1}frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}\\
            &=sum_{m=2}^N (-1)^m sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}tag1
            end{align}$$



            It is sufficient to show that the sequence $S_m=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $lim_{mtoinfty}S_mne0$). Proceeding we have the crude estimates



            $$begin{align}
            S_m-S_{m-1}&=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}-sum_{n=(m-1)^2}^{m^2-1} frac{1}{log(n)}\\
            &ge frac{2m+1}{2log(m+1)}-frac{2m-1}{2log(m-1)}\\
            &=frac{(m+1/2)log(m-1)-(m-1/2)log(m+1)}{log(m+1)log(m-1)}tag2
            end{align}$$



            We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$, $T_m=(m+1/2)log(m-1)-(m-1/2)log(m+1)$ as



            $$begin{align}
            T_m&=(m+1/2)log(m-1)-(m-1/2)log(m+1)&\\
            &=log(m)+mlogleft(1-frac2{m+1}right)+frac12logleft(1-frac1{m^2}right)\\
            &ge log(m)-frac{2m}{m-1}-frac1{m^2}tag3
            end{align}$$



            From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.



            And we are done!






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Just curious ... was there a reason that you revoked accepting this answer?
              $endgroup$
              – Mark Viola
              Dec 23 '18 at 17:00














            1












            1








            1





            $begingroup$

            We can write the partial sums of the series of interest as



            $$begin{align}
            sum_{n=4}^{(N+1)^2-1} frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}&=sum_{m=2}^N sum_{n=m^2}^{(m+1)^2-1}frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}\\
            &=sum_{m=2}^N (-1)^m sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}tag1
            end{align}$$



            It is sufficient to show that the sequence $S_m=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $lim_{mtoinfty}S_mne0$). Proceeding we have the crude estimates



            $$begin{align}
            S_m-S_{m-1}&=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}-sum_{n=(m-1)^2}^{m^2-1} frac{1}{log(n)}\\
            &ge frac{2m+1}{2log(m+1)}-frac{2m-1}{2log(m-1)}\\
            &=frac{(m+1/2)log(m-1)-(m-1/2)log(m+1)}{log(m+1)log(m-1)}tag2
            end{align}$$



            We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$, $T_m=(m+1/2)log(m-1)-(m-1/2)log(m+1)$ as



            $$begin{align}
            T_m&=(m+1/2)log(m-1)-(m-1/2)log(m+1)&\\
            &=log(m)+mlogleft(1-frac2{m+1}right)+frac12logleft(1-frac1{m^2}right)\\
            &ge log(m)-frac{2m}{m-1}-frac1{m^2}tag3
            end{align}$$



            From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.



            And we are done!






            share|cite|improve this answer











            $endgroup$



            We can write the partial sums of the series of interest as



            $$begin{align}
            sum_{n=4}^{(N+1)^2-1} frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}&=sum_{m=2}^N sum_{n=m^2}^{(m+1)^2-1}frac{(-1)^{lfloor sqrt nrfloor}}{log(n)}\\
            &=sum_{m=2}^N (-1)^m sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}tag1
            end{align}$$



            It is sufficient to show that the sequence $S_m=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $lim_{mtoinfty}S_mne0$). Proceeding we have the crude estimates



            $$begin{align}
            S_m-S_{m-1}&=sum_{n=m^2}^{(m+1)^2-1} frac{1}{log(n)}-sum_{n=(m-1)^2}^{m^2-1} frac{1}{log(n)}\\
            &ge frac{2m+1}{2log(m+1)}-frac{2m-1}{2log(m-1)}\\
            &=frac{(m+1/2)log(m-1)-(m-1/2)log(m+1)}{log(m+1)log(m-1)}tag2
            end{align}$$



            We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$, $T_m=(m+1/2)log(m-1)-(m-1/2)log(m+1)$ as



            $$begin{align}
            T_m&=(m+1/2)log(m-1)-(m-1/2)log(m+1)&\\
            &=log(m)+mlogleft(1-frac2{m+1}right)+frac12logleft(1-frac1{m^2}right)\\
            &ge log(m)-frac{2m}{m-1}-frac1{m^2}tag3
            end{align}$$



            From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.



            And we are done!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 21:32

























            answered Dec 19 '18 at 21:25









            Mark ViolaMark Viola

            133k1277174




            133k1277174












            • $begingroup$
              Just curious ... was there a reason that you revoked accepting this answer?
              $endgroup$
              – Mark Viola
              Dec 23 '18 at 17:00


















            • $begingroup$
              Just curious ... was there a reason that you revoked accepting this answer?
              $endgroup$
              – Mark Viola
              Dec 23 '18 at 17:00
















            $begingroup$
            Just curious ... was there a reason that you revoked accepting this answer?
            $endgroup$
            – Mark Viola
            Dec 23 '18 at 17:00




            $begingroup$
            Just curious ... was there a reason that you revoked accepting this answer?
            $endgroup$
            – Mark Viola
            Dec 23 '18 at 17:00


















            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%2f3046773%2fconvergence-test-sum-n-1-infty-1-left-lfloor-sqrtn-right-rfloor%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