Testing for convergence in Infinite series with factorial in numerator












1












$begingroup$


I have the following infinite series that I need to test for convergence/divergence:



$$sum_{n=1}^{infty} frac{n!}{1 times 3 times 5 times cdots times (2n-1)}$$



I can see that the denominator will eventually blow up and surpass the numerator, and so it would seem that the series would converge, but I am not sure how to test this algebraically given the factorial in the numerator and the sequence in denominator. The recursive function for factorial $n! = n times (n-1)!$ doesn't seem to simplify things in this case, as I cannot eliminate the $(2n-1)$ in the denominator. Is there a way to find a general equation for the denominator such that I could perform convergence tests (e.g. by taking the integral, limit comparison, etc.)






sequences-and-series factorial







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  • 2




    $begingroup$
    This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
    $endgroup$
    – Fixed Point
    May 2 '13 at 5:14












  • $begingroup$
    I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
    $endgroup$
    – dtg
    May 2 '13 at 5:32










  • $begingroup$
    No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
    $endgroup$
    – Fixed Point
    May 2 '13 at 6:01
















1












$begingroup$


I have the following infinite series that I need to test for convergence/divergence:



$$sum_{n=1}^{infty} frac{n!}{1 times 3 times 5 times cdots times (2n-1)}$$



I can see that the denominator will eventually blow up and surpass the numerator, and so it would seem that the series would converge, but I am not sure how to test this algebraically given the factorial in the numerator and the sequence in denominator. The recursive function for factorial $n! = n times (n-1)!$ doesn't seem to simplify things in this case, as I cannot eliminate the $(2n-1)$ in the denominator. Is there a way to find a general equation for the denominator such that I could perform convergence tests (e.g. by taking the integral, limit comparison, etc.)






sequences-and-series factorial







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
    $endgroup$
    – Fixed Point
    May 2 '13 at 5:14












  • $begingroup$
    I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
    $endgroup$
    – dtg
    May 2 '13 at 5:32










  • $begingroup$
    No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
    $endgroup$
    – Fixed Point
    May 2 '13 at 6:01














1












1








1





$begingroup$


I have the following infinite series that I need to test for convergence/divergence:



$$sum_{n=1}^{infty} frac{n!}{1 times 3 times 5 times cdots times (2n-1)}$$



I can see that the denominator will eventually blow up and surpass the numerator, and so it would seem that the series would converge, but I am not sure how to test this algebraically given the factorial in the numerator and the sequence in denominator. The recursive function for factorial $n! = n times (n-1)!$ doesn't seem to simplify things in this case, as I cannot eliminate the $(2n-1)$ in the denominator. Is there a way to find a general equation for the denominator such that I could perform convergence tests (e.g. by taking the integral, limit comparison, etc.)






sequences-and-series factorial







share|cite|improve this question











$endgroup$




I have the following infinite series that I need to test for convergence/divergence:



$$sum_{n=1}^{infty} frac{n!}{1 times 3 times 5 times cdots times (2n-1)}$$



I can see that the denominator will eventually blow up and surpass the numerator, and so it would seem that the series would converge, but I am not sure how to test this algebraically given the factorial in the numerator and the sequence in denominator. The recursive function for factorial $n! = n times (n-1)!$ doesn't seem to simplify things in this case, as I cannot eliminate the $(2n-1)$ in the denominator. Is there a way to find a general equation for the denominator such that I could perform convergence tests (e.g. by taking the integral, limit comparison, etc.)






sequences-and-series factorial




sequences-and-series factorial






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited May 2 '13 at 5:14









Michael Hardy

1




1










asked May 2 '13 at 5:11









dtgdtg

645102333




645102333








  • 2




    $begingroup$
    This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
    $endgroup$
    – Fixed Point
    May 2 '13 at 5:14












  • $begingroup$
    I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
    $endgroup$
    – dtg
    May 2 '13 at 5:32










  • $begingroup$
    No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
    $endgroup$
    – Fixed Point
    May 2 '13 at 6:01














  • 2




    $begingroup$
    This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
    $endgroup$
    – Fixed Point
    May 2 '13 at 5:14












  • $begingroup$
    I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
    $endgroup$
    – dtg
    May 2 '13 at 5:32










  • $begingroup$
    No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
    $endgroup$
    – Fixed Point
    May 2 '13 at 6:01








2




2




$begingroup$
This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
$endgroup$
– Fixed Point
May 2 '13 at 5:14






$begingroup$
This problem yells "ratio test" to me. And by the way, the numerator will also blow up. I hope you see that the bottom is (2n-1)! with half the terms missing. So twice as long with half the terms missing. What does your intuition say?
$endgroup$
– Fixed Point
May 2 '13 at 5:14














$begingroup$
I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
$endgroup$
– dtg
May 2 '13 at 5:32




$begingroup$
I see now that $(2n-1)!!$ stands for odd numbered factorials. I did not know this until now, and I probably would have been lost for hours without this bit of knowledge!
$endgroup$
– dtg
May 2 '13 at 5:32












$begingroup$
No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
$endgroup$
– Fixed Point
May 2 '13 at 6:01




$begingroup$
No worries ;-) that is how we all learn. I still lose hours because of what I don't know.
$endgroup$
– Fixed Point
May 2 '13 at 6:01










3 Answers
3






active

oldest

votes


















4












$begingroup$

Hint: Use the Ratio Test. No modification of the expression is needed.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    We have
    $$a_n = dfrac{n!}{(2n-1)!!} = dfrac{n!}{(2n)!} times 2^n n! = dfrac{2^n}{dbinom{2n}n}$$
    Use ratio test now to get that
    $$dfrac{a_{n+1}}{a_n} = dfrac{2^{n+1}}{dbinom{2n+2}{n+1}} cdot dfrac{dbinom{2n}n}{2^n} = dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = dfrac{n+1}{2n+1}$$
    We can also use Stirling. From Stirling, we have
    $$dbinom{2n}n sim dfrac{4^n}{sqrt{pi n}}$$
    Use this to conclude, about the convergence/divergence of the series.




    EDIT



    $$1 times 3 times 5 times cdots times(2n-1) = dfrac{left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right)}{ left(2 times 4 times cdots times (2n)right)}$$
    Now note that
    $$left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right) = (2n)!$$
    and
    $$left(2 times 4 times cdots times (2n)right) = 2^n left(1 times 2 times cdots times nright) = 2^n n!$$
    Hence,
    $$1 times 3 times 5 times cdots times(2n-1) = dfrac{(2n)!}{2^n cdot n!}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
      $endgroup$
      – dtg
      May 2 '13 at 5:38






    • 1




      $begingroup$
      Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
      $endgroup$
      – André Nicolas
      May 2 '13 at 5:48










    • $begingroup$
      Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
      $endgroup$
      – dtg
      May 2 '13 at 5:50










    • $begingroup$
      I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
      $endgroup$
      – dtg
      May 2 '13 at 5:53



















    0












    $begingroup$

    Note that
    $$
    frac{n!}{(2n-1)!!}=frac11frac23frac35frac47cdotsfrac{n}{2n-1}
    $$
    For $nge2$, $dfrac{n}{2n-1}ledfrac23$. Thus, for $nge1$, we have
    $$
    frac{n!}{(2n-1)!!}leleft(frac23right)^{n-1}
    $$
    We can then compare with a geometric series.






    share|cite|improve this answer











    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Hint: Use the Ratio Test. No modification of the expression is needed.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Hint: Use the Ratio Test. No modification of the expression is needed.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Hint: Use the Ratio Test. No modification of the expression is needed.






          share|cite|improve this answer









          $endgroup$



          Hint: Use the Ratio Test. No modification of the expression is needed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 2 '13 at 5:15









          André NicolasAndré Nicolas

          455k36432820




          455k36432820























              3












              $begingroup$

              We have
              $$a_n = dfrac{n!}{(2n-1)!!} = dfrac{n!}{(2n)!} times 2^n n! = dfrac{2^n}{dbinom{2n}n}$$
              Use ratio test now to get that
              $$dfrac{a_{n+1}}{a_n} = dfrac{2^{n+1}}{dbinom{2n+2}{n+1}} cdot dfrac{dbinom{2n}n}{2^n} = dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = dfrac{n+1}{2n+1}$$
              We can also use Stirling. From Stirling, we have
              $$dbinom{2n}n sim dfrac{4^n}{sqrt{pi n}}$$
              Use this to conclude, about the convergence/divergence of the series.




              EDIT



              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right)}{ left(2 times 4 times cdots times (2n)right)}$$
              Now note that
              $$left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right) = (2n)!$$
              and
              $$left(2 times 4 times cdots times (2n)right) = 2^n left(1 times 2 times cdots times nright) = 2^n n!$$
              Hence,
              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{(2n)!}{2^n cdot n!}$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
                $endgroup$
                – dtg
                May 2 '13 at 5:38






              • 1




                $begingroup$
                Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
                $endgroup$
                – André Nicolas
                May 2 '13 at 5:48










              • $begingroup$
                Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
                $endgroup$
                – dtg
                May 2 '13 at 5:50










              • $begingroup$
                I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
                $endgroup$
                – dtg
                May 2 '13 at 5:53
















              3












              $begingroup$

              We have
              $$a_n = dfrac{n!}{(2n-1)!!} = dfrac{n!}{(2n)!} times 2^n n! = dfrac{2^n}{dbinom{2n}n}$$
              Use ratio test now to get that
              $$dfrac{a_{n+1}}{a_n} = dfrac{2^{n+1}}{dbinom{2n+2}{n+1}} cdot dfrac{dbinom{2n}n}{2^n} = dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = dfrac{n+1}{2n+1}$$
              We can also use Stirling. From Stirling, we have
              $$dbinom{2n}n sim dfrac{4^n}{sqrt{pi n}}$$
              Use this to conclude, about the convergence/divergence of the series.




              EDIT



              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right)}{ left(2 times 4 times cdots times (2n)right)}$$
              Now note that
              $$left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right) = (2n)!$$
              and
              $$left(2 times 4 times cdots times (2n)right) = 2^n left(1 times 2 times cdots times nright) = 2^n n!$$
              Hence,
              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{(2n)!}{2^n cdot n!}$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
                $endgroup$
                – dtg
                May 2 '13 at 5:38






              • 1




                $begingroup$
                Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
                $endgroup$
                – André Nicolas
                May 2 '13 at 5:48










              • $begingroup$
                Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
                $endgroup$
                – dtg
                May 2 '13 at 5:50










              • $begingroup$
                I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
                $endgroup$
                – dtg
                May 2 '13 at 5:53














              3












              3








              3





              $begingroup$

              We have
              $$a_n = dfrac{n!}{(2n-1)!!} = dfrac{n!}{(2n)!} times 2^n n! = dfrac{2^n}{dbinom{2n}n}$$
              Use ratio test now to get that
              $$dfrac{a_{n+1}}{a_n} = dfrac{2^{n+1}}{dbinom{2n+2}{n+1}} cdot dfrac{dbinom{2n}n}{2^n} = dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = dfrac{n+1}{2n+1}$$
              We can also use Stirling. From Stirling, we have
              $$dbinom{2n}n sim dfrac{4^n}{sqrt{pi n}}$$
              Use this to conclude, about the convergence/divergence of the series.




              EDIT



              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right)}{ left(2 times 4 times cdots times (2n)right)}$$
              Now note that
              $$left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right) = (2n)!$$
              and
              $$left(2 times 4 times cdots times (2n)right) = 2^n left(1 times 2 times cdots times nright) = 2^n n!$$
              Hence,
              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{(2n)!}{2^n cdot n!}$$






              share|cite|improve this answer











              $endgroup$



              We have
              $$a_n = dfrac{n!}{(2n-1)!!} = dfrac{n!}{(2n)!} times 2^n n! = dfrac{2^n}{dbinom{2n}n}$$
              Use ratio test now to get that
              $$dfrac{a_{n+1}}{a_n} = dfrac{2^{n+1}}{dbinom{2n+2}{n+1}} cdot dfrac{dbinom{2n}n}{2^n} = dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = dfrac{n+1}{2n+1}$$
              We can also use Stirling. From Stirling, we have
              $$dbinom{2n}n sim dfrac{4^n}{sqrt{pi n}}$$
              Use this to conclude, about the convergence/divergence of the series.




              EDIT



              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right)}{ left(2 times 4 times cdots times (2n)right)}$$
              Now note that
              $$left( 1 times 3 times 5 times cdots times(2n-1) right) times left(2 times 4 times cdots times (2n)right) = (2n)!$$
              and
              $$left(2 times 4 times cdots times (2n)right) = 2^n left(1 times 2 times cdots times nright) = 2^n n!$$
              Hence,
              $$1 times 3 times 5 times cdots times(2n-1) = dfrac{(2n)!}{2^n cdot n!}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited May 2 '13 at 5:45

























              answered May 2 '13 at 5:14







              user17762



















              • $begingroup$
                Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
                $endgroup$
                – dtg
                May 2 '13 at 5:38






              • 1




                $begingroup$
                Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
                $endgroup$
                – André Nicolas
                May 2 '13 at 5:48










              • $begingroup$
                Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
                $endgroup$
                – dtg
                May 2 '13 at 5:50










              • $begingroup$
                I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
                $endgroup$
                – dtg
                May 2 '13 at 5:53


















              • $begingroup$
                Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
                $endgroup$
                – dtg
                May 2 '13 at 5:38






              • 1




                $begingroup$
                Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
                $endgroup$
                – André Nicolas
                May 2 '13 at 5:48










              • $begingroup$
                Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
                $endgroup$
                – dtg
                May 2 '13 at 5:50










              • $begingroup$
                I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
                $endgroup$
                – dtg
                May 2 '13 at 5:53
















              $begingroup$
              Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
              $endgroup$
              – dtg
              May 2 '13 at 5:38




              $begingroup$
              Yikes. It is not at all apparent how $frac{1}{(2n-1)!!}$ becomes $frac{2^{n}n!}{(2n)!}$
              $endgroup$
              – dtg
              May 2 '13 at 5:38




              1




              1




              $begingroup$
              Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
              $endgroup$
              – André Nicolas
              May 2 '13 at 5:48




              $begingroup$
              Soon it will be obvious. Put in the "missing" terms $2,4,6,8,dots,2n$. The product of what you just put in is $(2cdot 1)(2cdot 2)(2cdot 3)cdots(2cdot n)$, which is $(2^n)(n!)$.
              $endgroup$
              – André Nicolas
              May 2 '13 at 5:48












              $begingroup$
              Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
              $endgroup$
              – dtg
              May 2 '13 at 5:50




              $begingroup$
              Also, is the parenthesis thing a binomial coefficient? I am not sure how to solve those. Sounds like I need to go back and do more studying on these ones.
              $endgroup$
              – dtg
              May 2 '13 at 5:50












              $begingroup$
              I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
              $endgroup$
              – dtg
              May 2 '13 at 5:53




              $begingroup$
              I see now -- you have to divide out all the even numbered terms, which are represented by $(2^n)(n!)$. You guys make it sound as though it were so easy! Not so for people like me who suck at Math. Thanks.
              $endgroup$
              – dtg
              May 2 '13 at 5:53











              0












              $begingroup$

              Note that
              $$
              frac{n!}{(2n-1)!!}=frac11frac23frac35frac47cdotsfrac{n}{2n-1}
              $$
              For $nge2$, $dfrac{n}{2n-1}ledfrac23$. Thus, for $nge1$, we have
              $$
              frac{n!}{(2n-1)!!}leleft(frac23right)^{n-1}
              $$
              We can then compare with a geometric series.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Note that
                $$
                frac{n!}{(2n-1)!!}=frac11frac23frac35frac47cdotsfrac{n}{2n-1}
                $$
                For $nge2$, $dfrac{n}{2n-1}ledfrac23$. Thus, for $nge1$, we have
                $$
                frac{n!}{(2n-1)!!}leleft(frac23right)^{n-1}
                $$
                We can then compare with a geometric series.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that
                  $$
                  frac{n!}{(2n-1)!!}=frac11frac23frac35frac47cdotsfrac{n}{2n-1}
                  $$
                  For $nge2$, $dfrac{n}{2n-1}ledfrac23$. Thus, for $nge1$, we have
                  $$
                  frac{n!}{(2n-1)!!}leleft(frac23right)^{n-1}
                  $$
                  We can then compare with a geometric series.






                  share|cite|improve this answer











                  $endgroup$



                  Note that
                  $$
                  frac{n!}{(2n-1)!!}=frac11frac23frac35frac47cdotsfrac{n}{2n-1}
                  $$
                  For $nge2$, $dfrac{n}{2n-1}ledfrac23$. Thus, for $nge1$, we have
                  $$
                  frac{n!}{(2n-1)!!}leleft(frac23right)^{n-1}
                  $$
                  We can then compare with a geometric series.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited May 4 '13 at 0:51

























                  answered May 3 '13 at 10:42









                  robjohn♦robjohn

                  270k27312640




                  270k27312640






























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