How would you calculate this limit? $limlimits_{n...












3












$begingroup$


I decided to calculate $int_{0}^{pi/2}cos(x)dx$ using the sum definition of the integral. Obviously the answer is $1$ . I managed to calculate the resulting limit using the geometric series, taking the real part of the complex exponential function and several iterations of l'hopital's rule. Are you able to simplify this absolute mess, i.e. find a better way of arriving at the desired answer?



$$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)$$



Every answer is highly appreciated =)



PS: If you want to see my solution, feel free to tell me! =)






real-analysis limits definite-integrals







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  • 2




    $begingroup$
    Indeed, that will be good to see your solution.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 8:35






  • 1




    $begingroup$
    $1$ is the result!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:42










  • $begingroup$
    Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
    $endgroup$
    – Flammable Maths
    Jan 16 at 8:44








  • 1




    $begingroup$
    Use math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:57










  • $begingroup$
    Brooo flammy! I'm a huge fan!
    $endgroup$
    – clathratus
    Jan 21 at 16:59


















3












$begingroup$


I decided to calculate $int_{0}^{pi/2}cos(x)dx$ using the sum definition of the integral. Obviously the answer is $1$ . I managed to calculate the resulting limit using the geometric series, taking the real part of the complex exponential function and several iterations of l'hopital's rule. Are you able to simplify this absolute mess, i.e. find a better way of arriving at the desired answer?



$$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)$$



Every answer is highly appreciated =)



PS: If you want to see my solution, feel free to tell me! =)






real-analysis limits definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Indeed, that will be good to see your solution.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 8:35






  • 1




    $begingroup$
    $1$ is the result!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:42










  • $begingroup$
    Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
    $endgroup$
    – Flammable Maths
    Jan 16 at 8:44








  • 1




    $begingroup$
    Use math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:57










  • $begingroup$
    Brooo flammy! I'm a huge fan!
    $endgroup$
    – clathratus
    Jan 21 at 16:59
















3












3








3


0



$begingroup$


I decided to calculate $int_{0}^{pi/2}cos(x)dx$ using the sum definition of the integral. Obviously the answer is $1$ . I managed to calculate the resulting limit using the geometric series, taking the real part of the complex exponential function and several iterations of l'hopital's rule. Are you able to simplify this absolute mess, i.e. find a better way of arriving at the desired answer?



$$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)$$



Every answer is highly appreciated =)



PS: If you want to see my solution, feel free to tell me! =)






real-analysis limits definite-integrals







share|cite|improve this question











$endgroup$




I decided to calculate $int_{0}^{pi/2}cos(x)dx$ using the sum definition of the integral. Obviously the answer is $1$ . I managed to calculate the resulting limit using the geometric series, taking the real part of the complex exponential function and several iterations of l'hopital's rule. Are you able to simplify this absolute mess, i.e. find a better way of arriving at the desired answer?



$$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)$$



Every answer is highly appreciated =)



PS: If you want to see my solution, feel free to tell me! =)






real-analysis limits definite-integrals




real-analysis limits definite-integrals






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 16 at 9:26









rtybase

11k21533




11k21533










asked Jan 16 at 8:32









Flammable MathsFlammable Maths

1035




1035








  • 2




    $begingroup$
    Indeed, that will be good to see your solution.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 8:35






  • 1




    $begingroup$
    $1$ is the result!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:42










  • $begingroup$
    Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
    $endgroup$
    – Flammable Maths
    Jan 16 at 8:44








  • 1




    $begingroup$
    Use math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:57










  • $begingroup$
    Brooo flammy! I'm a huge fan!
    $endgroup$
    – clathratus
    Jan 21 at 16:59
















  • 2




    $begingroup$
    Indeed, that will be good to see your solution.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 8:35






  • 1




    $begingroup$
    $1$ is the result!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 16 at 8:42










  • $begingroup$
    Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
    $endgroup$
    – Flammable Maths
    Jan 16 at 8:44








  • 1




    $begingroup$
    Use math.stackexchange.com/questions/469885/…
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 8:57










  • $begingroup$
    Brooo flammy! I'm a huge fan!
    $endgroup$
    – clathratus
    Jan 21 at 16:59










2




2




$begingroup$
Indeed, that will be good to see your solution.
$endgroup$
– mathcounterexamples.net
Jan 16 at 8:35




$begingroup$
Indeed, that will be good to see your solution.
$endgroup$
– mathcounterexamples.net
Jan 16 at 8:35




1




1




$begingroup$
$1$ is the result!
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 8:42




$begingroup$
$1$ is the result!
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 8:42












$begingroup$
Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
$endgroup$
– Flammable Maths
Jan 16 at 8:44






$begingroup$
Here is it, I hope you can make sense of it! =) imgur.com/a/rWhICtl
$endgroup$
– Flammable Maths
Jan 16 at 8:44






1




1




$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 16 at 8:57




$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 16 at 8:57












$begingroup$
Brooo flammy! I'm a huge fan!
$endgroup$
– clathratus
Jan 21 at 16:59






$begingroup$
Brooo flammy! I'm a huge fan!
$endgroup$
– clathratus
Jan 21 at 16:59












3 Answers
3






active

oldest

votes


















4












$begingroup$

HINTS:



(1)
begin{equation}
cosleft(frac{pi}{2n}cdot kright) = Releft[expleft(frac{pi}{2n}cdot k i right) right]
end{equation}



(2)
begin{equation}
expleft(frac{pi}{2n}cdot k i right) = a^k, quad a = expleft(frac{pi}{2n}i right)
end{equation}



(3)
begin{equation}
sum_{k = 1}^{n} a^k = frac{aleft(a^{n} - 1right)}{a - 1}
end{equation}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Like I said, this is exactly my approach^^
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:30






  • 1




    $begingroup$
    Then why did you have to ask the question then?
    $endgroup$
    – DavidG
    Jan 16 at 9:38










  • $begingroup$
    To find a simplified approach.
    $endgroup$
    – Flammable Maths
    Jan 16 at 10:33










  • $begingroup$
    @FlammableMaths - Okay, where did you get stuck?
    $endgroup$
    – DavidG
    Jan 16 at 10:37



















4












$begingroup$

According to this question



$$1 + sumlimits_{k=1}^n cos{(k theta)}=frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)thetaright]}{2sinleft(frac{theta}{2}right)}$$



As a result



$$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)=
limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)frac{pi}{2n}right]}{2sinleft(frac{frac{pi}{2n}}{2}right)}-1right)=\
limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{sinleft(frac{pi}{2}+frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}-frac{1}{2}right)=
limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{cosleft(frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}right)=\
frac{limlimits_{n rightarrowinfty}cosleft(frac{pi}{4n}right)}{limlimits_{n rightarrowinfty} frac{sinleft(frac{pi}{4n}right)}{frac{pi}{4n}}}=frac{1}{1}=1$$
using the fact that $limlimits_{xrightarrow 0}frac{sin{x}}{x}=1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:29



















4












$begingroup$

Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see
$$begin{eqnarray}
sinfrac{theta}{2}sum_{k=1}^ncos ktheta &=&frac{1}{2}sum_{k=0}^nleft(sinfrac{2k+1}{2}theta-sinfrac{2k-1}{2}thetaright)\
&=&frac{1}{2}left(sinfrac{2n+1}{2}theta-sinfrac{1}{2}thetaright).
end{eqnarray}$$ This gives
$$begin{eqnarray}
limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)&=&limlimits_{n rightarrowinfty}frac{pi}{4nsinfrac{pi}{4n}}left(sinfrac{2n+1}{4n}pi-sinfrac{1}{4n}piright)=sinfrac{pi}{2}=1.
end{eqnarray}$$






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    HINTS:



    (1)
    begin{equation}
    cosleft(frac{pi}{2n}cdot kright) = Releft[expleft(frac{pi}{2n}cdot k i right) right]
    end{equation}



    (2)
    begin{equation}
    expleft(frac{pi}{2n}cdot k i right) = a^k, quad a = expleft(frac{pi}{2n}i right)
    end{equation}



    (3)
    begin{equation}
    sum_{k = 1}^{n} a^k = frac{aleft(a^{n} - 1right)}{a - 1}
    end{equation}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Like I said, this is exactly my approach^^
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:30






    • 1




      $begingroup$
      Then why did you have to ask the question then?
      $endgroup$
      – DavidG
      Jan 16 at 9:38










    • $begingroup$
      To find a simplified approach.
      $endgroup$
      – Flammable Maths
      Jan 16 at 10:33










    • $begingroup$
      @FlammableMaths - Okay, where did you get stuck?
      $endgroup$
      – DavidG
      Jan 16 at 10:37
















    4












    $begingroup$

    HINTS:



    (1)
    begin{equation}
    cosleft(frac{pi}{2n}cdot kright) = Releft[expleft(frac{pi}{2n}cdot k i right) right]
    end{equation}



    (2)
    begin{equation}
    expleft(frac{pi}{2n}cdot k i right) = a^k, quad a = expleft(frac{pi}{2n}i right)
    end{equation}



    (3)
    begin{equation}
    sum_{k = 1}^{n} a^k = frac{aleft(a^{n} - 1right)}{a - 1}
    end{equation}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Like I said, this is exactly my approach^^
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:30






    • 1




      $begingroup$
      Then why did you have to ask the question then?
      $endgroup$
      – DavidG
      Jan 16 at 9:38










    • $begingroup$
      To find a simplified approach.
      $endgroup$
      – Flammable Maths
      Jan 16 at 10:33










    • $begingroup$
      @FlammableMaths - Okay, where did you get stuck?
      $endgroup$
      – DavidG
      Jan 16 at 10:37














    4












    4








    4





    $begingroup$

    HINTS:



    (1)
    begin{equation}
    cosleft(frac{pi}{2n}cdot kright) = Releft[expleft(frac{pi}{2n}cdot k i right) right]
    end{equation}



    (2)
    begin{equation}
    expleft(frac{pi}{2n}cdot k i right) = a^k, quad a = expleft(frac{pi}{2n}i right)
    end{equation}



    (3)
    begin{equation}
    sum_{k = 1}^{n} a^k = frac{aleft(a^{n} - 1right)}{a - 1}
    end{equation}






    share|cite|improve this answer









    $endgroup$



    HINTS:



    (1)
    begin{equation}
    cosleft(frac{pi}{2n}cdot kright) = Releft[expleft(frac{pi}{2n}cdot k i right) right]
    end{equation}



    (2)
    begin{equation}
    expleft(frac{pi}{2n}cdot k i right) = a^k, quad a = expleft(frac{pi}{2n}i right)
    end{equation}



    (3)
    begin{equation}
    sum_{k = 1}^{n} a^k = frac{aleft(a^{n} - 1right)}{a - 1}
    end{equation}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 8:56









    DavidGDavidG

    2,1421724




    2,1421724












    • $begingroup$
      Like I said, this is exactly my approach^^
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:30






    • 1




      $begingroup$
      Then why did you have to ask the question then?
      $endgroup$
      – DavidG
      Jan 16 at 9:38










    • $begingroup$
      To find a simplified approach.
      $endgroup$
      – Flammable Maths
      Jan 16 at 10:33










    • $begingroup$
      @FlammableMaths - Okay, where did you get stuck?
      $endgroup$
      – DavidG
      Jan 16 at 10:37


















    • $begingroup$
      Like I said, this is exactly my approach^^
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:30






    • 1




      $begingroup$
      Then why did you have to ask the question then?
      $endgroup$
      – DavidG
      Jan 16 at 9:38










    • $begingroup$
      To find a simplified approach.
      $endgroup$
      – Flammable Maths
      Jan 16 at 10:33










    • $begingroup$
      @FlammableMaths - Okay, where did you get stuck?
      $endgroup$
      – DavidG
      Jan 16 at 10:37
















    $begingroup$
    Like I said, this is exactly my approach^^
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:30




    $begingroup$
    Like I said, this is exactly my approach^^
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:30




    1




    1




    $begingroup$
    Then why did you have to ask the question then?
    $endgroup$
    – DavidG
    Jan 16 at 9:38




    $begingroup$
    Then why did you have to ask the question then?
    $endgroup$
    – DavidG
    Jan 16 at 9:38












    $begingroup$
    To find a simplified approach.
    $endgroup$
    – Flammable Maths
    Jan 16 at 10:33




    $begingroup$
    To find a simplified approach.
    $endgroup$
    – Flammable Maths
    Jan 16 at 10:33












    $begingroup$
    @FlammableMaths - Okay, where did you get stuck?
    $endgroup$
    – DavidG
    Jan 16 at 10:37




    $begingroup$
    @FlammableMaths - Okay, where did you get stuck?
    $endgroup$
    – DavidG
    Jan 16 at 10:37











    4












    $begingroup$

    According to this question



    $$1 + sumlimits_{k=1}^n cos{(k theta)}=frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)thetaright]}{2sinleft(frac{theta}{2}right)}$$



    As a result



    $$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)frac{pi}{2n}right]}{2sinleft(frac{frac{pi}{2n}}{2}right)}-1right)=\
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{sinleft(frac{pi}{2}+frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}-frac{1}{2}right)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{cosleft(frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}right)=\
    frac{limlimits_{n rightarrowinfty}cosleft(frac{pi}{4n}right)}{limlimits_{n rightarrowinfty} frac{sinleft(frac{pi}{4n}right)}{frac{pi}{4n}}}=frac{1}{1}=1$$
    using the fact that $limlimits_{xrightarrow 0}frac{sin{x}}{x}=1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:29
















    4












    $begingroup$

    According to this question



    $$1 + sumlimits_{k=1}^n cos{(k theta)}=frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)thetaright]}{2sinleft(frac{theta}{2}right)}$$



    As a result



    $$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)frac{pi}{2n}right]}{2sinleft(frac{frac{pi}{2n}}{2}right)}-1right)=\
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{sinleft(frac{pi}{2}+frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}-frac{1}{2}right)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{cosleft(frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}right)=\
    frac{limlimits_{n rightarrowinfty}cosleft(frac{pi}{4n}right)}{limlimits_{n rightarrowinfty} frac{sinleft(frac{pi}{4n}right)}{frac{pi}{4n}}}=frac{1}{1}=1$$
    using the fact that $limlimits_{xrightarrow 0}frac{sin{x}}{x}=1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:29














    4












    4








    4





    $begingroup$

    According to this question



    $$1 + sumlimits_{k=1}^n cos{(k theta)}=frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)thetaright]}{2sinleft(frac{theta}{2}right)}$$



    As a result



    $$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)frac{pi}{2n}right]}{2sinleft(frac{frac{pi}{2n}}{2}right)}-1right)=\
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{sinleft(frac{pi}{2}+frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}-frac{1}{2}right)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{cosleft(frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}right)=\
    frac{limlimits_{n rightarrowinfty}cosleft(frac{pi}{4n}right)}{limlimits_{n rightarrowinfty} frac{sinleft(frac{pi}{4n}right)}{frac{pi}{4n}}}=frac{1}{1}=1$$
    using the fact that $limlimits_{xrightarrow 0}frac{sin{x}}{x}=1$.






    share|cite|improve this answer









    $endgroup$



    According to this question



    $$1 + sumlimits_{k=1}^n cos{(k theta)}=frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)thetaright]}{2sinleft(frac{theta}{2}right)}$$



    As a result



    $$limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{1}{2}+frac{sinleft[left(n+frac{1}{2}right)frac{pi}{2n}right]}{2sinleft(frac{frac{pi}{2n}}{2}right)}-1right)=\
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{sinleft(frac{pi}{2}+frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}-frac{1}{2}right)=
    limlimits_{n rightarrowinfty}frac{pi}{2n}left(frac{cosleft(frac{pi}{4n}right)}{2sinleft(frac{pi}{4n}right)}right)=\
    frac{limlimits_{n rightarrowinfty}cosleft(frac{pi}{4n}right)}{limlimits_{n rightarrowinfty} frac{sinleft(frac{pi}{4n}right)}{frac{pi}{4n}}}=frac{1}{1}=1$$
    using the fact that $limlimits_{xrightarrow 0}frac{sin{x}}{x}=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 9:25









    rtybasertybase

    11k21533




    11k21533








    • 1




      $begingroup$
      Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:29














    • 1




      $begingroup$
      Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
      $endgroup$
      – Flammable Maths
      Jan 16 at 9:29








    1




    1




    $begingroup$
    Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:29




    $begingroup$
    Omg! The dirichlet kernel... That's basically where I also got the idea from regarding the use of the geometric series. I should have just rewritten everything ^^' Awesome, great simplification =)
    $endgroup$
    – Flammable Maths
    Jan 16 at 9:29











    4












    $begingroup$

    Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see
    $$begin{eqnarray}
    sinfrac{theta}{2}sum_{k=1}^ncos ktheta &=&frac{1}{2}sum_{k=0}^nleft(sinfrac{2k+1}{2}theta-sinfrac{2k-1}{2}thetaright)\
    &=&frac{1}{2}left(sinfrac{2n+1}{2}theta-sinfrac{1}{2}thetaright).
    end{eqnarray}$$     This gives
    $$begin{eqnarray}
    limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)&=&limlimits_{n rightarrowinfty}frac{pi}{4nsinfrac{pi}{4n}}left(sinfrac{2n+1}{4n}pi-sinfrac{1}{4n}piright)=sinfrac{pi}{2}=1.
    end{eqnarray}$$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see
      $$begin{eqnarray}
      sinfrac{theta}{2}sum_{k=1}^ncos ktheta &=&frac{1}{2}sum_{k=0}^nleft(sinfrac{2k+1}{2}theta-sinfrac{2k-1}{2}thetaright)\
      &=&frac{1}{2}left(sinfrac{2n+1}{2}theta-sinfrac{1}{2}thetaright).
      end{eqnarray}$$       This gives
      $$begin{eqnarray}
      limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)&=&limlimits_{n rightarrowinfty}frac{pi}{4nsinfrac{pi}{4n}}left(sinfrac{2n+1}{4n}pi-sinfrac{1}{4n}piright)=sinfrac{pi}{2}=1.
      end{eqnarray}$$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see
        $$begin{eqnarray}
        sinfrac{theta}{2}sum_{k=1}^ncos ktheta &=&frac{1}{2}sum_{k=0}^nleft(sinfrac{2k+1}{2}theta-sinfrac{2k-1}{2}thetaright)\
        &=&frac{1}{2}left(sinfrac{2n+1}{2}theta-sinfrac{1}{2}thetaright).
        end{eqnarray}$$         This gives
        $$begin{eqnarray}
        limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)&=&limlimits_{n rightarrowinfty}frac{pi}{4nsinfrac{pi}{4n}}left(sinfrac{2n+1}{4n}pi-sinfrac{1}{4n}piright)=sinfrac{pi}{2}=1.
        end{eqnarray}$$






        share|cite|improve this answer











        $endgroup$



        Here's an approach without Euler's formula using telescoping. With the help of the sum-product formula, we can see
        $$begin{eqnarray}
        sinfrac{theta}{2}sum_{k=1}^ncos ktheta &=&frac{1}{2}sum_{k=0}^nleft(sinfrac{2k+1}{2}theta-sinfrac{2k-1}{2}thetaright)\
        &=&frac{1}{2}left(sinfrac{2n+1}{2}theta-sinfrac{1}{2}thetaright).
        end{eqnarray}$$         This gives
        $$begin{eqnarray}
        limlimits_{n rightarrowinfty}frac{pi}{2n}sumlimits_{k=1}^{n}cosleft(frac{pi}{2n}kright)&=&limlimits_{n rightarrowinfty}frac{pi}{4nsinfrac{pi}{4n}}left(sinfrac{2n+1}{4n}pi-sinfrac{1}{4n}piright)=sinfrac{pi}{2}=1.
        end{eqnarray}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 10:10

























        answered Jan 16 at 10:02









        SongSong

        14.2k1633




        14.2k1633






























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