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












2














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



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




















  • 4




    limit comparison?
    – Lord Shark the Unknown
    2 days ago










  • yea you're gonna need to use comparison test
    – user29418
    2 days ago










  • the given series diverges
    – Dr. Sonnhard Graubner
    2 days ago
















2














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



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




















  • 4




    limit comparison?
    – Lord Shark the Unknown
    2 days ago










  • yea you're gonna need to use comparison test
    – user29418
    2 days ago










  • the given series diverges
    – Dr. Sonnhard Graubner
    2 days ago














2












2








2







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



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















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



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 2 days ago









Dando18

4,66741235




4,66741235










asked 2 days ago









Mather

1517




1517








  • 4




    limit comparison?
    – Lord Shark the Unknown
    2 days ago










  • yea you're gonna need to use comparison test
    – user29418
    2 days ago










  • the given series diverges
    – Dr. Sonnhard Graubner
    2 days ago














  • 4




    limit comparison?
    – Lord Shark the Unknown
    2 days ago










  • yea you're gonna need to use comparison test
    – user29418
    2 days ago










  • the given series diverges
    – Dr. Sonnhard Graubner
    2 days ago








4




4




limit comparison?
– Lord Shark the Unknown
2 days ago




limit comparison?
– Lord Shark the Unknown
2 days ago












yea you're gonna need to use comparison test
– user29418
2 days ago




yea you're gonna need to use comparison test
– user29418
2 days ago












the given series diverges
– Dr. Sonnhard Graubner
2 days ago




the given series diverges
– Dr. Sonnhard Graubner
2 days ago










3 Answers
3






active

oldest

votes


















1














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





























    2














    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





















    • how did you know that it behaves like that
      – Mather
      2 days ago










    • 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.
      – Zachary
      2 days ago












    • oh nice fast and subtle , i would have give this one as a very good answer too !
      – Mather
      2 days ago










    • thank you @Zachary
      – Mather
      2 days ago










    • No problem! @Mather
      – Zachary
      yesterday



















    1














    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























    • how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
      – Mather
      2 days ago












    • and for $(2)$ how did you get this inequality
      – Mather
      2 days ago











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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


























      1














      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
























        1












        1








        1






        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












        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 2 days ago









        robjohn♦

        265k27303624




        265k27303624























            2














            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





















            • how did you know that it behaves like that
              – Mather
              2 days ago










            • 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.
              – Zachary
              2 days ago












            • oh nice fast and subtle , i would have give this one as a very good answer too !
              – Mather
              2 days ago










            • thank you @Zachary
              – Mather
              2 days ago










            • No problem! @Mather
              – Zachary
              yesterday
















            2














            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





















            • how did you know that it behaves like that
              – Mather
              2 days ago










            • 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.
              – Zachary
              2 days ago












            • oh nice fast and subtle , i would have give this one as a very good answer too !
              – Mather
              2 days ago










            • thank you @Zachary
              – Mather
              2 days ago










            • No problem! @Mather
              – Zachary
              yesterday














            2












            2








            2






            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












            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 2 days ago









            Zachary

            2,2971213




            2,2971213












            • how did you know that it behaves like that
              – Mather
              2 days ago










            • 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.
              – Zachary
              2 days ago












            • oh nice fast and subtle , i would have give this one as a very good answer too !
              – Mather
              2 days ago










            • thank you @Zachary
              – Mather
              2 days ago










            • No problem! @Mather
              – Zachary
              yesterday


















            • how did you know that it behaves like that
              – Mather
              2 days ago










            • 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.
              – Zachary
              2 days ago












            • oh nice fast and subtle , i would have give this one as a very good answer too !
              – Mather
              2 days ago










            • thank you @Zachary
              – Mather
              2 days ago










            • No problem! @Mather
              – Zachary
              yesterday
















            how did you know that it behaves like that
            – Mather
            2 days ago




            how did you know that it behaves like that
            – Mather
            2 days ago












            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.
            – Zachary
            2 days ago






            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.
            – Zachary
            2 days ago














            oh nice fast and subtle , i would have give this one as a very good answer too !
            – Mather
            2 days ago




            oh nice fast and subtle , i would have give this one as a very good answer too !
            – Mather
            2 days ago












            thank you @Zachary
            – Mather
            2 days ago




            thank you @Zachary
            – Mather
            2 days ago












            No problem! @Mather
            – Zachary
            yesterday




            No problem! @Mather
            – Zachary
            yesterday











            1














            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























            • how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              – Mather
              2 days ago












            • and for $(2)$ how did you get this inequality
              – Mather
              2 days ago
















            1














            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























            • how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              – Mather
              2 days ago












            • and for $(2)$ how did you get this inequality
              – Mather
              2 days ago














            1












            1








            1






            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














            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 2 days ago

























            answered 2 days ago









            Olivier Oloa

            108k17176293




            108k17176293












            • how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              – Mather
              2 days ago












            • and for $(2)$ how did you get this inequality
              – Mather
              2 days ago


















            • how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
              – Mather
              2 days ago












            • and for $(2)$ how did you get this inequality
              – Mather
              2 days ago
















            how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
            – Mather
            2 days ago






            how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
            – Mather
            2 days ago














            and for $(2)$ how did you get this inequality
            – Mather
            2 days ago




            and for $(2)$ how did you get this inequality
            – Mather
            2 days ago


















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