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







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$endgroup$












  • $begingroup$
    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






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edited Jan 1 at 12:25









Did

247k23222458




247k23222458










asked Jan 1 at 11:39









Omojola MichealOmojola Micheal

1,824324




1,824324












  • $begingroup$
    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

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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






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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

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    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

    155k22124227




    155k22124227












    • $begingroup$
      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

        8,6343516




        8,6343516






























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