Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies...












0












$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22
















0












$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22














0












0








0





$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







share|cite|improve this question











$endgroup$




Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions




real-analysis calculus sequences-and-series inequality monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 6:31







stackofhay42

















asked Dec 12 '18 at 5:37









stackofhay42stackofhay42

2267




2267








  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22














  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22








1




1




$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43






$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43














$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22




$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22










4 Answers
4






active

oldest

votes


















4












$begingroup$

Use principle of mathematical induction,



Basis for induction
$$a_1>2$$



Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.



Hence proved



Hope it is helpful






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    AM-GM gives:
    $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



    Edit after comments:




    • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

    • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


    So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



    For a direct proof that all members of the iteration are greater than $2$ you may consider




    • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


    Now use MVT:



    $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      how do i know $a neq 2$?
      $endgroup$
      – stackofhay42
      Dec 12 '18 at 6:31










    • $begingroup$
      @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
      $endgroup$
      – Yadati Kiran
      Dec 12 '18 at 6:47










    • $begingroup$
      Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
      $endgroup$
      – stackofhay42
      Dec 12 '18 at 6:51












    • $begingroup$
      @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
      $endgroup$
      – Martund
      Dec 12 '18 at 9:25










    • $begingroup$
      @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
      $endgroup$
      – Martund
      Dec 12 '18 at 9:27





















    2












    $begingroup$

    Note $a_n >0$, $n=1,2,3,.....$.



    Let $n ge 1$:



    $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



    $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



    $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



    (The first term $ge 0$, a square divided by a positive number.)



    Hence



    $a_{n+1} ge 2$, $n in mathbb{N}$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Following with your reasoning, suppose $x + 1/x = 2$. Then:
      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
      But this is impossible because...






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        What about $a_n<2$?
        $endgroup$
        – Rakesh Bhatt
        Dec 12 '18 at 10:14










      • $begingroup$
        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
        $endgroup$
        – Martín-Blas Pérez Pinilla
        Dec 12 '18 at 10:43












      • $begingroup$
        Why is this impossible?
        $endgroup$
        – Ekesh Kumar
        Dec 13 '18 at 23:45










      • $begingroup$
        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
        $endgroup$
        – Martín-Blas Pérez Pinilla
        Dec 14 '18 at 7:12











      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%2f3036275%2fprove-the-sequence-a-1-4-a-n-1-fraca-n2-frac2a-n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Use principle of mathematical induction,



      Basis for induction
      $$a_1>2$$



      Induction Hypothesis
      $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



      Inductive step
      $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
      which is true by condition of equality in AM-GM inequality and induction hypothesis.



      Hence proved



      Hope it is helpful






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Use principle of mathematical induction,



        Basis for induction
        $$a_1>2$$



        Induction Hypothesis
        $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



        Inductive step
        $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
        which is true by condition of equality in AM-GM inequality and induction hypothesis.



        Hence proved



        Hope it is helpful






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Use principle of mathematical induction,



          Basis for induction
          $$a_1>2$$



          Induction Hypothesis
          $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



          Inductive step
          $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
          which is true by condition of equality in AM-GM inequality and induction hypothesis.



          Hence proved



          Hope it is helpful






          share|cite|improve this answer









          $endgroup$



          Use principle of mathematical induction,



          Basis for induction
          $$a_1>2$$



          Induction Hypothesis
          $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



          Inductive step
          $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
          which is true by condition of equality in AM-GM inequality and induction hypothesis.



          Hence proved



          Hope it is helpful







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 6:08









          MartundMartund

          1,645213




          1,645213























              4












              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27


















              4












              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27
















              4












              4








              4





              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$



              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 12 '18 at 9:35

























              answered Dec 12 '18 at 5:49









              trancelocationtrancelocation

              12.3k1826




              12.3k1826












              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27




















              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27


















              $begingroup$
              how do i know $a neq 2$?
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:31




              $begingroup$
              how do i know $a neq 2$?
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:31












              $begingroup$
              @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
              $endgroup$
              – Yadati Kiran
              Dec 12 '18 at 6:47




              $begingroup$
              @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
              $endgroup$
              – Yadati Kiran
              Dec 12 '18 at 6:47












              $begingroup$
              Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:51






              $begingroup$
              Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:51














              $begingroup$
              @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:25




              $begingroup$
              @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:25












              $begingroup$
              @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:27






              $begingroup$
              @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:27













              2












              $begingroup$

              Note $a_n >0$, $n=1,2,3,.....$.



              Let $n ge 1$:



              $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



              $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



              $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



              (The first term $ge 0$, a square divided by a positive number.)



              Hence



              $a_{n+1} ge 2$, $n in mathbb{N}$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note $a_n >0$, $n=1,2,3,.....$.



                Let $n ge 1$:



                $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                (The first term $ge 0$, a square divided by a positive number.)



                Hence



                $a_{n+1} ge 2$, $n in mathbb{N}$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note $a_n >0$, $n=1,2,3,.....$.



                  Let $n ge 1$:



                  $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                  $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                  $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                  (The first term $ge 0$, a square divided by a positive number.)



                  Hence



                  $a_{n+1} ge 2$, $n in mathbb{N}$.






                  share|cite|improve this answer









                  $endgroup$



                  Note $a_n >0$, $n=1,2,3,.....$.



                  Let $n ge 1$:



                  $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                  $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                  $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                  (The first term $ge 0$, a square divided by a positive number.)



                  Hence



                  $a_{n+1} ge 2$, $n in mathbb{N}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 11:01









                  Peter SzilasPeter Szilas

                  11.4k2822




                  11.4k2822























                      0












                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12
















                      0












                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12














                      0












                      0








                      0





                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$



                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 12 '18 at 10:44

























                      answered Dec 12 '18 at 7:44









                      Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                      34.4k42871




                      34.4k42871












                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12


















                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12
















                      $begingroup$
                      What about $a_n<2$?
                      $endgroup$
                      – Rakesh Bhatt
                      Dec 12 '18 at 10:14




                      $begingroup$
                      What about $a_n<2$?
                      $endgroup$
                      – Rakesh Bhatt
                      Dec 12 '18 at 10:14












                      $begingroup$
                      @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 12 '18 at 10:43






                      $begingroup$
                      @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 12 '18 at 10:43














                      $begingroup$
                      Why is this impossible?
                      $endgroup$
                      – Ekesh Kumar
                      Dec 13 '18 at 23:45




                      $begingroup$
                      Why is this impossible?
                      $endgroup$
                      – Ekesh Kumar
                      Dec 13 '18 at 23:45












                      $begingroup$
                      @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 14 '18 at 7:12




                      $begingroup$
                      @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 14 '18 at 7:12


















                      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%2f3036275%2fprove-the-sequence-a-1-4-a-n-1-fraca-n2-frac2a-n%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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more