Limit of $a_n=(1+ frac{1}{n^2})^n$ as $nrightarrow infty$












0














I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







share|cite|improve this question


















  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11
















0














I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







share|cite|improve this question


















  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11














0












0








0


2





I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







share|cite|improve this question













I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis




calculus real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 6:14









Wesley Strik

1,493422




1,493422








  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11














  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11








1




1




Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
– Paramanand Singh
Nov 25 at 16:11




Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
– Paramanand Singh
Nov 25 at 16:11










3 Answers
3






active

oldest

votes


















4














We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






share|cite|improve this answer























  • DeepSea.Very nice.
    – Peter Szilas
    Nov 25 at 6:44










  • Yes indeed, thank you DeepSea, very satisfying.
    – Wesley Strik
    Nov 25 at 7:11



















3














Note that:
$$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






share|cite|improve this answer





























    2














    You could also notice that by expansion,



    $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



    $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



    $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



    And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






    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%2f3012491%2flimit-of-a-n-1-frac1n2n-as-n-rightarrow-infty%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









      4














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer























      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11
















      4














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer























      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11














      4












      4








      4






      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 25 at 6:28

























      answered Nov 25 at 6:18









      DeepSea

      70.9k54487




      70.9k54487












      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11


















      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11
















      DeepSea.Very nice.
      – Peter Szilas
      Nov 25 at 6:44




      DeepSea.Very nice.
      – Peter Szilas
      Nov 25 at 6:44












      Yes indeed, thank you DeepSea, very satisfying.
      – Wesley Strik
      Nov 25 at 7:11




      Yes indeed, thank you DeepSea, very satisfying.
      – Wesley Strik
      Nov 25 at 7:11











      3














      Note that:
      $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






      share|cite|improve this answer


























        3














        Note that:
        $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






        share|cite|improve this answer
























          3












          3








          3






          Note that:
          $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






          share|cite|improve this answer












          Note that:
          $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 6:40









          farruhota

          19.3k2736




          19.3k2736























              2














              You could also notice that by expansion,



              $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



              $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



              $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



              And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






              share|cite|improve this answer


























                2














                You could also notice that by expansion,



                $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  You could also notice that by expansion,



                  $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                  $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                  $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                  And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






                  share|cite|improve this answer












                  You could also notice that by expansion,



                  $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                  $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                  $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                  And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 6:37









                  KM101

                  4,846421




                  4,846421






























                      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%2f3012491%2flimit-of-a-n-1-frac1n2n-as-n-rightarrow-infty%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                      Read more

                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                      Image
                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                      Read more