Prove the sequence $a_n =frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit [duplicate]












1















This question already has an answer here:




  • What is the limit of $frac{prod Odd}{prod Even}?!$

    7 answers




I have several questions to ask:



1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.



$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;



bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$



To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?



2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit



$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.



$a_n = (1-1/2)(1-1/4)…(1-1/2n)$










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Nov 25 at 9:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
    – robjohn
    Nov 25 at 1:34
















1















This question already has an answer here:




  • What is the limit of $frac{prod Odd}{prod Even}?!$

    7 answers




I have several questions to ask:



1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.



$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;



bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$



To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?



2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit



$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.



$a_n = (1-1/2)(1-1/4)…(1-1/2n)$










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Nov 25 at 9:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
    – robjohn
    Nov 25 at 1:34














1












1








1








This question already has an answer here:




  • What is the limit of $frac{prod Odd}{prod Even}?!$

    7 answers




I have several questions to ask:



1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.



$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;



bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$



To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?



2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit



$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.



$a_n = (1-1/2)(1-1/4)…(1-1/2n)$










share|cite|improve this question
















This question already has an answer here:




  • What is the limit of $frac{prod Odd}{prod Even}?!$

    7 answers




I have several questions to ask:



1) Show increasing, find the upper bound if you can of
$sqrt{(n^2-1)}/n$.



$sqrt{(n^2-1)}/n= |n|sqrt{1-1/n^2}/n$ if $n$ is positive than $sqrt1$ else $-sqrt1$;



bound: $sqrt{(n^2-1)}/n le sqrt {n^2}/n = |n|/n=1$



To show if it is increasing should I do $frac{sqrt{(n+1)^2-1}}{n+1} ge frac{sqrt{(n)^2-1}}{n}$?



2)Prove the sequence $a_n = frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}$ has a limit



$a_n$ is a decreasing sequence, so to have a limit it must be bounded from below.



$a_n = (1-1/2)(1-1/4)…(1-1/2n)$





This question already has an answer here:




  • What is the limit of $frac{prod Odd}{prod Even}?!$

    7 answers








real-analysis sequences-and-series limits analysis






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edited Nov 24 at 22:40

























asked Nov 24 at 22:06









Sargis Iskandaryan

560112




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Nov 25 at 9:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Nosrati, Jyrki Lahtonen, Mostafa Ayaz, Did limits
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Nov 25 at 9:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
    – robjohn
    Nov 25 at 1:34














  • 1




    If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
    – robjohn
    Nov 25 at 1:34








1




1




If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn
Nov 25 at 1:34




If you have several questions to ask, it is better to ask separate questions, unless the questions are closely related.
– robjohn
Nov 25 at 1:34










4 Answers
4






active

oldest

votes


















4














Cross-multiplication yields, for $kge1$,
$$
left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
$$

Therefore,
$$
begin{align}
prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
&leprod_{k=1}^nfrac{2k-1}{2k+1}\
&=frac1{2n+1}
end{align}
$$

Thus,
$$
prod_{k=1}^inftyfrac{2k-1}{2k}=0
$$





For $nge1$,
$$
begin{align}
frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
&=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
&=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
&lt1
end{align}
$$

Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
$$
begin{align}
lim_{ntoinfty}frac{sqrt{n^2-1}}n
&=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
&=1
end{align}
$$






share|cite|improve this answer































    3














    HINT



    We have that



    $$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$



    then we can use Stirling's approximation.



    Refer to the related




    • What is the limit of $frac{prod Odd}{prod Even}?!$






    share|cite|improve this answer





















    • Can you help with the other two?
      – Sargis Iskandaryan
      Nov 24 at 22:22










    • @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
      – gimusi
      Nov 24 at 22:28










    • @SargisIskandaryan I don't understand the first question.
      – gimusi
      Nov 24 at 22:29



















    2














    $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
    is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

    Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.






    share|cite|improve this answer





























      0














      The sequence is decreasing and bounded below by 0. That's why it has a limit.






      share|cite|improve this answer




























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        Cross-multiplication yields, for $kge1$,
        $$
        left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
        $$

        Therefore,
        $$
        begin{align}
        prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
        &leprod_{k=1}^nfrac{2k-1}{2k+1}\
        &=frac1{2n+1}
        end{align}
        $$

        Thus,
        $$
        prod_{k=1}^inftyfrac{2k-1}{2k}=0
        $$





        For $nge1$,
        $$
        begin{align}
        frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
        &=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
        &=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
        &lt1
        end{align}
        $$

        Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
        $$
        begin{align}
        lim_{ntoinfty}frac{sqrt{n^2-1}}n
        &=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
        &=1
        end{align}
        $$






        share|cite|improve this answer




























          4














          Cross-multiplication yields, for $kge1$,
          $$
          left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
          $$

          Therefore,
          $$
          begin{align}
          prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
          &leprod_{k=1}^nfrac{2k-1}{2k+1}\
          &=frac1{2n+1}
          end{align}
          $$

          Thus,
          $$
          prod_{k=1}^inftyfrac{2k-1}{2k}=0
          $$





          For $nge1$,
          $$
          begin{align}
          frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
          &=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
          &=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
          &lt1
          end{align}
          $$

          Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
          $$
          begin{align}
          lim_{ntoinfty}frac{sqrt{n^2-1}}n
          &=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
          &=1
          end{align}
          $$






          share|cite|improve this answer


























            4












            4








            4






            Cross-multiplication yields, for $kge1$,
            $$
            left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
            $$

            Therefore,
            $$
            begin{align}
            prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
            &leprod_{k=1}^nfrac{2k-1}{2k+1}\
            &=frac1{2n+1}
            end{align}
            $$

            Thus,
            $$
            prod_{k=1}^inftyfrac{2k-1}{2k}=0
            $$





            For $nge1$,
            $$
            begin{align}
            frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
            &=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
            &=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
            &lt1
            end{align}
            $$

            Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
            $$
            begin{align}
            lim_{ntoinfty}frac{sqrt{n^2-1}}n
            &=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
            &=1
            end{align}
            $$






            share|cite|improve this answer














            Cross-multiplication yields, for $kge1$,
            $$
            left(frac{2k-1}{2k}right)^2lefrac{2k-1}{2k+1}
            $$

            Therefore,
            $$
            begin{align}
            prod_{k=1}^nleft(frac{2k-1}{2k}right)^2
            &leprod_{k=1}^nfrac{2k-1}{2k+1}\
            &=frac1{2n+1}
            end{align}
            $$

            Thus,
            $$
            prod_{k=1}^inftyfrac{2k-1}{2k}=0
            $$





            For $nge1$,
            $$
            begin{align}
            frac{frac{sqrt{n^2-1}}n}{frac{sqrt{(n+1)^2-1}}{n+1}}
            &=sqrt{frac{(n+1)^3(n-1)}{n^3(n+2)}}\
            &=sqrt{frac{n^4+2n^3-2n-1}{n^4+2n^3}}\[18pt]
            &lt1
            end{align}
            $$

            Thus, $frac{sqrt{n^2-1}}n$ is increasing. Furthermore,
            $$
            begin{align}
            lim_{ntoinfty}frac{sqrt{n^2-1}}n
            &=lim_{ntoinfty}sqrt{1-frac1{n^2}}\[6pt]
            &=1
            end{align}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 at 4:35

























            answered Nov 24 at 22:50









            robjohn

            264k27302623




            264k27302623























                3














                HINT



                We have that



                $$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$



                then we can use Stirling's approximation.



                Refer to the related




                • What is the limit of $frac{prod Odd}{prod Even}?!$






                share|cite|improve this answer





















                • Can you help with the other two?
                  – Sargis Iskandaryan
                  Nov 24 at 22:22










                • @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                  – gimusi
                  Nov 24 at 22:28










                • @SargisIskandaryan I don't understand the first question.
                  – gimusi
                  Nov 24 at 22:29
















                3














                HINT



                We have that



                $$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$



                then we can use Stirling's approximation.



                Refer to the related




                • What is the limit of $frac{prod Odd}{prod Even}?!$






                share|cite|improve this answer





















                • Can you help with the other two?
                  – Sargis Iskandaryan
                  Nov 24 at 22:22










                • @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                  – gimusi
                  Nov 24 at 22:28










                • @SargisIskandaryan I don't understand the first question.
                  – gimusi
                  Nov 24 at 22:29














                3












                3








                3






                HINT



                We have that



                $$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$



                then we can use Stirling's approximation.



                Refer to the related




                • What is the limit of $frac{prod Odd}{prod Even}?!$






                share|cite|improve this answer












                HINT



                We have that



                $$a_n=frac{1cdot 3cdot…cdot(2n-1)}{2cdot4cdot…cdot2n}=frac{(2n)!}{(2^nn!)^2}$$



                then we can use Stirling's approximation.



                Refer to the related




                • What is the limit of $frac{prod Odd}{prod Even}?!$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 at 22:09









                gimusi

                1




                1












                • Can you help with the other two?
                  – Sargis Iskandaryan
                  Nov 24 at 22:22










                • @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                  – gimusi
                  Nov 24 at 22:28










                • @SargisIskandaryan I don't understand the first question.
                  – gimusi
                  Nov 24 at 22:29


















                • Can you help with the other two?
                  – Sargis Iskandaryan
                  Nov 24 at 22:22










                • @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                  – gimusi
                  Nov 24 at 22:28










                • @SargisIskandaryan I don't understand the first question.
                  – gimusi
                  Nov 24 at 22:29
















                Can you help with the other two?
                – Sargis Iskandaryan
                Nov 24 at 22:22




                Can you help with the other two?
                – Sargis Iskandaryan
                Nov 24 at 22:22












                @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                – gimusi
                Nov 24 at 22:28




                @SargisIskandaryan The second one is trivial indeed $frac{n^n}{n!}=frac{ncdot nldots n}{1cdot 2cdot n}$ or by ratio test for example.
                – gimusi
                Nov 24 at 22:28












                @SargisIskandaryan I don't understand the first question.
                – gimusi
                Nov 24 at 22:29




                @SargisIskandaryan I don't understand the first question.
                – gimusi
                Nov 24 at 22:29











                2














                $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
                is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

                Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.






                share|cite|improve this answer


























                  2














                  $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
                  is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

                  Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
                    is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

                    Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.






                    share|cite|improve this answer












                    $$frac{(2n-1)!!}{(2n)!!}=frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}left(costhetaright)^{2n},dtheta $$
                    is quite clearly decreasing and convergent to zero (by the dominated convergence theorem).

                    Laplace's method gives $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pileft(n+frac{1}{4}right)}}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 22:24









                    Jack D'Aurizio

                    285k33278655




                    285k33278655























                        0














                        The sequence is decreasing and bounded below by 0. That's why it has a limit.






                        share|cite|improve this answer


























                          0














                          The sequence is decreasing and bounded below by 0. That's why it has a limit.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            The sequence is decreasing and bounded below by 0. That's why it has a limit.






                            share|cite|improve this answer












                            The sequence is decreasing and bounded below by 0. That's why it has a limit.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 22:10









                            Martin Erhardt

                            1859




                            1859















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