Show that $frac{(2n)!!}{(2n+1)!!}$ converges.












0












$begingroup$



Given a sequence ${x_n}, ninBbb N$:
$$
x_n = frac{(2n)!!}{(2n+1)!!}
$$
Show that $x_n$ converges.




I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence):
$$
begin{align}
x_n &= frac{(2n)!!}{(2n+1)!!} \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1) } \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{2(n)!}{2(n)!} \
&= frac{4^n (n!)^2}{(2n+1)!} \
&= frac{4^n (n!)^2}{(2n+1)cdot (2n)!} \
&=frac{4^n}{2n+1}cdot frac{(n!)^2}{(2n)!}
end{align}
$$



By Binomial coefficients:
$$
{2nchoose n} = frac{(2n)!}{n!(2n-n)!} = frac{(2n)!}{(n!)^2}
$$



Thus:
$$
x_n = frac{4^n}{2n+1}cdot frac{1}{{2n choose n}}
$$



Doesn't $frac{4^n}{2n+1}$ grow faster than $frac{1}{2nchoose n}$ is declining? Shouldn't $x_n$ diverge in that case?






calculus limits factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
    $endgroup$
    – 5xum
    Jan 3 at 13:05










  • $begingroup$
    The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    $endgroup$
    – xbh
    Jan 3 at 13:05










  • $begingroup$
    @5xum I made a mistake in the nominator.
    $endgroup$
    – roman
    Jan 3 at 13:08










  • $begingroup$
    @5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
    $endgroup$
    – S. Franssen
    Jan 3 at 13:12
















0












$begingroup$



Given a sequence ${x_n}, ninBbb N$:
$$
x_n = frac{(2n)!!}{(2n+1)!!}
$$
Show that $x_n$ converges.




I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence):
$$
begin{align}
x_n &= frac{(2n)!!}{(2n+1)!!} \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1) } \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{2(n)!}{2(n)!} \
&= frac{4^n (n!)^2}{(2n+1)!} \
&= frac{4^n (n!)^2}{(2n+1)cdot (2n)!} \
&=frac{4^n}{2n+1}cdot frac{(n!)^2}{(2n)!}
end{align}
$$



By Binomial coefficients:
$$
{2nchoose n} = frac{(2n)!}{n!(2n-n)!} = frac{(2n)!}{(n!)^2}
$$



Thus:
$$
x_n = frac{4^n}{2n+1}cdot frac{1}{{2n choose n}}
$$



Doesn't $frac{4^n}{2n+1}$ grow faster than $frac{1}{2nchoose n}$ is declining? Shouldn't $x_n$ diverge in that case?






calculus limits factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
    $endgroup$
    – 5xum
    Jan 3 at 13:05










  • $begingroup$
    The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    $endgroup$
    – xbh
    Jan 3 at 13:05










  • $begingroup$
    @5xum I made a mistake in the nominator.
    $endgroup$
    – roman
    Jan 3 at 13:08










  • $begingroup$
    @5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
    $endgroup$
    – S. Franssen
    Jan 3 at 13:12














0












0








0





$begingroup$



Given a sequence ${x_n}, ninBbb N$:
$$
x_n = frac{(2n)!!}{(2n+1)!!}
$$
Show that $x_n$ converges.




I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence):
$$
begin{align}
x_n &= frac{(2n)!!}{(2n+1)!!} \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1) } \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{2(n)!}{2(n)!} \
&= frac{4^n (n!)^2}{(2n+1)!} \
&= frac{4^n (n!)^2}{(2n+1)cdot (2n)!} \
&=frac{4^n}{2n+1}cdot frac{(n!)^2}{(2n)!}
end{align}
$$



By Binomial coefficients:
$$
{2nchoose n} = frac{(2n)!}{n!(2n-n)!} = frac{(2n)!}{(n!)^2}
$$



Thus:
$$
x_n = frac{4^n}{2n+1}cdot frac{1}{{2n choose n}}
$$



Doesn't $frac{4^n}{2n+1}$ grow faster than $frac{1}{2nchoose n}$ is declining? Shouldn't $x_n$ diverge in that case?






calculus limits factorial







share|cite|improve this question











$endgroup$





Given a sequence ${x_n}, ninBbb N$:
$$
x_n = frac{(2n)!!}{(2n+1)!!}
$$
Show that $x_n$ converges.




I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence):
$$
begin{align}
x_n &= frac{(2n)!!}{(2n+1)!!} \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1) } \
&= frac{2cdot 4cdot 6cdots (2n-2)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{2(n)!}{2(n)!} \
&= frac{4^n (n!)^2}{(2n+1)!} \
&= frac{4^n (n!)^2}{(2n+1)cdot (2n)!} \
&=frac{4^n}{2n+1}cdot frac{(n!)^2}{(2n)!}
end{align}
$$



By Binomial coefficients:
$$
{2nchoose n} = frac{(2n)!}{n!(2n-n)!} = frac{(2n)!}{(n!)^2}
$$



Thus:
$$
x_n = frac{4^n}{2n+1}cdot frac{1}{{2n choose n}}
$$



Doesn't $frac{4^n}{2n+1}$ grow faster than $frac{1}{2nchoose n}$ is declining? Shouldn't $x_n$ diverge in that case?






calculus limits factorial




calculus limits factorial






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 13:14







roman

















asked Jan 3 at 12:54









romanroman

2,34321224




2,34321224












  • $begingroup$
    I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
    $endgroup$
    – 5xum
    Jan 3 at 13:05










  • $begingroup$
    The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    $endgroup$
    – xbh
    Jan 3 at 13:05










  • $begingroup$
    @5xum I made a mistake in the nominator.
    $endgroup$
    – roman
    Jan 3 at 13:08










  • $begingroup$
    @5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
    $endgroup$
    – S. Franssen
    Jan 3 at 13:12


















  • $begingroup$
    I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
    $endgroup$
    – 5xum
    Jan 3 at 13:05










  • $begingroup$
    The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    $endgroup$
    – xbh
    Jan 3 at 13:05










  • $begingroup$
    @5xum I made a mistake in the nominator.
    $endgroup$
    – roman
    Jan 3 at 13:08










  • $begingroup$
    @5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
    $endgroup$
    – S. Franssen
    Jan 3 at 13:12
















$begingroup$
I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
$endgroup$
– 5xum
Jan 3 at 13:05




$begingroup$
I don't see how $$frac{2cdot 4cdot 6cdots (2n-1)cdot(2n)}{3cdot 5cdot 7cdots (2n-1)cdot(2n+1)} cdot frac{(2n)!}{(2n)!} = \= frac{4^n (n!)^2}{(2n+1)!} $$ would be true.
$endgroup$
– 5xum
Jan 3 at 13:05












$begingroup$
The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
$endgroup$
– xbh
Jan 3 at 13:05




$begingroup$
The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
$endgroup$
– xbh
Jan 3 at 13:05












$begingroup$
@5xum I made a mistake in the nominator.
$endgroup$
– roman
Jan 3 at 13:08




$begingroup$
@5xum I made a mistake in the nominator.
$endgroup$
– roman
Jan 3 at 13:08












$begingroup$
@5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
$endgroup$
– S. Franssen
Jan 3 at 13:12




$begingroup$
@5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling.
$endgroup$
– S. Franssen
Jan 3 at 13:12










2 Answers
2






active

oldest

votes


















10












$begingroup$

You have



$$0leq x_{n+1} = x_n cdot frac{2n+2}{2n+3} leq x_n$$
thus, your sequence is monotone decreasing and bounded from below and therefore convergent.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Using the Stirling approximation for factorials we can observe the following: $n! sim sqrt{2 pi n} left( frac{n}{e} right)^n$. This means that, while being a bit sloppy with the $sim$



    $
    begin{align*}
    x_n &= frac{(2n)!!}{(2n+1)!!} \
    &= frac{4^n (n!)^2}{(2n+1) (2n)!}\
    & sim frac{4^n 2 pi n left( frac{n}{e} right)^{2n}}{ (2n+1) 2sqrt{pi n} left(frac{2n}{e} right)^{2n}}\
    &= frac{sqrt{pi n}}{2 n + 1} rightarrow 0
    end{align*}
    $



    So $x_n rightarrow 0$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10












      $begingroup$

      You have



      $$0leq x_{n+1} = x_n cdot frac{2n+2}{2n+3} leq x_n$$
      thus, your sequence is monotone decreasing and bounded from below and therefore convergent.






      share|cite|improve this answer









      $endgroup$


















        10












        $begingroup$

        You have



        $$0leq x_{n+1} = x_n cdot frac{2n+2}{2n+3} leq x_n$$
        thus, your sequence is monotone decreasing and bounded from below and therefore convergent.






        share|cite|improve this answer









        $endgroup$
















          10












          10








          10





          $begingroup$

          You have



          $$0leq x_{n+1} = x_n cdot frac{2n+2}{2n+3} leq x_n$$
          thus, your sequence is monotone decreasing and bounded from below and therefore convergent.






          share|cite|improve this answer









          $endgroup$



          You have



          $$0leq x_{n+1} = x_n cdot frac{2n+2}{2n+3} leq x_n$$
          thus, your sequence is monotone decreasing and bounded from below and therefore convergent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 13:01









          Severin SchravenSeverin Schraven

          6,4051935




          6,4051935























              2












              $begingroup$

              Using the Stirling approximation for factorials we can observe the following: $n! sim sqrt{2 pi n} left( frac{n}{e} right)^n$. This means that, while being a bit sloppy with the $sim$



              $
              begin{align*}
              x_n &= frac{(2n)!!}{(2n+1)!!} \
              &= frac{4^n (n!)^2}{(2n+1) (2n)!}\
              & sim frac{4^n 2 pi n left( frac{n}{e} right)^{2n}}{ (2n+1) 2sqrt{pi n} left(frac{2n}{e} right)^{2n}}\
              &= frac{sqrt{pi n}}{2 n + 1} rightarrow 0
              end{align*}
              $



              So $x_n rightarrow 0$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Using the Stirling approximation for factorials we can observe the following: $n! sim sqrt{2 pi n} left( frac{n}{e} right)^n$. This means that, while being a bit sloppy with the $sim$



                $
                begin{align*}
                x_n &= frac{(2n)!!}{(2n+1)!!} \
                &= frac{4^n (n!)^2}{(2n+1) (2n)!}\
                & sim frac{4^n 2 pi n left( frac{n}{e} right)^{2n}}{ (2n+1) 2sqrt{pi n} left(frac{2n}{e} right)^{2n}}\
                &= frac{sqrt{pi n}}{2 n + 1} rightarrow 0
                end{align*}
                $



                So $x_n rightarrow 0$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Using the Stirling approximation for factorials we can observe the following: $n! sim sqrt{2 pi n} left( frac{n}{e} right)^n$. This means that, while being a bit sloppy with the $sim$



                  $
                  begin{align*}
                  x_n &= frac{(2n)!!}{(2n+1)!!} \
                  &= frac{4^n (n!)^2}{(2n+1) (2n)!}\
                  & sim frac{4^n 2 pi n left( frac{n}{e} right)^{2n}}{ (2n+1) 2sqrt{pi n} left(frac{2n}{e} right)^{2n}}\
                  &= frac{sqrt{pi n}}{2 n + 1} rightarrow 0
                  end{align*}
                  $



                  So $x_n rightarrow 0$






                  share|cite|improve this answer









                  $endgroup$



                  Using the Stirling approximation for factorials we can observe the following: $n! sim sqrt{2 pi n} left( frac{n}{e} right)^n$. This means that, while being a bit sloppy with the $sim$



                  $
                  begin{align*}
                  x_n &= frac{(2n)!!}{(2n+1)!!} \
                  &= frac{4^n (n!)^2}{(2n+1) (2n)!}\
                  & sim frac{4^n 2 pi n left( frac{n}{e} right)^{2n}}{ (2n+1) 2sqrt{pi n} left(frac{2n}{e} right)^{2n}}\
                  &= frac{sqrt{pi n}}{2 n + 1} rightarrow 0
                  end{align*}
                  $



                  So $x_n rightarrow 0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 13:19









                  S. FranssenS. Franssen

                  1186




                  1186






























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