Verification of the approach on finding the limit of $x_n$ as $ntoinfty$ given $x_{n+1} = sqrt[k]{5x_n}$...












2












$begingroup$



Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$




I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$



Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$



This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$



Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$



Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$



Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$



Is it the right way to approach this problem? Could it be simplified? Thank you!






calculus limits proof-verification







share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Let $x_n$ denote a sequence, $ninBbb N$:
    $$
    begin{cases}
    x_{n+1} = sqrt[k]{5x_n}\
    x_1 = sqrt[k]{5}\
    kinBbb N
    end{cases}
    $$
    Find:
    $$
    lim_{ntoinfty} sqrt[k]{x_n}
    $$




    I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
    $$
    x_1 = sqrt[k]{5}\
    x_2 = sqrt[k]{5sqrt[k]{5}}\
    x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
    dots
    $$



    Rewrite as:
    $$
    begin{align}
    x_1 &= 5^{1over k}\
    x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
    &cdots\
    x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
    end{align}
    $$



    This is hardly readable using powers in mathjax. Rewrite:
    $$
    log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
    $$



    Nominator is a regular geometric series:
    $$
    1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
    $$



    Thus:
    $$
    frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
    $$



    Since $a^x$ is continuous we may consider the limit of $(1)$:
    $$
    lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
    $$
    Which implies:
    $$
    lim_{ntoinfty}x_n = sqrt[k-1]{5}
    $$



    Is it the right way to approach this problem? Could it be simplified? Thank you!






    calculus limits proof-verification







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      Let $x_n$ denote a sequence, $ninBbb N$:
      $$
      begin{cases}
      x_{n+1} = sqrt[k]{5x_n}\
      x_1 = sqrt[k]{5}\
      kinBbb N
      end{cases}
      $$
      Find:
      $$
      lim_{ntoinfty} sqrt[k]{x_n}
      $$




      I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
      $$
      x_1 = sqrt[k]{5}\
      x_2 = sqrt[k]{5sqrt[k]{5}}\
      x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
      dots
      $$



      Rewrite as:
      $$
      begin{align}
      x_1 &= 5^{1over k}\
      x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
      &cdots\
      x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
      end{align}
      $$



      This is hardly readable using powers in mathjax. Rewrite:
      $$
      log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
      $$



      Nominator is a regular geometric series:
      $$
      1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
      $$



      Thus:
      $$
      frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
      $$



      Since $a^x$ is continuous we may consider the limit of $(1)$:
      $$
      lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
      $$
      Which implies:
      $$
      lim_{ntoinfty}x_n = sqrt[k-1]{5}
      $$



      Is it the right way to approach this problem? Could it be simplified? Thank you!






      calculus limits proof-verification







      share|cite|improve this question









      $endgroup$





      Let $x_n$ denote a sequence, $ninBbb N$:
      $$
      begin{cases}
      x_{n+1} = sqrt[k]{5x_n}\
      x_1 = sqrt[k]{5}\
      kinBbb N
      end{cases}
      $$
      Find:
      $$
      lim_{ntoinfty} sqrt[k]{x_n}
      $$




      I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
      $$
      x_1 = sqrt[k]{5}\
      x_2 = sqrt[k]{5sqrt[k]{5}}\
      x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
      dots
      $$



      Rewrite as:
      $$
      begin{align}
      x_1 &= 5^{1over k}\
      x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
      &cdots\
      x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
      end{align}
      $$



      This is hardly readable using powers in mathjax. Rewrite:
      $$
      log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
      $$



      Nominator is a regular geometric series:
      $$
      1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
      $$



      Thus:
      $$
      frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
      $$



      Since $a^x$ is continuous we may consider the limit of $(1)$:
      $$
      lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
      $$
      Which implies:
      $$
      lim_{ntoinfty}x_n = sqrt[k-1]{5}
      $$



      Is it the right way to approach this problem? Could it be simplified? Thank you!






      calculus limits proof-verification




      calculus limits proof-verification






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 11:26









      romanroman

      2,36221224




      2,36221224






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Induction is indeed a better suit here.
            $endgroup$
            – roman
            Jan 4 at 11:55











          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%2f3061547%2fverification-of-the-approach-on-finding-the-limit-of-x-n-as-n-to-infty-given%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Induction is indeed a better suit here.
            $endgroup$
            – roman
            Jan 4 at 11:55
















          2












          $begingroup$

          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Induction is indeed a better suit here.
            $endgroup$
            – roman
            Jan 4 at 11:55














          2












          2








          2





          $begingroup$

          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.






          share|cite|improve this answer









          $endgroup$



          We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
          $sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.



          We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
          Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.



          The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.



          Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
          $$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$



          It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 11:40









          mechanodroidmechanodroid

          28.7k62548




          28.7k62548












          • $begingroup$
            Thank you! Induction is indeed a better suit here.
            $endgroup$
            – roman
            Jan 4 at 11:55


















          • $begingroup$
            Thank you! Induction is indeed a better suit here.
            $endgroup$
            – roman
            Jan 4 at 11:55
















          $begingroup$
          Thank you! Induction is indeed a better suit here.
          $endgroup$
          – roman
          Jan 4 at 11:55




          $begingroup$
          Thank you! Induction is indeed a better suit here.
          $endgroup$
          – roman
          Jan 4 at 11:55


















          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%2f3061547%2fverification-of-the-approach-on-finding-the-limit-of-x-n-as-n-to-infty-given%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more