how to show $ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})$...












1












$begingroup$


$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$



I have tried a lot of stuff, didn't work at all . A hint will be good too.



I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    limit comparison?
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 15:46










  • $begingroup$
    yea you're gonna need to use comparison test
    $endgroup$
    – user29418
    Jan 3 at 15:48










  • $begingroup$
    the given series diverges
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 15:53
















1












$begingroup$


$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$



I have tried a lot of stuff, didn't work at all . A hint will be good too.



I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    limit comparison?
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 15:46










  • $begingroup$
    yea you're gonna need to use comparison test
    $endgroup$
    – user29418
    Jan 3 at 15:48










  • $begingroup$
    the given series diverges
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 15:53














1












1








1





$begingroup$


$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$



I have tried a lot of stuff, didn't work at all . A hint will be good too.



I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.






sequences-and-series convergence







share|cite|improve this question











$endgroup$




$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$



I have tried a lot of stuff, didn't work at all . A hint will be good too.



I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.






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 Jan 3 at 16:05









Dando18

4,71741235




4,71741235










asked Jan 3 at 15:43









Mather Mather

4028




4028








  • 4




    $begingroup$
    limit comparison?
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 15:46










  • $begingroup$
    yea you're gonna need to use comparison test
    $endgroup$
    – user29418
    Jan 3 at 15:48










  • $begingroup$
    the given series diverges
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 15:53














  • 4




    $begingroup$
    limit comparison?
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 15:46










  • $begingroup$
    yea you're gonna need to use comparison test
    $endgroup$
    – user29418
    Jan 3 at 15:48










  • $begingroup$
    the given series diverges
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 3 at 15:53








4




4




$begingroup$
limit comparison?
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46




$begingroup$
limit comparison?
$endgroup$
– Lord Shark the Unknown
Jan 3 at 15:46












$begingroup$
yea you're gonna need to use comparison test
$endgroup$
– user29418
Jan 3 at 15:48




$begingroup$
yea you're gonna need to use comparison test
$endgroup$
– user29418
Jan 3 at 15:48












$begingroup$
the given series diverges
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 15:53




$begingroup$
the given series diverges
$endgroup$
– Dr. Sonnhard Graubner
Jan 3 at 15:53










3 Answers
3






active

oldest

votes


















1












$begingroup$

Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For large $n$, your sequence behaves as
    $$frac{1}{nlog n}$$
    Because the following integral diverges,
    $$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
    By the integral test, your series also diverges.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      how did you know that it behaves like that
      $endgroup$
      – Mather
      Jan 3 at 17:04










    • $begingroup$
      Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
      $endgroup$
      – Zachary
      Jan 3 at 17:31












    • $begingroup$
      oh nice fast and subtle , i would have give this one as a very good answer too !
      $endgroup$
      – Mather
      Jan 3 at 17:33










    • $begingroup$
      thank you @Zachary
      $endgroup$
      – Mather
      Jan 3 at 17:34










    • $begingroup$
      No problem! @Mather
      $endgroup$
      – Zachary
      Jan 4 at 1:24



















    1












    $begingroup$

    Hint. One may observe that,
    $$
    lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
    $$    giving, for a certain $n_0ge10$, $nge n_0$,
    $$
    frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
    $$     then, as $n ge 10$,
    $$
    frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
    $$     yielding, for $Nge n_0$,
    $$
    frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
    $$     leading to the divergence of the given series.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
      $endgroup$
      – Mather
      Jan 3 at 17:06












    • $begingroup$
      and for $(2)$ how did you get this inequality
      $endgroup$
      – Mather
      Jan 3 at 17:07












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
    $$
    frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
    $$
    Therefore,
    $$
    sin(x)gefrac{2x}pitag2
    $$
    Since $e^x$ is convex,
    $$
    begin{align}
    frac{e^x-e^0}{x-0}
    &gelim_{tto0}frac{e^t-e^0}{t-0}\
    &=e^0\[6pt]
    &=1tag3
    end{align}
    $$
    Therefore,
    $$
    e^xge1+xtag4
    $$
    and that
    $$
    begin{align}
    x-sqrt{x^2-a}
    &=frac{a}{x+sqrt{x^2-a}}\
    &lefrac axtag5
    end{align}
    $$
    Therefore,
    $$
    sqrt{x^2-a}ge x-frac axtag6
    $$
    Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
    $$
    begin{align}
    frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
    &gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
    &=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
    &gefrac1pifrac1{nlog(n)}tag7
    end{align}
    $$
    The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
      $$
      frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
      $$
      Therefore,
      $$
      sin(x)gefrac{2x}pitag2
      $$
      Since $e^x$ is convex,
      $$
      begin{align}
      frac{e^x-e^0}{x-0}
      &gelim_{tto0}frac{e^t-e^0}{t-0}\
      &=e^0\[6pt]
      &=1tag3
      end{align}
      $$
      Therefore,
      $$
      e^xge1+xtag4
      $$
      and that
      $$
      begin{align}
      x-sqrt{x^2-a}
      &=frac{a}{x+sqrt{x^2-a}}\
      &lefrac axtag5
      end{align}
      $$
      Therefore,
      $$
      sqrt{x^2-a}ge x-frac axtag6
      $$
      Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
      $$
      begin{align}
      frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
      &gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
      &=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
      &gefrac1pifrac1{nlog(n)}tag7
      end{align}
      $$
      The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
        $$
        frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
        $$
        Therefore,
        $$
        sin(x)gefrac{2x}pitag2
        $$
        Since $e^x$ is convex,
        $$
        begin{align}
        frac{e^x-e^0}{x-0}
        &gelim_{tto0}frac{e^t-e^0}{t-0}\
        &=e^0\[6pt]
        &=1tag3
        end{align}
        $$
        Therefore,
        $$
        e^xge1+xtag4
        $$
        and that
        $$
        begin{align}
        x-sqrt{x^2-a}
        &=frac{a}{x+sqrt{x^2-a}}\
        &lefrac axtag5
        end{align}
        $$
        Therefore,
        $$
        sqrt{x^2-a}ge x-frac axtag6
        $$
        Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
        $$
        begin{align}
        frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
        &gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
        &=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
        &gefrac1pifrac1{nlog(n)}tag7
        end{align}
        $$
        The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.






        share|cite|improve this answer









        $endgroup$



        Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
        $$
        frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
        $$
        Therefore,
        $$
        sin(x)gefrac{2x}pitag2
        $$
        Since $e^x$ is convex,
        $$
        begin{align}
        frac{e^x-e^0}{x-0}
        &gelim_{tto0}frac{e^t-e^0}{t-0}\
        &=e^0\[6pt]
        &=1tag3
        end{align}
        $$
        Therefore,
        $$
        e^xge1+xtag4
        $$
        and that
        $$
        begin{align}
        x-sqrt{x^2-a}
        &=frac{a}{x+sqrt{x^2-a}}\
        &lefrac axtag5
        end{align}
        $$
        Therefore,
        $$
        sqrt{x^2-a}ge x-frac axtag6
        $$
        Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
        $$
        begin{align}
        frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
        &gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
        &=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
        &gefrac1pifrac1{nlog(n)}tag7
        end{align}
        $$
        The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 17:09









        robjohn♦robjohn

        270k27312640




        270k27312640























            1












            $begingroup$

            For large $n$, your sequence behaves as
            $$frac{1}{nlog n}$$
            Because the following integral diverges,
            $$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
            By the integral test, your series also diverges.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you know that it behaves like that
              $endgroup$
              – Mather
              Jan 3 at 17:04










            • $begingroup$
              Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
              $endgroup$
              – Zachary
              Jan 3 at 17:31












            • $begingroup$
              oh nice fast and subtle , i would have give this one as a very good answer too !
              $endgroup$
              – Mather
              Jan 3 at 17:33










            • $begingroup$
              thank you @Zachary
              $endgroup$
              – Mather
              Jan 3 at 17:34










            • $begingroup$
              No problem! @Mather
              $endgroup$
              – Zachary
              Jan 4 at 1:24
















            1












            $begingroup$

            For large $n$, your sequence behaves as
            $$frac{1}{nlog n}$$
            Because the following integral diverges,
            $$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
            By the integral test, your series also diverges.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              how did you know that it behaves like that
              $endgroup$
              – Mather
              Jan 3 at 17:04










            • $begingroup$
              Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
              $endgroup$
              – Zachary
              Jan 3 at 17:31












            • $begingroup$
              oh nice fast and subtle , i would have give this one as a very good answer too !
              $endgroup$
              – Mather
              Jan 3 at 17:33










            • $begingroup$
              thank you @Zachary
              $endgroup$
              – Mather
              Jan 3 at 17:34










            • $begingroup$
              No problem! @Mather
              $endgroup$
              – Zachary
              Jan 4 at 1:24














            1












            1








            1





            $begingroup$

            For large $n$, your sequence behaves as
            $$frac{1}{nlog n}$$
            Because the following integral diverges,
            $$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
            By the integral test, your series also diverges.






            share|cite|improve this answer









            $endgroup$



            For large $n$, your sequence behaves as
            $$frac{1}{nlog n}$$
            Because the following integral diverges,
            $$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
            By the integral test, your series also diverges.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 15:58









            ZacharyZachary

            2,3551214




            2,3551214












            • $begingroup$
              how did you know that it behaves like that
              $endgroup$
              – Mather
              Jan 3 at 17:04










            • $begingroup$
              Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
              $endgroup$
              – Zachary
              Jan 3 at 17:31












            • $begingroup$
              oh nice fast and subtle , i would have give this one as a very good answer too !
              $endgroup$
              – Mather
              Jan 3 at 17:33










            • $begingroup$
              thank you @Zachary
              $endgroup$
              – Mather
              Jan 3 at 17:34










            • $begingroup$
              No problem! @Mather
              $endgroup$
              – Zachary
              Jan 4 at 1:24


















            • $begingroup$
              how did you know that it behaves like that
              $endgroup$
              – Mather
              Jan 3 at 17:04










            • $begingroup$
              Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
              $endgroup$
              – Zachary
              Jan 3 at 17:31












            • $begingroup$
              oh nice fast and subtle , i would have give this one as a very good answer too !
              $endgroup$
              – Mather
              Jan 3 at 17:33










            • $begingroup$
              thank you @Zachary
              $endgroup$
              – Mather
              Jan 3 at 17:34










            • $begingroup$
              No problem! @Mather
              $endgroup$
              – Zachary
              Jan 4 at 1:24
















            $begingroup$
            how did you know that it behaves like that
            $endgroup$
            – Mather
            Jan 3 at 17:04




            $begingroup$
            how did you know that it behaves like that
            $endgroup$
            – Mather
            Jan 3 at 17:04












            $begingroup$
            Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
            $endgroup$
            – Zachary
            Jan 3 at 17:31






            $begingroup$
            Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
            $endgroup$
            – Zachary
            Jan 3 at 17:31














            $begingroup$
            oh nice fast and subtle , i would have give this one as a very good answer too !
            $endgroup$
            – Mather
            Jan 3 at 17:33




            $begingroup$
            oh nice fast and subtle , i would have give this one as a very good answer too !
            $endgroup$
            – Mather
            Jan 3 at 17:33












            $begingroup$
            thank you @Zachary
            $endgroup$
            – Mather
            Jan 3 at 17:34




            $begingroup$
            thank you @Zachary
            $endgroup$
            – Mather
            Jan 3 at 17:34












            $begingroup$
            No problem! @Mather
            $endgroup$
            – Zachary
            Jan 4 at 1:24




            $begingroup$
            No problem! @Mather
            $endgroup$
            – Zachary
            Jan 4 at 1:24











            1












            $begingroup$

            Hint. One may observe that,
            $$
            lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
            $$            giving, for a certain $n_0ge10$, $nge n_0$,
            $$
            frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
            $$             then, as $n ge 10$,
            $$
            frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
            $$             yielding, for $Nge n_0$,
            $$
            frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
            $$             leading to the divergence of the given series.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              $endgroup$
              – Mather
              Jan 3 at 17:06












            • $begingroup$
              and for $(2)$ how did you get this inequality
              $endgroup$
              – Mather
              Jan 3 at 17:07
















            1












            $begingroup$

            Hint. One may observe that,
            $$
            lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
            $$            giving, for a certain $n_0ge10$, $nge n_0$,
            $$
            frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
            $$             then, as $n ge 10$,
            $$
            frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
            $$             yielding, for $Nge n_0$,
            $$
            frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
            $$             leading to the divergence of the given series.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              $endgroup$
              – Mather
              Jan 3 at 17:06












            • $begingroup$
              and for $(2)$ how did you get this inequality
              $endgroup$
              – Mather
              Jan 3 at 17:07














            1












            1








            1





            $begingroup$

            Hint. One may observe that,
            $$
            lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
            $$            giving, for a certain $n_0ge10$, $nge n_0$,
            $$
            frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
            $$             then, as $n ge 10$,
            $$
            frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
            $$             yielding, for $Nge n_0$,
            $$
            frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
            $$             leading to the divergence of the given series.






            share|cite|improve this answer











            $endgroup$



            Hint. One may observe that,
            $$
            lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
            $$            giving, for a certain $n_0ge10$, $nge n_0$,
            $$
            frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
            $$             then, as $n ge 10$,
            $$
            frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
            $$             yielding, for $Nge n_0$,
            $$
            frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
            $$             leading to the divergence of the given series.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 16:10

























            answered Jan 3 at 15:58









            Olivier OloaOlivier Oloa

            109k17178294




            109k17178294












            • $begingroup$
              how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              $endgroup$
              – Mather
              Jan 3 at 17:06












            • $begingroup$
              and for $(2)$ how did you get this inequality
              $endgroup$
              – Mather
              Jan 3 at 17:07


















            • $begingroup$
              how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              $endgroup$
              – Mather
              Jan 3 at 17:06












            • $begingroup$
              and for $(2)$ how did you get this inequality
              $endgroup$
              – Mather
              Jan 3 at 17:07
















            $begingroup$
            how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
            $endgroup$
            – Mather
            Jan 3 at 17:06






            $begingroup$
            how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
            $endgroup$
            – Mather
            Jan 3 at 17:06














            $begingroup$
            and for $(2)$ how did you get this inequality
            $endgroup$
            – Mather
            Jan 3 at 17:07




            $begingroup$
            and for $(2)$ how did you get this inequality
            $endgroup$
            – Mather
            Jan 3 at 17:07


















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