Help understand part of the proof. Radius of convergence is $frac{1}{limsup |a_n|^{1/n}}$












4












$begingroup$


Can you help me understand the highlighted parts of the proof. Thanks :)



Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.



Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.



Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.




For the first highlighted part, I don't understand where the geometric series part comes from.
$sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?



And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?



Thanks!






complex-analysis power-series limsup-and-liminf







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Can you help me understand the highlighted parts of the proof. Thanks :)



    Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.



    Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.



    Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
    Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.




    For the first highlighted part, I don't understand where the geometric series part comes from.
    $sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?



    And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?



    Thanks!






    complex-analysis power-series limsup-and-liminf







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Can you help me understand the highlighted parts of the proof. Thanks :)



      Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.



      Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.



      Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
      Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.




      For the first highlighted part, I don't understand where the geometric series part comes from.
      $sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?



      And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?



      Thanks!






      complex-analysis power-series limsup-and-liminf







      share|cite|improve this question











      $endgroup$




      Can you help me understand the highlighted parts of the proof. Thanks :)



      Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.



      Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.



      Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
      Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.




      For the first highlighted part, I don't understand where the geometric series part comes from.
      $sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?



      And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?



      Thanks!






      complex-analysis power-series limsup-and-liminf




      complex-analysis power-series limsup-and-liminf






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 19 '14 at 16:48







      user181662

















      asked Oct 18 '14 at 21:07









      user181662user181662

      674




      674






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.



          To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore






          share|cite|improve this answer









          $endgroup$













            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%2f979993%2fhelp-understand-part-of-the-proof-radius-of-convergence-is-frac1-limsup-a%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












            $begingroup$

            The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.



            To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.



              To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.



                To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore






                share|cite|improve this answer









                $endgroup$



                The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.



                To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 14:53









                RM777RM777

                38312




                38312






























                    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%2f979993%2fhelp-understand-part-of-the-proof-radius-of-convergence-is-frac1-limsup-a%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.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more