Showing that $lim_{n rightarrow infty} sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))=infty$












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Does anyone know how to show $ lim_{nrightarrowinfty} sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))=infty$? My thoughts were that you should you a series approximation of some sort, but I don't know how to get through it.






real-analysis sequences-and-series







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  • 1




    $begingroup$
    maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
    $endgroup$
    – mm8511
    Dec 9 '18 at 19:45












  • $begingroup$
    So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
    $endgroup$
    – Vercingetorix
    Dec 9 '18 at 20:18










  • $begingroup$
    The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
    $endgroup$
    – mm8511
    Dec 9 '18 at 20:59


















0












$begingroup$


Does anyone know how to show $ lim_{nrightarrowinfty} sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))=infty$? My thoughts were that you should you a series approximation of some sort, but I don't know how to get through it.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
    $endgroup$
    – mm8511
    Dec 9 '18 at 19:45












  • $begingroup$
    So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
    $endgroup$
    – Vercingetorix
    Dec 9 '18 at 20:18










  • $begingroup$
    The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
    $endgroup$
    – mm8511
    Dec 9 '18 at 20:59
















0












0








0





$begingroup$


Does anyone know how to show $ lim_{nrightarrowinfty} sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))=infty$? My thoughts were that you should you a series approximation of some sort, but I don't know how to get through it.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$




Does anyone know how to show $ lim_{nrightarrowinfty} sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))=infty$? My thoughts were that you should you a series approximation of some sort, but I don't know how to get through it.






real-analysis sequences-and-series




real-analysis sequences-and-series






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asked Dec 9 '18 at 19:01









VercingetorixVercingetorix

285




285








  • 1




    $begingroup$
    maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
    $endgroup$
    – mm8511
    Dec 9 '18 at 19:45












  • $begingroup$
    So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
    $endgroup$
    – Vercingetorix
    Dec 9 '18 at 20:18










  • $begingroup$
    The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
    $endgroup$
    – mm8511
    Dec 9 '18 at 20:59
















  • 1




    $begingroup$
    maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
    $endgroup$
    – mm8511
    Dec 9 '18 at 19:45












  • $begingroup$
    So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
    $endgroup$
    – Vercingetorix
    Dec 9 '18 at 20:18










  • $begingroup$
    The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
    $endgroup$
    – mm8511
    Dec 9 '18 at 20:59










1




1




$begingroup$
maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
$endgroup$
– mm8511
Dec 9 '18 at 19:45






$begingroup$
maybe you could change the upper limit of the sum to n (so that you have a lower bound). Then, note that $(n/k)^2$ is rational by definition, so it's never a multiple of $pi$, which implies that the $1-cos((n/k)^2)>t$ for some constant t. I think it's pretty clear that this new limit is $>log n$. Word of caution: my analysis is sloppy, and I could be way off.
$endgroup$
– mm8511
Dec 9 '18 at 19:45














$begingroup$
So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
$endgroup$
– Vercingetorix
Dec 9 '18 at 20:18




$begingroup$
So what you are saying is that $sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2})) >sum_{k=1}^{n}k^{1/2} n^{-1} t(n)$, since $t$ must depend on $n$, right? How is this limit $>log(n)$?
$endgroup$
– Vercingetorix
Dec 9 '18 at 20:18












$begingroup$
The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
$endgroup$
– mm8511
Dec 9 '18 at 20:59






$begingroup$
The solution below seems correct, I said $log n$ since $sum^{n} k^{1/2}/n > sum^{sqrt n}_{k=1} (k/ sqrt{n}) approx 1/2 log n$
$endgroup$
– mm8511
Dec 9 '18 at 20:59












2 Answers
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$begingroup$

Define $$f(x)=
begin{cases}
c,;;xin [dfrac{pi}{2},dfrac{3pi}{2}]\
0,;;otherwise
end{cases}$$
where $capprox 0$ and $c> 0$. Then $1-cos(x)geq f(x)$ for $xin [0,dfrac{3pi}{2}]$.



$dfrac{n^2}{k^2}<dfrac{pi}{2}Leftrightarrow k>sqrt{dfrac{2}{pi}}n$



We have
$$sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))geq
sum_{k=1}^{infty}k^{1/2} n^{-1} f(n^2 k^{-2})geq
sum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1} f(n^2 k^{-2})geq$$
$$geq csum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1}approx
dfrac{c}{n} int _1 ^{sqrt{frac{2}{pi}}n} x^{1/2}dx geq bar{c}sqrt{n}
to infty$$






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    0












    $begingroup$

    Let
    $s(n, m)
    = sum_{k=1}^{m}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))
    $



    $1 - cos(n^2 k^{-2})
    = 2sin^2(frac{n^2}{2k^2})
    $
    so
    $s(n,m)
    = 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})
    $    .



    If
    $m > n$ then



    $begin{array}\
    s(n,m)
    &= 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
    &> 2sum_{k=n}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
    &> 2sum_{k=n}^{m}k^{1/2} n^{-1}(frac{n^2}{2k^2})^2(frac{4}{pi^2})\
    &> frac{8}{pi^2}sum_{k=n}^{m}k^{1/2} n^{-1}frac{n^4}{k^4}\
    &= frac{8}{pi^2}sum_{k=n}^{m}frac{n^3}{k^{7/2}}\
    &= frac{8n^3}{pi^2}sum_{k=n}^{m}frac{1}{k^{7/2}}\
    &approx frac{8n^3}{pi^2}int_{n}^{m}x^{-7/2}dx\
    &= frac{8n^3}{pi^2}dfrac{x^{-5/2}}{-5/2}_{n}^{m}\
    &= frac{8n^3}{pi^2}dfrac{m^{-5/2}-n^{-5/2}}{-5/2}\
    &= frac{16n^3}{5pi^2}(n^{-5/2}-m^{-5/2})\
    &= frac{16n^{1/2}}{5pi^2}(1-(m/n)^{-5/2})\
    &= frac{16n^{1/2}}{5pi^2}(1-(n/m)^{5/2})\
    end{array}
    $



    If $m > 2n$ then
    $s(n, m)
    gt frac{16n^{1/2}}{5pi^2}(1-(1/2)^{-5/2})
    gt frac{8n^{1/2}}{5pi^2}
    to infty$
    as $n to infty$.






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      Define $$f(x)=
      begin{cases}
      c,;;xin [dfrac{pi}{2},dfrac{3pi}{2}]\
      0,;;otherwise
      end{cases}$$
      where $capprox 0$ and $c> 0$. Then $1-cos(x)geq f(x)$ for $xin [0,dfrac{3pi}{2}]$.



      $dfrac{n^2}{k^2}<dfrac{pi}{2}Leftrightarrow k>sqrt{dfrac{2}{pi}}n$



      We have
      $$sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))geq
      sum_{k=1}^{infty}k^{1/2} n^{-1} f(n^2 k^{-2})geq
      sum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1} f(n^2 k^{-2})geq$$
      $$geq csum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1}approx
      dfrac{c}{n} int _1 ^{sqrt{frac{2}{pi}}n} x^{1/2}dx geq bar{c}sqrt{n}
      to infty$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Define $$f(x)=
        begin{cases}
        c,;;xin [dfrac{pi}{2},dfrac{3pi}{2}]\
        0,;;otherwise
        end{cases}$$
        where $capprox 0$ and $c> 0$. Then $1-cos(x)geq f(x)$ for $xin [0,dfrac{3pi}{2}]$.



        $dfrac{n^2}{k^2}<dfrac{pi}{2}Leftrightarrow k>sqrt{dfrac{2}{pi}}n$



        We have
        $$sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))geq
        sum_{k=1}^{infty}k^{1/2} n^{-1} f(n^2 k^{-2})geq
        sum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1} f(n^2 k^{-2})geq$$
        $$geq csum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1}approx
        dfrac{c}{n} int _1 ^{sqrt{frac{2}{pi}}n} x^{1/2}dx geq bar{c}sqrt{n}
        to infty$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Define $$f(x)=
          begin{cases}
          c,;;xin [dfrac{pi}{2},dfrac{3pi}{2}]\
          0,;;otherwise
          end{cases}$$
          where $capprox 0$ and $c> 0$. Then $1-cos(x)geq f(x)$ for $xin [0,dfrac{3pi}{2}]$.



          $dfrac{n^2}{k^2}<dfrac{pi}{2}Leftrightarrow k>sqrt{dfrac{2}{pi}}n$



          We have
          $$sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))geq
          sum_{k=1}^{infty}k^{1/2} n^{-1} f(n^2 k^{-2})geq
          sum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1} f(n^2 k^{-2})geq$$
          $$geq csum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1}approx
          dfrac{c}{n} int _1 ^{sqrt{frac{2}{pi}}n} x^{1/2}dx geq bar{c}sqrt{n}
          to infty$$






          share|cite|improve this answer









          $endgroup$



          Define $$f(x)=
          begin{cases}
          c,;;xin [dfrac{pi}{2},dfrac{3pi}{2}]\
          0,;;otherwise
          end{cases}$$
          where $capprox 0$ and $c> 0$. Then $1-cos(x)geq f(x)$ for $xin [0,dfrac{3pi}{2}]$.



          $dfrac{n^2}{k^2}<dfrac{pi}{2}Leftrightarrow k>sqrt{dfrac{2}{pi}}n$



          We have
          $$sum_{k=1}^{infty}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))geq
          sum_{k=1}^{infty}k^{1/2} n^{-1} f(n^2 k^{-2})geq
          sum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1} f(n^2 k^{-2})geq$$
          $$geq csum_{k=1}^{sqrt{frac{2}{pi}}n}k^{1/2} n^{-1}approx
          dfrac{c}{n} int _1 ^{sqrt{frac{2}{pi}}n} x^{1/2}dx geq bar{c}sqrt{n}
          to infty$$







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          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 19:52









          Fat ninjaFat ninja

          756




          756























              0












              $begingroup$

              Let
              $s(n, m)
              = sum_{k=1}^{m}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))
              $



              $1 - cos(n^2 k^{-2})
              = 2sin^2(frac{n^2}{2k^2})
              $
              so
              $s(n,m)
              = 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})
              $              .



              If
              $m > n$ then



              $begin{array}\
              s(n,m)
              &= 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
              &> 2sum_{k=n}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
              &> 2sum_{k=n}^{m}k^{1/2} n^{-1}(frac{n^2}{2k^2})^2(frac{4}{pi^2})\
              &> frac{8}{pi^2}sum_{k=n}^{m}k^{1/2} n^{-1}frac{n^4}{k^4}\
              &= frac{8}{pi^2}sum_{k=n}^{m}frac{n^3}{k^{7/2}}\
              &= frac{8n^3}{pi^2}sum_{k=n}^{m}frac{1}{k^{7/2}}\
              &approx frac{8n^3}{pi^2}int_{n}^{m}x^{-7/2}dx\
              &= frac{8n^3}{pi^2}dfrac{x^{-5/2}}{-5/2}_{n}^{m}\
              &= frac{8n^3}{pi^2}dfrac{m^{-5/2}-n^{-5/2}}{-5/2}\
              &= frac{16n^3}{5pi^2}(n^{-5/2}-m^{-5/2})\
              &= frac{16n^{1/2}}{5pi^2}(1-(m/n)^{-5/2})\
              &= frac{16n^{1/2}}{5pi^2}(1-(n/m)^{5/2})\
              end{array}
              $



              If $m > 2n$ then
              $s(n, m)
              gt frac{16n^{1/2}}{5pi^2}(1-(1/2)^{-5/2})
              gt frac{8n^{1/2}}{5pi^2}
              to infty$
              as $n to infty$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let
                $s(n, m)
                = sum_{k=1}^{m}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))
                $



                $1 - cos(n^2 k^{-2})
                = 2sin^2(frac{n^2}{2k^2})
                $
                so
                $s(n,m)
                = 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})
                $                .



                If
                $m > n$ then



                $begin{array}\
                s(n,m)
                &= 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                &> 2sum_{k=n}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                &> 2sum_{k=n}^{m}k^{1/2} n^{-1}(frac{n^2}{2k^2})^2(frac{4}{pi^2})\
                &> frac{8}{pi^2}sum_{k=n}^{m}k^{1/2} n^{-1}frac{n^4}{k^4}\
                &= frac{8}{pi^2}sum_{k=n}^{m}frac{n^3}{k^{7/2}}\
                &= frac{8n^3}{pi^2}sum_{k=n}^{m}frac{1}{k^{7/2}}\
                &approx frac{8n^3}{pi^2}int_{n}^{m}x^{-7/2}dx\
                &= frac{8n^3}{pi^2}dfrac{x^{-5/2}}{-5/2}_{n}^{m}\
                &= frac{8n^3}{pi^2}dfrac{m^{-5/2}-n^{-5/2}}{-5/2}\
                &= frac{16n^3}{5pi^2}(n^{-5/2}-m^{-5/2})\
                &= frac{16n^{1/2}}{5pi^2}(1-(m/n)^{-5/2})\
                &= frac{16n^{1/2}}{5pi^2}(1-(n/m)^{5/2})\
                end{array}
                $



                If $m > 2n$ then
                $s(n, m)
                gt frac{16n^{1/2}}{5pi^2}(1-(1/2)^{-5/2})
                gt frac{8n^{1/2}}{5pi^2}
                to infty$
                as $n to infty$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let
                  $s(n, m)
                  = sum_{k=1}^{m}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))
                  $



                  $1 - cos(n^2 k^{-2})
                  = 2sin^2(frac{n^2}{2k^2})
                  $
                  so
                  $s(n,m)
                  = 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})
                  $                  .



                  If
                  $m > n$ then



                  $begin{array}\
                  s(n,m)
                  &= 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                  &> 2sum_{k=n}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                  &> 2sum_{k=n}^{m}k^{1/2} n^{-1}(frac{n^2}{2k^2})^2(frac{4}{pi^2})\
                  &> frac{8}{pi^2}sum_{k=n}^{m}k^{1/2} n^{-1}frac{n^4}{k^4}\
                  &= frac{8}{pi^2}sum_{k=n}^{m}frac{n^3}{k^{7/2}}\
                  &= frac{8n^3}{pi^2}sum_{k=n}^{m}frac{1}{k^{7/2}}\
                  &approx frac{8n^3}{pi^2}int_{n}^{m}x^{-7/2}dx\
                  &= frac{8n^3}{pi^2}dfrac{x^{-5/2}}{-5/2}_{n}^{m}\
                  &= frac{8n^3}{pi^2}dfrac{m^{-5/2}-n^{-5/2}}{-5/2}\
                  &= frac{16n^3}{5pi^2}(n^{-5/2}-m^{-5/2})\
                  &= frac{16n^{1/2}}{5pi^2}(1-(m/n)^{-5/2})\
                  &= frac{16n^{1/2}}{5pi^2}(1-(n/m)^{5/2})\
                  end{array}
                  $



                  If $m > 2n$ then
                  $s(n, m)
                  gt frac{16n^{1/2}}{5pi^2}(1-(1/2)^{-5/2})
                  gt frac{8n^{1/2}}{5pi^2}
                  to infty$
                  as $n to infty$.






                  share|cite|improve this answer









                  $endgroup$



                  Let
                  $s(n, m)
                  = sum_{k=1}^{m}k^{1/2} n^{-1} (1 - cos(n^2 k^{-2}))
                  $



                  $1 - cos(n^2 k^{-2})
                  = 2sin^2(frac{n^2}{2k^2})
                  $
                  so
                  $s(n,m)
                  = 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})
                  $                  .



                  If
                  $m > n$ then



                  $begin{array}\
                  s(n,m)
                  &= 2sum_{k=1}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                  &> 2sum_{k=n}^{m}k^{1/2} n^{-1}sin^2(frac{n^2}{2k^2})\
                  &> 2sum_{k=n}^{m}k^{1/2} n^{-1}(frac{n^2}{2k^2})^2(frac{4}{pi^2})\
                  &> frac{8}{pi^2}sum_{k=n}^{m}k^{1/2} n^{-1}frac{n^4}{k^4}\
                  &= frac{8}{pi^2}sum_{k=n}^{m}frac{n^3}{k^{7/2}}\
                  &= frac{8n^3}{pi^2}sum_{k=n}^{m}frac{1}{k^{7/2}}\
                  &approx frac{8n^3}{pi^2}int_{n}^{m}x^{-7/2}dx\
                  &= frac{8n^3}{pi^2}dfrac{x^{-5/2}}{-5/2}_{n}^{m}\
                  &= frac{8n^3}{pi^2}dfrac{m^{-5/2}-n^{-5/2}}{-5/2}\
                  &= frac{16n^3}{5pi^2}(n^{-5/2}-m^{-5/2})\
                  &= frac{16n^{1/2}}{5pi^2}(1-(m/n)^{-5/2})\
                  &= frac{16n^{1/2}}{5pi^2}(1-(n/m)^{5/2})\
                  end{array}
                  $



                  If $m > 2n$ then
                  $s(n, m)
                  gt frac{16n^{1/2}}{5pi^2}(1-(1/2)^{-5/2})
                  gt frac{8n^{1/2}}{5pi^2}
                  to infty$
                  as $n to infty$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 22:16









                  marty cohenmarty cohen

                  73.4k549128




                  73.4k549128






























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