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












0















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
























  • 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.
    – 5xum
    yesterday










  • The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    – xbh
    yesterday










  • @5xum I made a mistake in the nominator.
    – roman
    yesterday










  • @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.
    – S. Franssen
    yesterday
















0















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
























  • 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.
    – 5xum
    yesterday










  • The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    – xbh
    yesterday










  • @5xum I made a mistake in the nominator.
    – roman
    yesterday










  • @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.
    – S. Franssen
    yesterday














0












0








0








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
















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 yesterday

























asked yesterday









roman

1,97121221




1,97121221












  • 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.
    – 5xum
    yesterday










  • The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    – xbh
    yesterday










  • @5xum I made a mistake in the nominator.
    – roman
    yesterday










  • @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.
    – S. Franssen
    yesterday


















  • 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.
    – 5xum
    yesterday










  • The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
    – xbh
    yesterday










  • @5xum I made a mistake in the nominator.
    – roman
    yesterday










  • @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.
    – S. Franssen
    yesterday
















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.
– 5xum
yesterday




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.
– 5xum
yesterday












The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
– xbh
yesterday




The asymptotic behavior of $binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think.
– xbh
yesterday












@5xum I made a mistake in the nominator.
– roman
yesterday




@5xum I made a mistake in the nominator.
– roman
yesterday












@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.
– S. Franssen
yesterday




@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.
– S. Franssen
yesterday










2 Answers
2






active

oldest

votes


















10














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





























    2














    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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060536%2fshow-that-frac2n2n1-converges%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10














      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


























        10














        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
























          10












          10








          10






          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












          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 yesterday









          Severin Schraven

          5,8281934




          5,8281934























              2














              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


























                2














                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
























                  2












                  2








                  2






                  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












                  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 yesterday









                  S. Franssen

                  1186




                  1186






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060536%2fshow-that-frac2n2n1-converges%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      An IMO inspired problem

                      Image
                      14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                      Read more

                      Management

                      Image
                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                      Read more

                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

                      Image
                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
                      Read more