How to show $lim_{ntoinfty}nleft{sum_{k=1}^nfrac{1}{(n+k)^2}right}=frac{1}{2}$












1












$begingroup$


Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$



Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.




Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$




That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is



... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.






real-analysis limits summation riemann-sum







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Ah yes, the famous Corollary 8.3. I love that corollary
    $endgroup$
    – mathworker21
    Dec 5 '18 at 10:49






  • 3




    $begingroup$
    Should we all know what Corollary 8.3 is?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 10:49










  • $begingroup$
    @mathworker21 Ooops. I added it now. Thank you.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:58










  • $begingroup$
    See math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 10:59
















1












$begingroup$


Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$



Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.




Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$




That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is



... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.






real-analysis limits summation riemann-sum







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Ah yes, the famous Corollary 8.3. I love that corollary
    $endgroup$
    – mathworker21
    Dec 5 '18 at 10:49






  • 3




    $begingroup$
    Should we all know what Corollary 8.3 is?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 10:49










  • $begingroup$
    @mathworker21 Ooops. I added it now. Thank you.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:58










  • $begingroup$
    See math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 10:59














1












1








1





$begingroup$


Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$



Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.




Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$




That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is



... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.






real-analysis limits summation riemann-sum







share|cite|improve this question











$endgroup$




Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$



Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.




Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$




That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is



... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.






real-analysis limits summation riemann-sum




real-analysis limits summation riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 11:13









Lorenzo B.

1,8602520




1,8602520










asked Dec 5 '18 at 10:47









kaisakaisa

2019




2019








  • 6




    $begingroup$
    Ah yes, the famous Corollary 8.3. I love that corollary
    $endgroup$
    – mathworker21
    Dec 5 '18 at 10:49






  • 3




    $begingroup$
    Should we all know what Corollary 8.3 is?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 10:49










  • $begingroup$
    @mathworker21 Ooops. I added it now. Thank you.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:58










  • $begingroup$
    See math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 10:59














  • 6




    $begingroup$
    Ah yes, the famous Corollary 8.3. I love that corollary
    $endgroup$
    – mathworker21
    Dec 5 '18 at 10:49






  • 3




    $begingroup$
    Should we all know what Corollary 8.3 is?
    $endgroup$
    – José Carlos Santos
    Dec 5 '18 at 10:49










  • $begingroup$
    @mathworker21 Ooops. I added it now. Thank you.
    $endgroup$
    – kaisa
    Dec 5 '18 at 10:58










  • $begingroup$
    See math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Dec 5 '18 at 10:59








6




6




$begingroup$
Ah yes, the famous Corollary 8.3. I love that corollary
$endgroup$
– mathworker21
Dec 5 '18 at 10:49




$begingroup$
Ah yes, the famous Corollary 8.3. I love that corollary
$endgroup$
– mathworker21
Dec 5 '18 at 10:49




3




3




$begingroup$
Should we all know what Corollary 8.3 is?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 10:49




$begingroup$
Should we all know what Corollary 8.3 is?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 10:49












$begingroup$
@mathworker21 Ooops. I added it now. Thank you.
$endgroup$
– kaisa
Dec 5 '18 at 10:58




$begingroup$
@mathworker21 Ooops. I added it now. Thank you.
$endgroup$
– kaisa
Dec 5 '18 at 10:58












$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 10:59




$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 10:59










2 Answers
2






active

oldest

votes


















1












$begingroup$

HINT



By Riemann's sum



$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$



Refer also to the related




  • Perfect understanding of Riemann Sums






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
    $endgroup$
    – kaisa
    Dec 5 '18 at 11:09





















0












$begingroup$

Another approach is to consider that the same result is obtained by the series



$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$



Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.



And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$



Which simplifies the summation ...






share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    HINT



    By Riemann's sum



    $$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$



    Refer also to the related




    • Perfect understanding of Riemann Sums






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
      $endgroup$
      – kaisa
      Dec 5 '18 at 11:09


















    1












    $begingroup$

    HINT



    By Riemann's sum



    $$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$



    Refer also to the related




    • Perfect understanding of Riemann Sums






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
      $endgroup$
      – kaisa
      Dec 5 '18 at 11:09
















    1












    1








    1





    $begingroup$

    HINT



    By Riemann's sum



    $$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$



    Refer also to the related




    • Perfect understanding of Riemann Sums






    share|cite|improve this answer









    $endgroup$



    HINT



    By Riemann's sum



    $$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$



    Refer also to the related




    • Perfect understanding of Riemann Sums







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 10:52









    gimusigimusi

    92.9k84494




    92.9k84494












    • $begingroup$
      But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
      $endgroup$
      – kaisa
      Dec 5 '18 at 11:09




















    • $begingroup$
      But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
      $endgroup$
      – kaisa
      Dec 5 '18 at 11:09


















    $begingroup$
    But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
    $endgroup$
    – kaisa
    Dec 5 '18 at 11:09






    $begingroup$
    But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
    $endgroup$
    – kaisa
    Dec 5 '18 at 11:09













    0












    $begingroup$

    Another approach is to consider that the same result is obtained by the series



    $$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$



    Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.



    And then to use
    $$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$



    Which simplifies the summation ...






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Another approach is to consider that the same result is obtained by the series



      $$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$



      Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.



      And then to use
      $$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$



      Which simplifies the summation ...






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Another approach is to consider that the same result is obtained by the series



        $$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$



        Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.



        And then to use
        $$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$



        Which simplifies the summation ...






        share|cite|improve this answer









        $endgroup$



        Another approach is to consider that the same result is obtained by the series



        $$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$



        Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.



        And then to use
        $$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$



        Which simplifies the summation ...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 11:24









        DamienDamien

        59714




        59714






























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