Convergence of Shanks Transformation












2












$begingroup$


If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.



It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.



    It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.



      It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.










      share|cite|improve this question











      $endgroup$




      If sequence $A_n rightarrow 0$ as $n rightarrow infty$, and the Shanks Transformation of $A_n$ defined as $$Sleft(A_nright)=frac{A_{n+1}A_{n-1}-A_n^2}{A_{n+1}+A_{n-1}-2A_n}$$ also converges, prove $Sleft(A_nright)$ converges to zero.



      It's problem 8-1 in Carl M. Bender's book, Advanced Mathematical Methods for Scientists and Engineers. I tried the definition of limit but couldn't finish the proof.







      sequences-and-series limits convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 20 at 22:43









      Davide Giraudo

      128k17156268




      128k17156268










      asked Jan 17 at 6:11









      BruceBruce

      114




      114






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
            $endgroup$
            – Bruce
            Feb 2 at 14:49










          • $begingroup$
            @Bruce You can get divergence by working with sequences such as $1/A_n$.
            $endgroup$
            – J.G.
            Feb 2 at 15:07














          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%2f3076642%2fconvergence-of-shanks-transformation%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









          0












          $begingroup$

          Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
            $endgroup$
            – Bruce
            Feb 2 at 14:49










          • $begingroup$
            @Bruce You can get divergence by working with sequences such as $1/A_n$.
            $endgroup$
            – J.G.
            Feb 2 at 15:07


















          0












          $begingroup$

          Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
            $endgroup$
            – Bruce
            Feb 2 at 14:49










          • $begingroup$
            @Bruce You can get divergence by working with sequences such as $1/A_n$.
            $endgroup$
            – J.G.
            Feb 2 at 15:07
















          0












          0








          0





          $begingroup$

          Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.






          share|cite|improve this answer









          $endgroup$



          Equivalently $S(A_n)=A_{n-1}-frac{(Delta A_{n-1})^2}{Delta^2 A_{n-1}}$ with the forward difference operator $Delta u_n:=u_{n+1}-u_n$. (Note the problem statement should have included $Delta^2 A_nne 0$.) Thus $$frac{S(A_n)}{A_{n-1}}=1-frac{Delta A_{n-1}}{A_{n-1}}cdotfrac{Delta A_{n-1}}{Delta^2A_{n-1}}.$$By Stolz–Cesàro this has limit $1-1=0$, so $S_n$ has limit $0times 0=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 6:53









          J.G.J.G.

          33.3k23252




          33.3k23252












          • $begingroup$
            Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
            $endgroup$
            – Bruce
            Feb 2 at 14:49










          • $begingroup$
            @Bruce You can get divergence by working with sequences such as $1/A_n$.
            $endgroup$
            – J.G.
            Feb 2 at 15:07




















          • $begingroup$
            Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
            $endgroup$
            – Bruce
            Feb 2 at 14:49










          • $begingroup$
            @Bruce You can get divergence by working with sequences such as $1/A_n$.
            $endgroup$
            – J.G.
            Feb 2 at 15:07


















          $begingroup$
          Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
          $endgroup$
          – Bruce
          Feb 2 at 14:49




          $begingroup$
          Thanks. I haven't heart about Stolz–Cesàro theorem before but wiki says that it can be viewed as a l'Hôpital's rule for sequences, which makes this easier to grasp. However, the statement of Stolz–Cesàro theorem says that the sequence in the denominator is strickly monotonic and divergence, while in our case, the sequence $A_{n-1}$ is neither monotonic nor divergence. So why can we apply this theorem?
          $endgroup$
          – Bruce
          Feb 2 at 14:49












          $begingroup$
          @Bruce You can get divergence by working with sequences such as $1/A_n$.
          $endgroup$
          – J.G.
          Feb 2 at 15:07






          $begingroup$
          @Bruce You can get divergence by working with sequences such as $1/A_n$.
          $endgroup$
          – J.G.
          Feb 2 at 15:07




















          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%2f3076642%2fconvergence-of-shanks-transformation%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

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅