How to calculate $lim_{n to infty} sum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}$?












2












$begingroup$


How to calculate $displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}$?



My try:



begin{align}
lim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}
&=displaystyle lim_{n to infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}}
\&=displaystylelim_{n to infty}int_{0}^{1}frac{dx}{sqrt{1+frac{1}{n}-x^2}}
\&=displaystylelim_{n to infty}arctan sqrt{n}
\&=frac{pi}{2}
end{align}



But,

I'm not sure whether this's right because I'm not sure whether the second equality is right.



Any helps and new ideas will be highly appreciated!






real-analysis calculus limits







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  • $begingroup$
    As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
    $endgroup$
    – Will M.
    Dec 21 '18 at 5:39
















2












$begingroup$


How to calculate $displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}$?



My try:



begin{align}
lim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}
&=displaystyle lim_{n to infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}}
\&=displaystylelim_{n to infty}int_{0}^{1}frac{dx}{sqrt{1+frac{1}{n}-x^2}}
\&=displaystylelim_{n to infty}arctan sqrt{n}
\&=frac{pi}{2}
end{align}



But,

I'm not sure whether this's right because I'm not sure whether the second equality is right.



Any helps and new ideas will be highly appreciated!






real-analysis calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
    $endgroup$
    – Will M.
    Dec 21 '18 at 5:39














2












2








2





$begingroup$


How to calculate $displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}$?



My try:



begin{align}
lim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}
&=displaystyle lim_{n to infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}}
\&=displaystylelim_{n to infty}int_{0}^{1}frac{dx}{sqrt{1+frac{1}{n}-x^2}}
\&=displaystylelim_{n to infty}arctan sqrt{n}
\&=frac{pi}{2}
end{align}



But,

I'm not sure whether this's right because I'm not sure whether the second equality is right.



Any helps and new ideas will be highly appreciated!






real-analysis calculus limits







share|cite|improve this question











$endgroup$




How to calculate $displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}$?



My try:



begin{align}
lim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}
&=displaystyle lim_{n to infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}}
\&=displaystylelim_{n to infty}int_{0}^{1}frac{dx}{sqrt{1+frac{1}{n}-x^2}}
\&=displaystylelim_{n to infty}arctan sqrt{n}
\&=frac{pi}{2}
end{align}



But,

I'm not sure whether this's right because I'm not sure whether the second equality is right.



Any helps and new ideas will be highly appreciated!






real-analysis calculus limits




real-analysis calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 9:13









Lorenzo B.

1,8602520




1,8602520










asked Dec 21 '18 at 5:21









ZeroZero

39510




39510












  • $begingroup$
    As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
    $endgroup$
    – Will M.
    Dec 21 '18 at 5:39


















  • $begingroup$
    As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
    $endgroup$
    – Will M.
    Dec 21 '18 at 5:39
















$begingroup$
As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
$endgroup$
– Will M.
Dec 21 '18 at 5:39




$begingroup$
As written is not correct because in the limit $k/n$ changed to $x,$ however, I believe the final answer is correct.
$endgroup$
– Will M.
Dec 21 '18 at 5:39










2 Answers
2






active

oldest

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2












$begingroup$

$$frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} preceq frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}} leq frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}} $$
(the symbole $preceq$ means: It is lower than form a $nin mathbb{N}$ to later)



But $$lim_{nto infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}}=int_0^1arcsin(x)dx=frac{pi}{2} $$ and $$lim_{nto infty} frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} =arcsin left(frac{1}{sqrt {1+epsilon}}right)$$ and
$$lim_{epsilon to 0^+} arcsinleft(frac{1}{sqrt {1+epsilon}}right)=frac{pi}{2}.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    To use integral method rigorously, I came up with a new solution.



    Notice that(due to the monotonicity)
    $$ displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}leint_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}+frac{1}{sqrt{n}}$$



    Then we have
    $$displaystylelim_{ntoinfty}displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylelim_{ntoinfty}displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}ledisplaystylelim_{ntoinfty}int_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}$$



    Considering
    $$displaystyleintfrac{dx}{sqrt {n^2+n-x^2}}=arctanfrac{x}{sqrt {n^2+n-x^2}}$$



    Then we can arrive at
    $$displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}=frac{pi}{2}$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      2












      $begingroup$

      $$frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} preceq frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}} leq frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}} $$
      (the symbole $preceq$ means: It is lower than form a $nin mathbb{N}$ to later)



      But $$lim_{nto infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}}=int_0^1arcsin(x)dx=frac{pi}{2} $$ and $$lim_{nto infty} frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} =arcsin left(frac{1}{sqrt {1+epsilon}}right)$$ and
      $$lim_{epsilon to 0^+} arcsinleft(frac{1}{sqrt {1+epsilon}}right)=frac{pi}{2}.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $$frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} preceq frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}} leq frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}} $$
        (the symbole $preceq$ means: It is lower than form a $nin mathbb{N}$ to later)



        But $$lim_{nto infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}}=int_0^1arcsin(x)dx=frac{pi}{2} $$ and $$lim_{nto infty} frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} =arcsin left(frac{1}{sqrt {1+epsilon}}right)$$ and
        $$lim_{epsilon to 0^+} arcsinleft(frac{1}{sqrt {1+epsilon}}right)=frac{pi}{2}.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $$frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} preceq frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}} leq frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}} $$
          (the symbole $preceq$ means: It is lower than form a $nin mathbb{N}$ to later)



          But $$lim_{nto infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}}=int_0^1arcsin(x)dx=frac{pi}{2} $$ and $$lim_{nto infty} frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} =arcsin left(frac{1}{sqrt {1+epsilon}}right)$$ and
          $$lim_{epsilon to 0^+} arcsinleft(frac{1}{sqrt {1+epsilon}}right)=frac{pi}{2}.$$






          share|cite|improve this answer











          $endgroup$



          $$frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} preceq frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+frac{1}{n}-(tfrac{k}{n})^2}} leq frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}} $$
          (the symbole $preceq$ means: It is lower than form a $nin mathbb{N}$ to later)



          But $$lim_{nto infty} frac{1}{n}displaystylesum_{k=1}^{n}frac{1}{sqrt {1-(tfrac{k}{n})^2}}=int_0^1arcsin(x)dx=frac{pi}{2} $$ and $$lim_{nto infty} frac{1}{n}sum_{k=1}^{n}frac{1}{sqrt {1+epsilon-(tfrac{k}{n})^2}} =arcsin left(frac{1}{sqrt {1+epsilon}}right)$$ and
          $$lim_{epsilon to 0^+} arcsinleft(frac{1}{sqrt {1+epsilon}}right)=frac{pi}{2}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 10:25









          amWhy

          1




          1










          answered Dec 21 '18 at 9:36









          DarmanDarman

          538112




          538112























              1












              $begingroup$

              To use integral method rigorously, I came up with a new solution.



              Notice that(due to the monotonicity)
              $$ displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}leint_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}+frac{1}{sqrt{n}}$$



              Then we have
              $$displaystylelim_{ntoinfty}displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylelim_{ntoinfty}displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}ledisplaystylelim_{ntoinfty}int_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}$$



              Considering
              $$displaystyleintfrac{dx}{sqrt {n^2+n-x^2}}=arctanfrac{x}{sqrt {n^2+n-x^2}}$$



              Then we can arrive at
              $$displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}=frac{pi}{2}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                To use integral method rigorously, I came up with a new solution.



                Notice that(due to the monotonicity)
                $$ displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}leint_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}+frac{1}{sqrt{n}}$$



                Then we have
                $$displaystylelim_{ntoinfty}displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylelim_{ntoinfty}displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}ledisplaystylelim_{ntoinfty}int_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}$$



                Considering
                $$displaystyleintfrac{dx}{sqrt {n^2+n-x^2}}=arctanfrac{x}{sqrt {n^2+n-x^2}}$$



                Then we can arrive at
                $$displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}=frac{pi}{2}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  To use integral method rigorously, I came up with a new solution.



                  Notice that(due to the monotonicity)
                  $$ displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}leint_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}+frac{1}{sqrt{n}}$$



                  Then we have
                  $$displaystylelim_{ntoinfty}displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylelim_{ntoinfty}displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}ledisplaystylelim_{ntoinfty}int_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}$$



                  Considering
                  $$displaystyleintfrac{dx}{sqrt {n^2+n-x^2}}=arctanfrac{x}{sqrt {n^2+n-x^2}}$$



                  Then we can arrive at
                  $$displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}=frac{pi}{2}$$






                  share|cite|improve this answer









                  $endgroup$



                  To use integral method rigorously, I came up with a new solution.



                  Notice that(due to the monotonicity)
                  $$ displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}leint_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}+frac{1}{sqrt{n}}$$



                  Then we have
                  $$displaystylelim_{ntoinfty}displaystyleint_{0}^{n}frac{dx}{sqrt {n^2+n-x^2}} le displaystylelim_{ntoinfty}displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}ledisplaystylelim_{ntoinfty}int_{1}^{n}frac{dx}{sqrt {n^2+n-x^2}}$$



                  Considering
                  $$displaystyleintfrac{dx}{sqrt {n^2+n-x^2}}=arctanfrac{x}{sqrt {n^2+n-x^2}}$$



                  Then we can arrive at
                  $$displaystylelim_{n to infty} displaystylesum_{k=1}^{n}frac{1}{sqrt {n^2+n-k^2}}=frac{pi}{2}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 9:30









                  ZeroZero

                  39510




                  39510






























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