Using the Cauchy-Hadamard Theorem to find a radius of convergence












0















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question
























  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40


















0















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question
























  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40
















0












0








0








Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question
















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion




sequences-and-series convergence power-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 11:28

























asked Nov 29 '18 at 11:22









joseph

4329




4329












  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40




















  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40


















$0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
– mathworker21
Nov 29 '18 at 11:31






$0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
– mathworker21
Nov 29 '18 at 11:31














I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
– joseph
Nov 29 '18 at 11:40






I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
– joseph
Nov 29 '18 at 11:40












1 Answer
1






active

oldest

votes


















0














begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
&=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
&=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
&=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
&=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



where I have used Stirling approximation.






share|cite|improve this answer























    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%2f3018505%2fusing-the-cauchy-hadamard-theorem-to-find-a-radius-of-convergence%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














    begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
    &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



    and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
    &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
    &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
    &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



    where I have used Stirling approximation.






    share|cite|improve this answer




























      0














      begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
      &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



      and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
      &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
      &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
      &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



      where I have used Stirling approximation.






      share|cite|improve this answer


























        0












        0








        0






        begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
        &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



        and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
        &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
        &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
        &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



        where I have used Stirling approximation.






        share|cite|improve this answer














        begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
        &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



        and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
        &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
        &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
        &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



        where I have used Stirling approximation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '18 at 23:49

























        answered Nov 29 '18 at 12:16









        Siong Thye Goh

        99.5k1464117




        99.5k1464117






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3018505%2fusing-the-cauchy-hadamard-theorem-to-find-a-radius-of-convergence%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