Why does the criterion for convergence of a power series not imply every series with bounded terms converges?












7












$begingroup$


I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:




Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by



$$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$



Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.




My bogus conclusion:




Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.




My reasoning:



Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.






complex-analysis convergence power-series







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:




    Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by



    $$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$



    Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.




    My bogus conclusion:




    Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.




    My reasoning:



    Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.






    complex-analysis convergence power-series







    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:




      Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by



      $$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$



      Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.




      My bogus conclusion:




      Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.




      My reasoning:



      Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.






      complex-analysis convergence power-series







      share|cite|improve this question











      $endgroup$




      I am reading Complex Made Simple by David C. Ullrich. There is a result from which I am deducing bogus conclusions, so I must be misunderstanding it somehow:




      Lemma 1.0. Suppose $(c_n)_{n = 0}^{infty}$ is a sequence of complex numbers, and define $R in [0, infty]$ by



      $$R = sup {r ge 0: text{the sequence } (c_nr^n) text{ is bounded}}.$$



      Then the power series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely and uniformly on every compact subset of the disk $D(z_0, R)$ and diverges at every point $z$ with $|z-z_0|>R$.




      My bogus conclusion:




      Let $c_n$ be a sequence of complex numbers and suppose that $c_n r^n$ is bounded. Then $sum_{n=0}^{infty} c_n r^n$ converges.




      My reasoning:



      Let $c_n$ be any sequence of complex numbers. The series $sum_{n=0}^{infty}c_n(z-z_0)^n$ converges absolutely whenever $|z - z_0|<R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $r < R$, so $sum_{n=0}^{infty}c_nr^n$ converges whenever $c_n r^n$ is bounded.






      complex-analysis convergence power-series




      complex-analysis convergence power-series






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      share|cite|improve this question













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      edited Dec 22 '18 at 23:09









      Eric Wofsey

      182k12209337




      182k12209337










      asked Dec 22 '18 at 22:51









      OviOvi

      12.4k1038112




      12.4k1038112






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
            $endgroup$
            – Ovi
            Dec 29 '18 at 14:45










          • $begingroup$
            (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
            $endgroup$
            – Ovi
            Dec 29 '18 at 14:45










          • $begingroup$
            Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
            $endgroup$
            – Ross Millikan
            Dec 29 '18 at 15:26










          • $begingroup$
            Thank you! ${}{}{}$
            $endgroup$
            – Ovi
            Dec 29 '18 at 16:06



















          7












          $begingroup$

          The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
          a simple example is $c_n = (-1)^n$ and $r=1$.






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
                $endgroup$
                – Ross Millikan
                Dec 29 '18 at 15:26










              • $begingroup$
                Thank you! ${}{}{}$
                $endgroup$
                – Ovi
                Dec 29 '18 at 16:06
















              3












              $begingroup$

              The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
                $endgroup$
                – Ross Millikan
                Dec 29 '18 at 15:26










              • $begingroup$
                Thank you! ${}{}{}$
                $endgroup$
                – Ovi
                Dec 29 '18 at 16:06














              3












              3








              3





              $begingroup$

              The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.






              share|cite|improve this answer









              $endgroup$



              The problem comes in the last step. Just because $sum_{n=1}^infty c_nr^n$ converges with $r lt R$ you cannot conclude that $sum_{n=1}^infty c_nR^n$ converges. As an example, let $c_n=1$ for all $n$. We note that $R=1$ here. $c_nR^n=1$, so is bounded. For any $r lt 1$, $sum_{n=1}^infty c_nr^n$ converges absolutely, but $sum_{n=1}^infty c_nR^n$ does not converge.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 22 '18 at 23:02









              Ross MillikanRoss Millikan

              293k23197371




              293k23197371












              • $begingroup$
                Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
                $endgroup$
                – Ross Millikan
                Dec 29 '18 at 15:26










              • $begingroup$
                Thank you! ${}{}{}$
                $endgroup$
                – Ovi
                Dec 29 '18 at 16:06


















              • $begingroup$
                Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
                $endgroup$
                – Ovi
                Dec 29 '18 at 14:45










              • $begingroup$
                Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
                $endgroup$
                – Ross Millikan
                Dec 29 '18 at 15:26










              • $begingroup$
                Thank you! ${}{}{}$
                $endgroup$
                – Ovi
                Dec 29 '18 at 16:06
















              $begingroup$
              Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
              $endgroup$
              – Ovi
              Dec 29 '18 at 14:45




              $begingroup$
              Hi! I $think$ I understand, but could I please run my whole reasoning by you to check? My reasoning is this: Let $c_n$ be a sequence and let's say $R=1$. Take $r=0.5$. Then $c_n (0.5)^n$ is bounded. Let $z_0=0$. The theorem tells us that $sum_{n=0}^{infty} c_nz^n$ converges absolutely for any $z$ with $|z|<1$; pick $z = 0.5$, Then $sum_{n=0}^{infty} c_n (0.5)^n$ converges. But here I made the mistake; the conclusion we can draw from here is that $sum_{n=0}^{infty} c_nr^n$ converges for "almost every" $r$ for which $c_nr^n$ is bounded (for any $r$ with $|r|<1$). (continued)
              $endgroup$
              – Ovi
              Dec 29 '18 at 14:45












              $begingroup$
              (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
              $endgroup$
              – Ovi
              Dec 29 '18 at 14:45




              $begingroup$
              (continued) But we cannot draw the conclusion that "If $c_n r^n$ is bounded, then $sum_{n=0}^{infty} c_n r^n$ converges, because it may be the case that $c_n R^n$ is bounded, but $sum_{n=0}^{infty} c_n R^n$ does not converge.
              $endgroup$
              – Ovi
              Dec 29 '18 at 14:45












              $begingroup$
              Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
              $endgroup$
              – Ross Millikan
              Dec 29 '18 at 15:26




              $begingroup$
              Yes, that is correct. The sum will converge for every $r$ with modulus strictly less than $R$, but not necessarily on the circle $|r|=R$.
              $endgroup$
              – Ross Millikan
              Dec 29 '18 at 15:26












              $begingroup$
              Thank you! ${}{}{}$
              $endgroup$
              – Ovi
              Dec 29 '18 at 16:06




              $begingroup$
              Thank you! ${}{}{}$
              $endgroup$
              – Ovi
              Dec 29 '18 at 16:06











              7












              $begingroup$

              The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
              a simple example is $c_n = (-1)^n$ and $r=1$.






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
                a simple example is $c_n = (-1)^n$ and $r=1$.






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
                  a simple example is $c_n = (-1)^n$ and $r=1$.






                  share|cite|improve this answer









                  $endgroup$



                  The fact that $c_n r^n$ is bounded, means that $r$ is in the set we're taking the sup of, so $r le R$. But the convergence of $sum_{n=0}^infty c_n s^n$ is only guaranteed for $s < R$. It could very well be that $r=R$;
                  a simple example is $c_n = (-1)^n$ and $r=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 22:59









                  Henno BrandsmaHenno Brandsma

                  106k347114




                  106k347114























                      5












                      $begingroup$

                      The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.






                      share|cite|improve this answer









                      $endgroup$


















                        5












                        $begingroup$

                        The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.






                        share|cite|improve this answer









                        $endgroup$
















                          5












                          5








                          5





                          $begingroup$

                          The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.






                          share|cite|improve this answer









                          $endgroup$



                          The condition $r<R$ is not equivalent to $c_nr^n$ being bounded. It is possible that $c_nr^n$ is bounded for $r=R$ as well, and in that case we cannot conclude that the series converges.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 22 '18 at 22:58









                          Eric WofseyEric Wofsey

                          182k12209337




                          182k12209337






























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