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

3528




3528












  • $begingroup$
    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











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    2 Answers
    2






    active

    oldest

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    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

            134k1278176




            134k1278176












            • $begingroup$
              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


















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