Find $lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right)$












4












$begingroup$


Find $lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right)$



My Attempt:



$forall x: |cos(x)|leq 1$, Therefore:



$$lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq lim_{n to infty} left(n - left| sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right| right) leq lim_{n to infty} left( sum_{k=1}^{n} |cos frac{sqrt{k}}{n} | right) leq lim_{n to infty} left(n - sum_{k=1} ^{n} k right) = 0$$



but I can't find a way to bound the limit such that I could prove that:



$$0 leq lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq 0$$



which would finish the proof.






real-analysis calculus limits







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  • 1




    $begingroup$
    Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
    $endgroup$
    – Mindlack
    Jan 18 at 17:47


















4












$begingroup$


Find $lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right)$



My Attempt:



$forall x: |cos(x)|leq 1$, Therefore:



$$lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq lim_{n to infty} left(n - left| sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right| right) leq lim_{n to infty} left( sum_{k=1}^{n} |cos frac{sqrt{k}}{n} | right) leq lim_{n to infty} left(n - sum_{k=1} ^{n} k right) = 0$$



but I can't find a way to bound the limit such that I could prove that:



$$0 leq lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq 0$$



which would finish the proof.






real-analysis calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
    $endgroup$
    – Mindlack
    Jan 18 at 17:47
















4












4








4


1



$begingroup$


Find $lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right)$



My Attempt:



$forall x: |cos(x)|leq 1$, Therefore:



$$lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq lim_{n to infty} left(n - left| sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right| right) leq lim_{n to infty} left( sum_{k=1}^{n} |cos frac{sqrt{k}}{n} | right) leq lim_{n to infty} left(n - sum_{k=1} ^{n} k right) = 0$$



but I can't find a way to bound the limit such that I could prove that:



$$0 leq lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq 0$$



which would finish the proof.






real-analysis calculus limits







share|cite|improve this question











$endgroup$




Find $lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right)$



My Attempt:



$forall x: |cos(x)|leq 1$, Therefore:



$$lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq lim_{n to infty} left(n - left| sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right| right) leq lim_{n to infty} left( sum_{k=1}^{n} |cos frac{sqrt{k}}{n} | right) leq lim_{n to infty} left(n - sum_{k=1} ^{n} k right) = 0$$



but I can't find a way to bound the limit such that I could prove that:



$$0 leq lim_{n to infty} left(n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} right) leq 0$$



which would finish the proof.






real-analysis calculus limits




real-analysis calculus limits






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edited Jan 18 at 17:52









mathcounterexamples.net

26.9k22157




26.9k22157










asked Jan 18 at 17:32









JnevenJneven

905322




905322








  • 1




    $begingroup$
    Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
    $endgroup$
    – Mindlack
    Jan 18 at 17:47
















  • 1




    $begingroup$
    Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
    $endgroup$
    – Mindlack
    Jan 18 at 17:47










1




1




$begingroup$
Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
$endgroup$
– Mindlack
Jan 18 at 17:47






$begingroup$
Hint: $ |1-cos(x)-x^2/2| leq x^4/24$ if $|x| leq 1$.
$endgroup$
– Mindlack
Jan 18 at 17:47












2 Answers
2






active

oldest

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7












$begingroup$

Using the estimate $cos(x)=1-frac12,x^2+O(x^4)$, we get
$$ lim_{ntoinfty} left(n-sum_{k=1}^{n} cosfrac{sqrt{k}}{n} right)
= lim_{ntoinfty} sum_{k=1}^n left( 1-cosfrac{sqrt k}nright)
= lim_{ntoinfty} sum_{k=1}^n left( frac k{2n^2} + OBig(frac{k^2}{n^4}Big) right). $$
The sum splits into
$$ sum_{k=1}^n frac k{2n^2} = frac1{2n^2} frac{n(n+1)}2 $$
converging to $frac14$, and the remainder term bounded by
$$ frac1{n^4}frac{n(n+1)(2n+1)}6 , $$
which converges to $0$. Therefore, your original limit is $frac14$.






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    4












    $begingroup$

    You have
    $$u_n = n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}right)$$



    And as $sum_{k=1} ^{n} k = frac{n(n+1)}{2}$:



    $$u_n - frac{n+1}{4n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}-frac{k}{2n^2}right )$$



    And using Taylor's theorem, you get for all $x in mathbb R$



    $$leftvert 1-cos(x)-x^2/2 rightvert leq x^4/24.$$



    Therefore
    $$leftvert u_n - frac{n+1}{4n} rightvert le sum_{k=1} ^{n}leftvert1-cos frac{sqrt{k}}{n}-frac{k}{4n^2}rightvert le frac{1}{24n^4}sum_{k=1} ^{n} k^2 tag{1}$$



    As $$sum_{k=1} ^{n} k^2 = frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $(1)$ converges to $0$.



    Finaly $(u_n)$ converges to $1/4$ as $limlimits_{n to infty} frac{n+1}{4n} =1/4$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      7












      $begingroup$

      Using the estimate $cos(x)=1-frac12,x^2+O(x^4)$, we get
      $$ lim_{ntoinfty} left(n-sum_{k=1}^{n} cosfrac{sqrt{k}}{n} right)
      = lim_{ntoinfty} sum_{k=1}^n left( 1-cosfrac{sqrt k}nright)
      = lim_{ntoinfty} sum_{k=1}^n left( frac k{2n^2} + OBig(frac{k^2}{n^4}Big) right). $$
      The sum splits into
      $$ sum_{k=1}^n frac k{2n^2} = frac1{2n^2} frac{n(n+1)}2 $$
      converging to $frac14$, and the remainder term bounded by
      $$ frac1{n^4}frac{n(n+1)(2n+1)}6 , $$
      which converges to $0$. Therefore, your original limit is $frac14$.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        Using the estimate $cos(x)=1-frac12,x^2+O(x^4)$, we get
        $$ lim_{ntoinfty} left(n-sum_{k=1}^{n} cosfrac{sqrt{k}}{n} right)
        = lim_{ntoinfty} sum_{k=1}^n left( 1-cosfrac{sqrt k}nright)
        = lim_{ntoinfty} sum_{k=1}^n left( frac k{2n^2} + OBig(frac{k^2}{n^4}Big) right). $$
        The sum splits into
        $$ sum_{k=1}^n frac k{2n^2} = frac1{2n^2} frac{n(n+1)}2 $$
        converging to $frac14$, and the remainder term bounded by
        $$ frac1{n^4}frac{n(n+1)(2n+1)}6 , $$
        which converges to $0$. Therefore, your original limit is $frac14$.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          Using the estimate $cos(x)=1-frac12,x^2+O(x^4)$, we get
          $$ lim_{ntoinfty} left(n-sum_{k=1}^{n} cosfrac{sqrt{k}}{n} right)
          = lim_{ntoinfty} sum_{k=1}^n left( 1-cosfrac{sqrt k}nright)
          = lim_{ntoinfty} sum_{k=1}^n left( frac k{2n^2} + OBig(frac{k^2}{n^4}Big) right). $$
          The sum splits into
          $$ sum_{k=1}^n frac k{2n^2} = frac1{2n^2} frac{n(n+1)}2 $$
          converging to $frac14$, and the remainder term bounded by
          $$ frac1{n^4}frac{n(n+1)(2n+1)}6 , $$
          which converges to $0$. Therefore, your original limit is $frac14$.






          share|cite|improve this answer









          $endgroup$



          Using the estimate $cos(x)=1-frac12,x^2+O(x^4)$, we get
          $$ lim_{ntoinfty} left(n-sum_{k=1}^{n} cosfrac{sqrt{k}}{n} right)
          = lim_{ntoinfty} sum_{k=1}^n left( 1-cosfrac{sqrt k}nright)
          = lim_{ntoinfty} sum_{k=1}^n left( frac k{2n^2} + OBig(frac{k^2}{n^4}Big) right). $$
          The sum splits into
          $$ sum_{k=1}^n frac k{2n^2} = frac1{2n^2} frac{n(n+1)}2 $$
          converging to $frac14$, and the remainder term bounded by
          $$ frac1{n^4}frac{n(n+1)(2n+1)}6 , $$
          which converges to $0$. Therefore, your original limit is $frac14$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 18:05









          W-t-PW-t-P

          1,287611




          1,287611























              4












              $begingroup$

              You have
              $$u_n = n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}right)$$



              And as $sum_{k=1} ^{n} k = frac{n(n+1)}{2}$:



              $$u_n - frac{n+1}{4n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}-frac{k}{2n^2}right )$$



              And using Taylor's theorem, you get for all $x in mathbb R$



              $$leftvert 1-cos(x)-x^2/2 rightvert leq x^4/24.$$



              Therefore
              $$leftvert u_n - frac{n+1}{4n} rightvert le sum_{k=1} ^{n}leftvert1-cos frac{sqrt{k}}{n}-frac{k}{4n^2}rightvert le frac{1}{24n^4}sum_{k=1} ^{n} k^2 tag{1}$$



              As $$sum_{k=1} ^{n} k^2 = frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $(1)$ converges to $0$.



              Finaly $(u_n)$ converges to $1/4$ as $limlimits_{n to infty} frac{n+1}{4n} =1/4$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                You have
                $$u_n = n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}right)$$



                And as $sum_{k=1} ^{n} k = frac{n(n+1)}{2}$:



                $$u_n - frac{n+1}{4n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}-frac{k}{2n^2}right )$$



                And using Taylor's theorem, you get for all $x in mathbb R$



                $$leftvert 1-cos(x)-x^2/2 rightvert leq x^4/24.$$



                Therefore
                $$leftvert u_n - frac{n+1}{4n} rightvert le sum_{k=1} ^{n}leftvert1-cos frac{sqrt{k}}{n}-frac{k}{4n^2}rightvert le frac{1}{24n^4}sum_{k=1} ^{n} k^2 tag{1}$$



                As $$sum_{k=1} ^{n} k^2 = frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $(1)$ converges to $0$.



                Finaly $(u_n)$ converges to $1/4$ as $limlimits_{n to infty} frac{n+1}{4n} =1/4$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  You have
                  $$u_n = n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}right)$$



                  And as $sum_{k=1} ^{n} k = frac{n(n+1)}{2}$:



                  $$u_n - frac{n+1}{4n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}-frac{k}{2n^2}right )$$



                  And using Taylor's theorem, you get for all $x in mathbb R$



                  $$leftvert 1-cos(x)-x^2/2 rightvert leq x^4/24.$$



                  Therefore
                  $$leftvert u_n - frac{n+1}{4n} rightvert le sum_{k=1} ^{n}leftvert1-cos frac{sqrt{k}}{n}-frac{k}{4n^2}rightvert le frac{1}{24n^4}sum_{k=1} ^{n} k^2 tag{1}$$



                  As $$sum_{k=1} ^{n} k^2 = frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $(1)$ converges to $0$.



                  Finaly $(u_n)$ converges to $1/4$ as $limlimits_{n to infty} frac{n+1}{4n} =1/4$.






                  share|cite|improve this answer









                  $endgroup$



                  You have
                  $$u_n = n - sum_{k=1} ^{n} cos frac{sqrt{k}}{n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}right)$$



                  And as $sum_{k=1} ^{n} k = frac{n(n+1)}{2}$:



                  $$u_n - frac{n+1}{4n} = sum_{k=1} ^{n}left(1-cos frac{sqrt{k}}{n}-frac{k}{2n^2}right )$$



                  And using Taylor's theorem, you get for all $x in mathbb R$



                  $$leftvert 1-cos(x)-x^2/2 rightvert leq x^4/24.$$



                  Therefore
                  $$leftvert u_n - frac{n+1}{4n} rightvert le sum_{k=1} ^{n}leftvert1-cos frac{sqrt{k}}{n}-frac{k}{4n^2}rightvert le frac{1}{24n^4}sum_{k=1} ^{n} k^2 tag{1}$$



                  As $$sum_{k=1} ^{n} k^2 = frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $(1)$ converges to $0$.



                  Finaly $(u_n)$ converges to $1/4$ as $limlimits_{n to infty} frac{n+1}{4n} =1/4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 18:07









                  mathcounterexamples.netmathcounterexamples.net

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