why is the limit of this expression equal to 1? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding the limit of the following expressionReforming series expression for limit of e$lim_x rightarrow inftyleft(fracpi2-tan^-1xright)^Largefrac1x$ Why aren't these two limits equal when they should be?What is the value of this limit?limit of an expressionUsing a definite integral find the value of $lim_nrightarrow infty (frac1n+frac1n+1+…+frac12n)$Why is the following limit operation valid?Is this expression on limit valid and/or meaningful?Why does this limit equal 0?A Problem on the Limit of an Integral

University's motivation for having tenure-track positions

How to handle characters who are more educated than the author?

What is the padding with red substance inside of steak packaging?

why is the limit of this expression equal to 1?

Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?

First use of “packing” as in carrying a gun

What information about me do stores get via my credit card?

Does Parliament need to approve the new Brexit delay to 31 October 2019?

How to type a long/em dash `—`

Why are there uneven bright areas in this photo of black hole?

How can a C program poll for user input while simultaneously performing other actions in a Linux environment?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

Can each chord in a progression create its own key?

What's the point in a preamp?

Button changing its text & action. Good or terrible?

Am I ethically obligated to go into work on an off day if the reason is sudden?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Drawing arrows from one table cell reference to another

My body leaves; my core can stay

How do you keep chess fun when your opponent constantly beats you?

Sort list of array linked objects by keys and values

How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?



why is the limit of this expression equal to 1?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding the limit of the following expressionReforming series expression for limit of e$lim_x rightarrow inftyleft(fracpi2-tan^-1xright)^Largefrac1x$ Why aren't these two limits equal when they should be?What is the value of this limit?limit of an expressionUsing a definite integral find the value of $lim_nrightarrow infty (frac1n+frac1n+1+…+frac12n)$Why is the following limit operation valid?Is this expression on limit valid and/or meaningful?Why does this limit equal 0?A Problem on the Limit of an Integral










1












$begingroup$


I found something which I find confusing.



$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$



It was something I encountered while learning probability on this webpage.










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    I found something which I find confusing.



    $$
    lim_nrightarrow infty fracn!n^k(n-k)! =1
    $$



    It was something I encountered while learning probability on this webpage.










    share|cite|improve this question











    $endgroup$














      1












      1








      1


      2



      $begingroup$


      I found something which I find confusing.



      $$
      lim_nrightarrow infty fracn!n^k(n-k)! =1
      $$



      It was something I encountered while learning probability on this webpage.










      share|cite|improve this question











      $endgroup$




      I found something which I find confusing.



      $$
      lim_nrightarrow infty fracn!n^k(n-k)! =1
      $$



      It was something I encountered while learning probability on this webpage.







      limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago







      billyandr

















      asked 2 hours ago









      billyandrbillyandr

      155




      155




















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you so much. I didn't know it was right there under my eyes.
            $endgroup$
            – billyandr
            1 hour ago










          • $begingroup$
            You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
            $endgroup$
            – trancelocation
            1 hour ago



















          2












          $begingroup$

          $$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          Use Stirling approximation and continue with Taylor series to get
          $$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
          $$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            This has already a slight touch of overkill, hasn't it? :-)
            $endgroup$
            – trancelocation
            1 hour ago










          • $begingroup$
            @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
            $endgroup$
            – Claude Leibovici
            1 hour ago












          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%2f3185830%2fwhy-is-the-limit-of-this-expression-equal-to-1%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









          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you so much. I didn't know it was right there under my eyes.
            $endgroup$
            – billyandr
            1 hour ago










          • $begingroup$
            You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
            $endgroup$
            – trancelocation
            1 hour ago
















          5












          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you so much. I didn't know it was right there under my eyes.
            $endgroup$
            – billyandr
            1 hour ago










          • $begingroup$
            You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
            $endgroup$
            – trancelocation
            1 hour ago














          5












          5








          5





          $begingroup$

          It is rather obvious if you cancel the factorials:



          $$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$






          share|cite|improve this answer









          $endgroup$



          It is rather obvious if you cancel the factorials:



          $$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          trancelocationtrancelocation

          14.1k1829




          14.1k1829











          • $begingroup$
            Thank you so much. I didn't know it was right there under my eyes.
            $endgroup$
            – billyandr
            1 hour ago










          • $begingroup$
            You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
            $endgroup$
            – trancelocation
            1 hour ago

















          • $begingroup$
            Thank you so much. I didn't know it was right there under my eyes.
            $endgroup$
            – billyandr
            1 hour ago










          • $begingroup$
            You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
            $endgroup$
            – trancelocation
            1 hour ago
















          $begingroup$
          Thank you so much. I didn't know it was right there under my eyes.
          $endgroup$
          – billyandr
          1 hour ago




          $begingroup$
          Thank you so much. I didn't know it was right there under my eyes.
          $endgroup$
          – billyandr
          1 hour ago












          $begingroup$
          You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
          $endgroup$
          – trancelocation
          1 hour ago





          $begingroup$
          You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
          $endgroup$
          – trancelocation
          1 hour ago












          2












          $begingroup$

          $$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          Use Stirling approximation and continue with Taylor series to get
          $$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
          $$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            This has already a slight touch of overkill, hasn't it? :-)
            $endgroup$
            – trancelocation
            1 hour ago










          • $begingroup$
            @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
            $endgroup$
            – Claude Leibovici
            1 hour ago
















          2












          $begingroup$

          $$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          Use Stirling approximation and continue with Taylor series to get
          $$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
          $$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            This has already a slight touch of overkill, hasn't it? :-)
            $endgroup$
            – trancelocation
            1 hour ago










          • $begingroup$
            @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
            $endgroup$
            – Claude Leibovici
            1 hour ago














          2












          2








          2





          $begingroup$

          $$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          Use Stirling approximation and continue with Taylor series to get
          $$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
          $$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$






          share|cite|improve this answer









          $endgroup$



          $$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$



          Use Stirling approximation and continue with Taylor series to get
          $$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
          $$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135











          • $begingroup$
            This has already a slight touch of overkill, hasn't it? :-)
            $endgroup$
            – trancelocation
            1 hour ago










          • $begingroup$
            @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
            $endgroup$
            – Claude Leibovici
            1 hour ago

















          • $begingroup$
            This has already a slight touch of overkill, hasn't it? :-)
            $endgroup$
            – trancelocation
            1 hour ago










          • $begingroup$
            @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
            $endgroup$
            – Claude Leibovici
            1 hour ago
















          $begingroup$
          This has already a slight touch of overkill, hasn't it? :-)
          $endgroup$
          – trancelocation
          1 hour ago




          $begingroup$
          This has already a slight touch of overkill, hasn't it? :-)
          $endgroup$
          – trancelocation
          1 hour ago












          $begingroup$
          @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
          $endgroup$
          – Claude Leibovici
          1 hour ago





          $begingroup$
          @trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
          $endgroup$
          – Claude Leibovici
          1 hour ago


















          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%2f3185830%2fwhy-is-the-limit-of-this-expression-equal-to-1%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

          Möglingen Índice Localización Historia Demografía Referencias Enlaces externos Menú de navegación48°53′18″N 9°07′45″E / 48.888333333333, 9.129166666666748°53′18″N 9°07′45″E / 48.888333333333, 9.1291666666667Sitio web oficial Mapa de Möglingen«Gemeinden in Deutschland nach Fläche, Bevölkerung und Postleitzahl am 30.09.2016»Möglingen

          Virtualbox - Configuration error: Querying “UUID” failed (VERR_CFGM_VALUE_NOT_FOUND)“VERR_SUPLIB_WORLD_WRITABLE” error when trying to installing OS in virtualboxVirtual Box Kernel errorFailed to open a seesion for the virtual machineFailed to open a session for the virtual machineUbuntu 14.04 LTS Virtualbox errorcan't use VM VirtualBoxusing virtualboxI can't run Linux-64 Bit on VirtualBoxUnable to insert the virtual optical disk (VBoxguestaddition) in virtual machine for ubuntu server in win 10VirtuaBox in Ubuntu 18.04 Issues with Win10.ISO Installation

          Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org