How do i show$lim_nrightarrow infty frac[(n+1)(n+2)…(2n)]^frac1nn=frac4e$ without using integration?Proving that $lim_n rightarrowinfty int_0^fracpi2 sin(t^n) dt =0$Prove that $lim_xtoinfty frace^xx^n=infty$ without using L'hôpital's ruleHow to show that $lim_nrightarrow infty (1+fracxn)^n=e^x$$lim_nrightarrowinftyfrac1nsum_k=1^nfrackk^2+1$Show that $lim_ntoinftyfrac2^nn^ln(n)=infty$Find the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$Show that $lim_t rightarrow infty tF(t) = 0.$Show that $lim_x rightarrow infty f(x)$ exists by the given condition.How can I evaluate $lim_n rightarrow infty int_n^infty fracn^2 arctan frac1xx^2+n^2 dx$?Why is $lim_x rightarrow -inftyfracxsqrtx^2 = -1$ and not $1$?

Multi tool use
Multi tool use

Is it a good idea to copy a trader when investing?

Why do unstable nuclei form?

Is every story set in the future "science fiction"?

What dice to use in a game that revolves around triangles?

How can Sam Wilson fulfill his future role?

What is the minimum required technology to reanimate someone who has been cryogenically frozen?

Why use steam instead of just hot air?

Is it a Munchausen Number?

How can I make parentheses stick to formula?

Are on’yomi words loanwords?

"Estrontium" on poster

What does the "DS" in "DS-..." US visa application forms stand for?

Probability of taking balls without replacement from a bag question

What's the "magic similar to the Knock spell" referenced in the Dungeon of the Mad Mage adventure?

Hexagonal Grid Filling

What is the status of the three crises in the history of mathematics?

Names of the Six Tastes

Output the date in the Mel calendar

Why did Missandei say this?

Why is there a cap on 401k contributions?

Thawing Glaciers return to hand interaction

Is it safe to keep the GPU on 100% utilization for a very long time?

Why does the electron wavefunction not collapse within atoms at room temperature in gas, liquids or solids due to decoherence?

Passport stamps art, can it be done?



How do i show$lim_nrightarrow infty frac[(n+1)(n+2)…(2n)]^frac1nn=frac4e$ without using integration?


Proving that $lim_n rightarrowinfty int_0^fracpi2 sin(t^n) dt =0$Prove that $lim_xtoinfty frace^xx^n=infty$ without using L'hôpital's ruleHow to show that $lim_nrightarrow infty (1+fracxn)^n=e^x$$lim_nrightarrowinftyfrac1nsum_k=1^nfrackk^2+1$Show that $lim_ntoinftyfrac2^nn^ln(n)=infty$Find the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$Show that $lim_t rightarrow infty tF(t) = 0.$Show that $lim_x rightarrow infty f(x)$ exists by the given condition.How can I evaluate $lim_n rightarrow infty int_n^infty fracn^2 arctan frac1xx^2+n^2 dx$?Why is $lim_x rightarrow -inftyfracxsqrtx^2 = -1$ and not $1$?













1












$begingroup$


Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.










    share|cite|improve this question











    $endgroup$














      1












      1








      1


      2



      $begingroup$


      Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.










      share|cite|improve this question











      $endgroup$




      Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.







      real-analysis integration limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 30 at 12:01









      YuiTo Cheng

      3,21161145




      3,21161145










      asked Apr 30 at 6:47







      user655791



























          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          By Stirling,



          $$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$






          share|cite|improve this answer









          $endgroup$




















            3












            $begingroup$

            If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.



            Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
              $endgroup$
              – Yves Daoust
              Apr 30 at 7:14











            • $begingroup$
              I justified....
              $endgroup$
              – Shamim Akhtar
              Apr 30 at 7:32










            • $begingroup$
              No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
              $endgroup$
              – Yves Daoust
              Apr 30 at 7:34



















            -1












            $begingroup$

            By taking logs of both sides we can get:
            $$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
            $$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
            $$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
            And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
            $$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$






            share|cite|improve this answer









            $endgroup$













              Your Answer








              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%2f3207894%2fhow-do-i-show-lim-n-rightarrow-infty-fracn1n2-2n-frac1n%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown
























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              By Stirling,



              $$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$






              share|cite|improve this answer









              $endgroup$

















                5












                $begingroup$

                By Stirling,



                $$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$






                share|cite|improve this answer









                $endgroup$















                  5












                  5








                  5





                  $begingroup$

                  By Stirling,



                  $$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$






                  share|cite|improve this answer









                  $endgroup$



                  By Stirling,



                  $$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 30 at 7:23









                  Yves DaoustYves Daoust

                  136k676237




                  136k676237





















                      3












                      $begingroup$

                      If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.



                      Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$






                      share|cite|improve this answer











                      $endgroup$












                      • $begingroup$
                        The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:14











                      • $begingroup$
                        I justified....
                        $endgroup$
                        – Shamim Akhtar
                        Apr 30 at 7:32










                      • $begingroup$
                        No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:34
















                      3












                      $begingroup$

                      If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.



                      Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$






                      share|cite|improve this answer











                      $endgroup$












                      • $begingroup$
                        The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:14











                      • $begingroup$
                        I justified....
                        $endgroup$
                        – Shamim Akhtar
                        Apr 30 at 7:32










                      • $begingroup$
                        No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:34














                      3












                      3








                      3





                      $begingroup$

                      If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.



                      Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$






                      share|cite|improve this answer











                      $endgroup$



                      If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.



                      Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 30 at 7:32

























                      answered Apr 30 at 7:01









                      Shamim AkhtarShamim Akhtar

                      54919




                      54919











                      • $begingroup$
                        The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:14











                      • $begingroup$
                        I justified....
                        $endgroup$
                        – Shamim Akhtar
                        Apr 30 at 7:32










                      • $begingroup$
                        No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:34

















                      • $begingroup$
                        The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:14











                      • $begingroup$
                        I justified....
                        $endgroup$
                        – Shamim Akhtar
                        Apr 30 at 7:32










                      • $begingroup$
                        No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                        $endgroup$
                        – Yves Daoust
                        Apr 30 at 7:34
















                      $begingroup$
                      The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                      $endgroup$
                      – Yves Daoust
                      Apr 30 at 7:14





                      $begingroup$
                      The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
                      $endgroup$
                      – Yves Daoust
                      Apr 30 at 7:14













                      $begingroup$
                      I justified....
                      $endgroup$
                      – Shamim Akhtar
                      Apr 30 at 7:32




                      $begingroup$
                      I justified....
                      $endgroup$
                      – Shamim Akhtar
                      Apr 30 at 7:32












                      $begingroup$
                      No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                      $endgroup$
                      – Yves Daoust
                      Apr 30 at 7:34





                      $begingroup$
                      No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
                      $endgroup$
                      – Yves Daoust
                      Apr 30 at 7:34












                      -1












                      $begingroup$

                      By taking logs of both sides we can get:
                      $$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
                      $$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
                      $$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
                      And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
                      $$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$






                      share|cite|improve this answer









                      $endgroup$

















                        -1












                        $begingroup$

                        By taking logs of both sides we can get:
                        $$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
                        $$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
                        $$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
                        And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
                        $$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$






                        share|cite|improve this answer









                        $endgroup$















                          -1












                          -1








                          -1





                          $begingroup$

                          By taking logs of both sides we can get:
                          $$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
                          $$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
                          $$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
                          And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
                          $$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$






                          share|cite|improve this answer









                          $endgroup$



                          By taking logs of both sides we can get:
                          $$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
                          $$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
                          $$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
                          And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
                          $$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 30 at 15:14









                          Henry LeeHenry Lee

                          2,288319




                          2,288319



























                              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.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3207894%2fhow-do-i-show-lim-n-rightarrow-infty-fracn1n2-2n-frac1n%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







                              wlYE,iCkbH1raNSaMAe,obuYrI8pMF,84 0tA F6D rOR0501Y5iERuXonZEZXP
                              ZOmzE hvDH8xH6,1v827lKE4v4j7imdqPCtmGYTpO GJJ3Ha3W,1urVborZn4W7hhFKQ04WBpsXhgj,L

                              Popular posts from this blog

                              RemoteApp sporadic failureWindows 2008 RemoteAPP client disconnects within a matter of minutesWhat is the minimum version of RDP supported by Server 2012 RDS?How to configure a Remoteapp server to increase stabilityMicrosoft RemoteApp Active SessionRDWeb TS connection broken for some users post RemoteApp certificate changeRemote Desktop Licensing, RemoteAPPRDS 2012 R2 some users are not able to logon after changed date and time on Connection BrokersWhat happens during Remote Desktop logon, and is there any logging?After installing RDS on WinServer 2016 I still can only connect with two users?RD Connection via RDGW to Session host is not connecting

                              Vilaño, A Laracha Índice Patrimonio | Lugares e parroquias | Véxase tamén | Menú de navegación43°14′52″N 8°36′03″O / 43.24775, -8.60070

                              Cegueira Índice Epidemioloxía | Deficiencia visual | Tipos de cegueira | Principais causas de cegueira | Tratamento | Técnicas de adaptación e axudas | Vida dos cegos | Primeiros auxilios | Crenzas respecto das persoas cegas | Crenzas das persoas cegas | O neno deficiente visual | Aspectos psicolóxicos da cegueira | Notas | Véxase tamén | Menú de navegación54.054.154.436928256blindnessDicionario da Real Academia GalegaPortal das Palabras"International Standards: Visual Standards — Aspects and Ranges of Vision Loss with Emphasis on Population Surveys.""Visual impairment and blindness""Presentan un plan para previr a cegueira"o orixinalACCDV Associació Catalana de Cecs i Disminuïts Visuals - PMFTrachoma"Effect of gene therapy on visual function in Leber's congenital amaurosis"1844137110.1056/NEJMoa0802268Cans guía - os mellores amigos dos cegosArquivadoEscola de cans guía para cegos en Mortágua, PortugalArquivado"Tecnología para ciegos y deficientes visuales. Recopilación de recursos gratuitos en la Red""Colorino""‘COL.diesis’, escuchar los sonidos del color""COL.diesis: Transforming Colour into Melody and Implementing the Result in a Colour Sensor Device"o orixinal"Sistema de desarrollo de sinestesia color-sonido para invidentes utilizando un protocolo de audio""Enseñanza táctil - geometría y color. Juegos didácticos para niños ciegos y videntes""Sistema Constanz"L'ocupació laboral dels cecs a l'Estat espanyol està pràcticament equiparada a la de les persones amb visió, entrevista amb Pedro ZuritaONCE (Organización Nacional de Cegos de España)Prevención da cegueiraDescrición de deficiencias visuais (Disc@pnet)Braillín, un boneco atractivo para calquera neno, con ou sen discapacidade, que permite familiarizarse co sistema de escritura e lectura brailleAxudas Técnicas36838ID00897494007150-90057129528256DOID:1432HP:0000618D001766C10.597.751.941.162C97109C0155020