Prove...












7












$begingroup$


I have to prove that the following limit is equal to $sqrt{pi/2}$:
$$lim_{ntoinfty}frac{2cdot4cdot6cdot...cdot(2n-2)(2n)}{1cdot3cdot5cdot...cdot(2n-1)}frac{1}{sqrt{2n+1}}=sqrtfrac{pi}{2}$$



In order to calculate this limit, we know that:
$$I_n=int_0^{frac{pi}{2}}sin^nx dxquad I_{2n}=frac{1cdot3cdot..cdot(2n-3)(2n-1)}{2cdot4cdot..cdot(2n-2)(2n)}frac{pi}{2}quad I_{2n+1}=frac{2cdot4cdot..cdot(2n-2)(2n)}{1cdot3cdot..cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$lim_{ntoinfty}frac{1}{I_{2n}sqrt{2n+1}}frac{pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!






integration sequences-and-series limits







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$endgroup$








  • 1




    $begingroup$
    You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
    $endgroup$
    – Paramanand Singh
    Jan 6 at 6:05
















7












$begingroup$


I have to prove that the following limit is equal to $sqrt{pi/2}$:
$$lim_{ntoinfty}frac{2cdot4cdot6cdot...cdot(2n-2)(2n)}{1cdot3cdot5cdot...cdot(2n-1)}frac{1}{sqrt{2n+1}}=sqrtfrac{pi}{2}$$



In order to calculate this limit, we know that:
$$I_n=int_0^{frac{pi}{2}}sin^nx dxquad I_{2n}=frac{1cdot3cdot..cdot(2n-3)(2n-1)}{2cdot4cdot..cdot(2n-2)(2n)}frac{pi}{2}quad I_{2n+1}=frac{2cdot4cdot..cdot(2n-2)(2n)}{1cdot3cdot..cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$lim_{ntoinfty}frac{1}{I_{2n}sqrt{2n+1}}frac{pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!






integration sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
    $endgroup$
    – Paramanand Singh
    Jan 6 at 6:05














7












7








7





$begingroup$


I have to prove that the following limit is equal to $sqrt{pi/2}$:
$$lim_{ntoinfty}frac{2cdot4cdot6cdot...cdot(2n-2)(2n)}{1cdot3cdot5cdot...cdot(2n-1)}frac{1}{sqrt{2n+1}}=sqrtfrac{pi}{2}$$



In order to calculate this limit, we know that:
$$I_n=int_0^{frac{pi}{2}}sin^nx dxquad I_{2n}=frac{1cdot3cdot..cdot(2n-3)(2n-1)}{2cdot4cdot..cdot(2n-2)(2n)}frac{pi}{2}quad I_{2n+1}=frac{2cdot4cdot..cdot(2n-2)(2n)}{1cdot3cdot..cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$lim_{ntoinfty}frac{1}{I_{2n}sqrt{2n+1}}frac{pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!






integration sequences-and-series limits







share|cite|improve this question











$endgroup$




I have to prove that the following limit is equal to $sqrt{pi/2}$:
$$lim_{ntoinfty}frac{2cdot4cdot6cdot...cdot(2n-2)(2n)}{1cdot3cdot5cdot...cdot(2n-1)}frac{1}{sqrt{2n+1}}=sqrtfrac{pi}{2}$$



In order to calculate this limit, we know that:
$$I_n=int_0^{frac{pi}{2}}sin^nx dxquad I_{2n}=frac{1cdot3cdot..cdot(2n-3)(2n-1)}{2cdot4cdot..cdot(2n-2)(2n)}frac{pi}{2}quad I_{2n+1}=frac{2cdot4cdot..cdot(2n-2)(2n)}{1cdot3cdot..cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$lim_{ntoinfty}frac{1}{I_{2n}sqrt{2n+1}}frac{pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!






integration sequences-and-series limits




integration sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 6:07









Paramanand Singh

49.5k556162




49.5k556162










asked Jan 5 at 10:15









GibbsGibbs

131111




131111








  • 1




    $begingroup$
    You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
    $endgroup$
    – Paramanand Singh
    Jan 6 at 6:05














  • 1




    $begingroup$
    You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
    $endgroup$
    – Paramanand Singh
    Jan 6 at 6:05








1




1




$begingroup$
You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
$endgroup$
– Paramanand Singh
Jan 6 at 6:05




$begingroup$
You are proceeding in correct direction. Just prove that $I_{2n}/I_{2n+1}to 1$. This is done via Squeeze theorem using $I_{2n+1}leq I_{2n}leq I_{2n-1}$ and $I_{2n+1}=dfrac{2n}{2n+1}cdot I_{2n-1}$.
$endgroup$
– Paramanand Singh
Jan 6 at 6:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

Have a look here: Wallis' integrals and here.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Amazingly useful link. Thank you a lot!!
    $endgroup$
    – Gibbs
    Jan 5 at 10:36



















3












$begingroup$

Using the double factorial notation we need to find $$lim_{ntoinfty} frac {(2n)!!}{(2n-1)!!sqrt {2n+1}}$$



Now using the relation between double factorial and the factorial, the limit changes to $$lim_{ntoinfty} frac {2^{2n}(n!)^2}{(2n)!sqrt {2n+1}}$$



Using Stirling's approximation for factorials we get $$lim_{ntoinfty} frac {2^{2n}cdot (2pi n)cdot left(frac ne right)^{2n}}{sqrt {2pi}cdotsqrt {2n} cdotleft(frac {2n}{e}right)^{2n} cdot sqrt {2n+1}}$$



Hence limit changes to $$lim_{ntoinfty} frac {nsqrt {2pi}}{sqrt {2n} cdot sqrt {2n+1}}$$



Which easily evaluates to $sqrt {frac {pi}{2}}$






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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    Have a look here: Wallis' integrals and here.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Amazingly useful link. Thank you a lot!!
      $endgroup$
      – Gibbs
      Jan 5 at 10:36
















    2












    $begingroup$

    Have a look here: Wallis' integrals and here.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Amazingly useful link. Thank you a lot!!
      $endgroup$
      – Gibbs
      Jan 5 at 10:36














    2












    2








    2





    $begingroup$

    Have a look here: Wallis' integrals and here.






    share|cite|improve this answer









    $endgroup$



    Have a look here: Wallis' integrals and here.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 10:28









    Olivier OloaOlivier Oloa

    108k17177294




    108k17177294








    • 1




      $begingroup$
      Amazingly useful link. Thank you a lot!!
      $endgroup$
      – Gibbs
      Jan 5 at 10:36














    • 1




      $begingroup$
      Amazingly useful link. Thank you a lot!!
      $endgroup$
      – Gibbs
      Jan 5 at 10:36








    1




    1




    $begingroup$
    Amazingly useful link. Thank you a lot!!
    $endgroup$
    – Gibbs
    Jan 5 at 10:36




    $begingroup$
    Amazingly useful link. Thank you a lot!!
    $endgroup$
    – Gibbs
    Jan 5 at 10:36











    3












    $begingroup$

    Using the double factorial notation we need to find $$lim_{ntoinfty} frac {(2n)!!}{(2n-1)!!sqrt {2n+1}}$$



    Now using the relation between double factorial and the factorial, the limit changes to $$lim_{ntoinfty} frac {2^{2n}(n!)^2}{(2n)!sqrt {2n+1}}$$



    Using Stirling's approximation for factorials we get $$lim_{ntoinfty} frac {2^{2n}cdot (2pi n)cdot left(frac ne right)^{2n}}{sqrt {2pi}cdotsqrt {2n} cdotleft(frac {2n}{e}right)^{2n} cdot sqrt {2n+1}}$$



    Hence limit changes to $$lim_{ntoinfty} frac {nsqrt {2pi}}{sqrt {2n} cdot sqrt {2n+1}}$$



    Which easily evaluates to $sqrt {frac {pi}{2}}$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Using the double factorial notation we need to find $$lim_{ntoinfty} frac {(2n)!!}{(2n-1)!!sqrt {2n+1}}$$



      Now using the relation between double factorial and the factorial, the limit changes to $$lim_{ntoinfty} frac {2^{2n}(n!)^2}{(2n)!sqrt {2n+1}}$$



      Using Stirling's approximation for factorials we get $$lim_{ntoinfty} frac {2^{2n}cdot (2pi n)cdot left(frac ne right)^{2n}}{sqrt {2pi}cdotsqrt {2n} cdotleft(frac {2n}{e}right)^{2n} cdot sqrt {2n+1}}$$



      Hence limit changes to $$lim_{ntoinfty} frac {nsqrt {2pi}}{sqrt {2n} cdot sqrt {2n+1}}$$



      Which easily evaluates to $sqrt {frac {pi}{2}}$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Using the double factorial notation we need to find $$lim_{ntoinfty} frac {(2n)!!}{(2n-1)!!sqrt {2n+1}}$$



        Now using the relation between double factorial and the factorial, the limit changes to $$lim_{ntoinfty} frac {2^{2n}(n!)^2}{(2n)!sqrt {2n+1}}$$



        Using Stirling's approximation for factorials we get $$lim_{ntoinfty} frac {2^{2n}cdot (2pi n)cdot left(frac ne right)^{2n}}{sqrt {2pi}cdotsqrt {2n} cdotleft(frac {2n}{e}right)^{2n} cdot sqrt {2n+1}}$$



        Hence limit changes to $$lim_{ntoinfty} frac {nsqrt {2pi}}{sqrt {2n} cdot sqrt {2n+1}}$$



        Which easily evaluates to $sqrt {frac {pi}{2}}$






        share|cite|improve this answer









        $endgroup$



        Using the double factorial notation we need to find $$lim_{ntoinfty} frac {(2n)!!}{(2n-1)!!sqrt {2n+1}}$$



        Now using the relation between double factorial and the factorial, the limit changes to $$lim_{ntoinfty} frac {2^{2n}(n!)^2}{(2n)!sqrt {2n+1}}$$



        Using Stirling's approximation for factorials we get $$lim_{ntoinfty} frac {2^{2n}cdot (2pi n)cdot left(frac ne right)^{2n}}{sqrt {2pi}cdotsqrt {2n} cdotleft(frac {2n}{e}right)^{2n} cdot sqrt {2n+1}}$$



        Hence limit changes to $$lim_{ntoinfty} frac {nsqrt {2pi}}{sqrt {2n} cdot sqrt {2n+1}}$$



        Which easily evaluates to $sqrt {frac {pi}{2}}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 4:00









        DigammaDigamma

        6,1621440




        6,1621440






























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