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!






calculus sequences-and-series limits convergence power-series







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$endgroup$

















    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!






    calculus sequences-and-series limits convergence power-series







    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!






      calculus sequences-and-series limits convergence power-series







      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




      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
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          $begingroup$

          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











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          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


















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