Find the interval of convergence of the series $ sum^{infty}limits_{k=0} ((-1)^k+3)^k(x-1)^k $












4












$begingroup$


I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}



PROOF



Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}



QUESTION:



Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
    $endgroup$
    – Masacroso
    Jan 1 at 11:48












  • $begingroup$
    @Masacroso: Oh oh, that has been my thought too!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:54










  • $begingroup$
    @Did: Got your advice!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:40
















4












$begingroup$


I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}



PROOF



Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}



QUESTION:



Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
    $endgroup$
    – Masacroso
    Jan 1 at 11:48












  • $begingroup$
    @Masacroso: Oh oh, that has been my thought too!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:54










  • $begingroup$
    @Did: Got your advice!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:40














4












4








4


1



$begingroup$


I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}



PROOF



Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}



QUESTION:



Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}



PROOF



Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}



QUESTION:



Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 12:25









Did

249k23226466




249k23226466










asked Jan 1 at 11:39









Omojola MichealOmojola Micheal

1,999424




1,999424












  • $begingroup$
    if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
    $endgroup$
    – Masacroso
    Jan 1 at 11:48












  • $begingroup$
    @Masacroso: Oh oh, that has been my thought too!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:54










  • $begingroup$
    @Did: Got your advice!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:40


















  • $begingroup$
    if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
    $endgroup$
    – Masacroso
    Jan 1 at 11:48












  • $begingroup$
    @Masacroso: Oh oh, that has been my thought too!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 11:54










  • $begingroup$
    @Did: Got your advice!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:40
















$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48






$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48














$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54




$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54












$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40




$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry, I edited my work. That equation truly, does not hold!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:18






  • 1




    $begingroup$
    So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
    $endgroup$
    – Did
    Jan 1 at 12:23












  • $begingroup$
    But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:23










  • $begingroup$
    @Did: Sorry, I don't get who you're referring to!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:24










  • $begingroup$
    @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
    $endgroup$
    – Did
    Jan 1 at 12:27





















0












$begingroup$

Hint



$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$



and now



$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$



which converges if $|4(x-1)| < 1$ etc. so the result is



$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$



for $frac 34lt xlt frac 54$



Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found



enter image description here






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%2f3058414%2ffind-the-interval-of-convergence-of-the-series-sum-infty-limits-k-0-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, I edited my work. That equation truly, does not hold!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:18






    • 1




      $begingroup$
      So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
      $endgroup$
      – Did
      Jan 1 at 12:23












    • $begingroup$
      But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:23










    • $begingroup$
      @Did: Sorry, I don't get who you're referring to!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:24










    • $begingroup$
      @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
      $endgroup$
      – Did
      Jan 1 at 12:27


















    1












    $begingroup$

    Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, I edited my work. That equation truly, does not hold!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:18






    • 1




      $begingroup$
      So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
      $endgroup$
      – Did
      Jan 1 at 12:23












    • $begingroup$
      But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:23










    • $begingroup$
      @Did: Sorry, I don't get who you're referring to!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:24










    • $begingroup$
      @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
      $endgroup$
      – Did
      Jan 1 at 12:27
















    1












    1








    1





    $begingroup$

    Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$






    share|cite|improve this answer











    $endgroup$



    Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 1 at 12:33

























    answered Jan 1 at 12:00









    José Carlos SantosJosé Carlos Santos

    170k23132238




    170k23132238












    • $begingroup$
      Sorry, I edited my work. That equation truly, does not hold!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:18






    • 1




      $begingroup$
      So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
      $endgroup$
      – Did
      Jan 1 at 12:23












    • $begingroup$
      But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:23










    • $begingroup$
      @Did: Sorry, I don't get who you're referring to!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:24










    • $begingroup$
      @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
      $endgroup$
      – Did
      Jan 1 at 12:27




















    • $begingroup$
      Sorry, I edited my work. That equation truly, does not hold!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:18






    • 1




      $begingroup$
      So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
      $endgroup$
      – Did
      Jan 1 at 12:23












    • $begingroup$
      But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:23










    • $begingroup$
      @Did: Sorry, I don't get who you're referring to!
      $endgroup$
      – Omojola Micheal
      Jan 1 at 12:24










    • $begingroup$
      @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
      $endgroup$
      – Did
      Jan 1 at 12:27


















    $begingroup$
    Sorry, I edited my work. That equation truly, does not hold!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:18




    $begingroup$
    Sorry, I edited my work. That equation truly, does not hold!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:18




    1




    1




    $begingroup$
    So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
    $endgroup$
    – Did
    Jan 1 at 12:23






    $begingroup$
    So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
    $endgroup$
    – Did
    Jan 1 at 12:23














    $begingroup$
    But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:23




    $begingroup$
    But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:23












    $begingroup$
    @Did: Sorry, I don't get who you're referring to!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:24




    $begingroup$
    @Did: Sorry, I don't get who you're referring to!
    $endgroup$
    – Omojola Micheal
    Jan 1 at 12:24












    $begingroup$
    @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
    $endgroup$
    – Did
    Jan 1 at 12:27






    $begingroup$
    @Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
    $endgroup$
    – Did
    Jan 1 at 12:27













    0












    $begingroup$

    Hint



    $$
    sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
    $$



    and now



    $$
    sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
    $$



    which converges if $|4(x-1)| < 1$ etc. so the result is



    $$
    sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
    $$



    for $frac 34lt xlt frac 54$



    Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found



    enter image description here






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint



      $$
      sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
      $$



      and now



      $$
      sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
      $$



      which converges if $|4(x-1)| < 1$ etc. so the result is



      $$
      sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
      $$



      for $frac 34lt xlt frac 54$



      Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found



      enter image description here






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint



        $$
        sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
        $$



        and now



        $$
        sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
        $$



        which converges if $|4(x-1)| < 1$ etc. so the result is



        $$
        sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
        $$



        for $frac 34lt xlt frac 54$



        Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found



        enter image description here






        share|cite|improve this answer











        $endgroup$



        Hint



        $$
        sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
        $$



        and now



        $$
        sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
        $$



        which converges if $|4(x-1)| < 1$ etc. so the result is



        $$
        sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
        $$



        for $frac 34lt xlt frac 54$



        Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 13:24

























        answered Jan 1 at 12:24









        CesareoCesareo

        9,4673517




        9,4673517






























            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%2f3058414%2ffind-the-interval-of-convergence-of-the-series-sum-infty-limits-k-0-1%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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more