Radius of convergence of two power series and the sum of the power serieses












0












$begingroup$


$R_1$ Is the Radius of convergence of $sum_{k=1}^{infty}a_nx^n$.
$R_2$ Is the Radius of convergence of $sum_{k=1}^{infty}b_nx^n$.

What can we say of the Radius of convergence of $sum_{k=1}^{infty}(a_n+b_n)x^n$,we'll call it $R$?





If $R_1 neq R_2$ I know the answer is $R=M=min(R_1,R_2)$ Because it is obvious that for all $x in (-M,M)$ the sum converges, and suppose $R>M$ (And without loss of generality $R_2>R_1$), than take new number $L>0$ Such that $L<R$ and $R_1<L<R_2$, than since $L<R$ we get $sum_{k=1}^{infty}(a_n+b_n)L^n$ converges, and that's a contradiction because $sum_{k=1}^{infty}a_nx^n$ diverges (because L is bigger than $R_1$) but $sum_{k=1}^{infty}b_nx^n$ converges (because L is bigger than $R_2$). so $R=M$.



But what if $R_1=R_2$? we can take for example $a_n=1/n$, $b_n$=$-1/n$ and then we'll get $R=infty$. We can take $a_n=b_n=1/n$ and than simply $R=R_1=R_2$. Is there anything we can say about this case? is it always $R=R_1=R_2$ or $R=infty$ in this case?



thanks!










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $R_1$ Is the Radius of convergence of $sum_{k=1}^{infty}a_nx^n$.
    $R_2$ Is the Radius of convergence of $sum_{k=1}^{infty}b_nx^n$.

    What can we say of the Radius of convergence of $sum_{k=1}^{infty}(a_n+b_n)x^n$,we'll call it $R$?





    If $R_1 neq R_2$ I know the answer is $R=M=min(R_1,R_2)$ Because it is obvious that for all $x in (-M,M)$ the sum converges, and suppose $R>M$ (And without loss of generality $R_2>R_1$), than take new number $L>0$ Such that $L<R$ and $R_1<L<R_2$, than since $L<R$ we get $sum_{k=1}^{infty}(a_n+b_n)L^n$ converges, and that's a contradiction because $sum_{k=1}^{infty}a_nx^n$ diverges (because L is bigger than $R_1$) but $sum_{k=1}^{infty}b_nx^n$ converges (because L is bigger than $R_2$). so $R=M$.



    But what if $R_1=R_2$? we can take for example $a_n=1/n$, $b_n$=$-1/n$ and then we'll get $R=infty$. We can take $a_n=b_n=1/n$ and than simply $R=R_1=R_2$. Is there anything we can say about this case? is it always $R=R_1=R_2$ or $R=infty$ in this case?



    thanks!










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $R_1$ Is the Radius of convergence of $sum_{k=1}^{infty}a_nx^n$.
      $R_2$ Is the Radius of convergence of $sum_{k=1}^{infty}b_nx^n$.

      What can we say of the Radius of convergence of $sum_{k=1}^{infty}(a_n+b_n)x^n$,we'll call it $R$?





      If $R_1 neq R_2$ I know the answer is $R=M=min(R_1,R_2)$ Because it is obvious that for all $x in (-M,M)$ the sum converges, and suppose $R>M$ (And without loss of generality $R_2>R_1$), than take new number $L>0$ Such that $L<R$ and $R_1<L<R_2$, than since $L<R$ we get $sum_{k=1}^{infty}(a_n+b_n)L^n$ converges, and that's a contradiction because $sum_{k=1}^{infty}a_nx^n$ diverges (because L is bigger than $R_1$) but $sum_{k=1}^{infty}b_nx^n$ converges (because L is bigger than $R_2$). so $R=M$.



      But what if $R_1=R_2$? we can take for example $a_n=1/n$, $b_n$=$-1/n$ and then we'll get $R=infty$. We can take $a_n=b_n=1/n$ and than simply $R=R_1=R_2$. Is there anything we can say about this case? is it always $R=R_1=R_2$ or $R=infty$ in this case?



      thanks!










      share|cite|improve this question











      $endgroup$




      $R_1$ Is the Radius of convergence of $sum_{k=1}^{infty}a_nx^n$.
      $R_2$ Is the Radius of convergence of $sum_{k=1}^{infty}b_nx^n$.

      What can we say of the Radius of convergence of $sum_{k=1}^{infty}(a_n+b_n)x^n$,we'll call it $R$?





      If $R_1 neq R_2$ I know the answer is $R=M=min(R_1,R_2)$ Because it is obvious that for all $x in (-M,M)$ the sum converges, and suppose $R>M$ (And without loss of generality $R_2>R_1$), than take new number $L>0$ Such that $L<R$ and $R_1<L<R_2$, than since $L<R$ we get $sum_{k=1}^{infty}(a_n+b_n)L^n$ converges, and that's a contradiction because $sum_{k=1}^{infty}a_nx^n$ diverges (because L is bigger than $R_1$) but $sum_{k=1}^{infty}b_nx^n$ converges (because L is bigger than $R_2$). so $R=M$.



      But what if $R_1=R_2$? we can take for example $a_n=1/n$, $b_n$=$-1/n$ and then we'll get $R=infty$. We can take $a_n=b_n=1/n$ and than simply $R=R_1=R_2$. Is there anything we can say about this case? is it always $R=R_1=R_2$ or $R=infty$ in this case?



      thanks!







      calculus sequences-and-series limits convergence power-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 14:24









      gt6989b

      34k22455




      34k22455










      asked Dec 10 '18 at 14:22









      OmerOmer

      3618




      3618






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Yes: you can say that $Rgeqslant R_1=R_2$. But that's all you can say.



          Take for instance, the series $displaystylesum_{n=0}^inftyleft(1+2^{-n}right)z^n$ and the series $displaystylesum_{n=0}^infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
            $endgroup$
            – Omer
            Dec 10 '18 at 14:28






          • 1




            $begingroup$
            I've added one such example to my answer.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:29






          • 1




            $begingroup$
            No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:47










          • $begingroup$
            Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
            $endgroup$
            – Omer
            Dec 10 '18 at 14:48











          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%2f3033970%2fradius-of-convergence-of-two-power-series-and-the-sum-of-the-power-serieses%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









          1












          $begingroup$

          Yes: you can say that $Rgeqslant R_1=R_2$. But that's all you can say.



          Take for instance, the series $displaystylesum_{n=0}^inftyleft(1+2^{-n}right)z^n$ and the series $displaystylesum_{n=0}^infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
            $endgroup$
            – Omer
            Dec 10 '18 at 14:28






          • 1




            $begingroup$
            I've added one such example to my answer.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:29






          • 1




            $begingroup$
            No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:47










          • $begingroup$
            Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
            $endgroup$
            – Omer
            Dec 10 '18 at 14:48
















          1












          $begingroup$

          Yes: you can say that $Rgeqslant R_1=R_2$. But that's all you can say.



          Take for instance, the series $displaystylesum_{n=0}^inftyleft(1+2^{-n}right)z^n$ and the series $displaystylesum_{n=0}^infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
            $endgroup$
            – Omer
            Dec 10 '18 at 14:28






          • 1




            $begingroup$
            I've added one such example to my answer.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:29






          • 1




            $begingroup$
            No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:47










          • $begingroup$
            Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
            $endgroup$
            – Omer
            Dec 10 '18 at 14:48














          1












          1








          1





          $begingroup$

          Yes: you can say that $Rgeqslant R_1=R_2$. But that's all you can say.



          Take for instance, the series $displaystylesum_{n=0}^inftyleft(1+2^{-n}right)z^n$ and the series $displaystylesum_{n=0}^infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.






          share|cite|improve this answer











          $endgroup$



          Yes: you can say that $Rgeqslant R_1=R_2$. But that's all you can say.



          Take for instance, the series $displaystylesum_{n=0}^inftyleft(1+2^{-n}right)z^n$ and the series $displaystylesum_{n=0}^infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 14:29

























          answered Dec 10 '18 at 14:25









          José Carlos SantosJosé Carlos Santos

          159k22126231




          159k22126231












          • $begingroup$
            Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
            $endgroup$
            – Omer
            Dec 10 '18 at 14:28






          • 1




            $begingroup$
            I've added one such example to my answer.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:29






          • 1




            $begingroup$
            No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:47










          • $begingroup$
            Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
            $endgroup$
            – Omer
            Dec 10 '18 at 14:48


















          • $begingroup$
            Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
            $endgroup$
            – Omer
            Dec 10 '18 at 14:28






          • 1




            $begingroup$
            I've added one such example to my answer.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:29






          • 1




            $begingroup$
            No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
            $endgroup$
            – José Carlos Santos
            Dec 10 '18 at 14:47










          • $begingroup$
            Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
            $endgroup$
            – Omer
            Dec 10 '18 at 14:48
















          $begingroup$
          Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
          $endgroup$
          – Omer
          Dec 10 '18 at 14:28




          $begingroup$
          Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity?
          $endgroup$
          – Omer
          Dec 10 '18 at 14:28




          1




          1




          $begingroup$
          I've added one such example to my answer.
          $endgroup$
          – José Carlos Santos
          Dec 10 '18 at 14:29




          $begingroup$
          I've added one such example to my answer.
          $endgroup$
          – José Carlos Santos
          Dec 10 '18 at 14:29




          1




          1




          $begingroup$
          No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
          $endgroup$
          – José Carlos Santos
          Dec 10 '18 at 14:47




          $begingroup$
          No. The radius of convergence of $sum_{n=0}^inftyfrac{z^n}{2^n}$ is $2$.
          $endgroup$
          – José Carlos Santos
          Dec 10 '18 at 14:47












          $begingroup$
          Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
          $endgroup$
          – Omer
          Dec 10 '18 at 14:48




          $begingroup$
          Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot!
          $endgroup$
          – Omer
          Dec 10 '18 at 14:48


















          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%2f3033970%2fradius-of-convergence-of-two-power-series-and-the-sum-of-the-power-serieses%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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Aardman Animations

          Are they similar matrix