Hint requested for: If $sum_{n=0}^{infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and...












2












$begingroup$


I would like to prove




If $displaystylesum_{n=0}^{infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $0<a<|x_0|$. (Sorry not enough characters to put $0<a$ in the title)




I can easily look at the proof in my book, but I really want to learn how to think about analysis and be able to prove these things on my own.



I tried to make it easier by assuming that $x_0>0$ and $forall (n in mathbb{N}) a_n>0$. In this case, we can apply the $M$-Test with $M_n = a_n x_0^n$.



However, I'm not sure how to modify this to work in the general case.






real-analysis analysis power-series uniform-convergence absolute-convergence







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  • $begingroup$
    What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 16:00






  • 1




    $begingroup$
    OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
    $endgroup$
    – D_S
    Dec 28 '18 at 16:42


















2












$begingroup$


I would like to prove




If $displaystylesum_{n=0}^{infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $0<a<|x_0|$. (Sorry not enough characters to put $0<a$ in the title)




I can easily look at the proof in my book, but I really want to learn how to think about analysis and be able to prove these things on my own.



I tried to make it easier by assuming that $x_0>0$ and $forall (n in mathbb{N}) a_n>0$. In this case, we can apply the $M$-Test with $M_n = a_n x_0^n$.



However, I'm not sure how to modify this to work in the general case.






real-analysis analysis power-series uniform-convergence absolute-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 16:00






  • 1




    $begingroup$
    OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
    $endgroup$
    – D_S
    Dec 28 '18 at 16:42
















2












2








2





$begingroup$


I would like to prove




If $displaystylesum_{n=0}^{infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $0<a<|x_0|$. (Sorry not enough characters to put $0<a$ in the title)




I can easily look at the proof in my book, but I really want to learn how to think about analysis and be able to prove these things on my own.



I tried to make it easier by assuming that $x_0>0$ and $forall (n in mathbb{N}) a_n>0$. In this case, we can apply the $M$-Test with $M_n = a_n x_0^n$.



However, I'm not sure how to modify this to work in the general case.






real-analysis analysis power-series uniform-convergence absolute-convergence







share|cite|improve this question











$endgroup$




I would like to prove




If $displaystylesum_{n=0}^{infty} a_n x^n$ converges for some $x_0$, then it converges uniformly and absolutely on $[-a, a]$ with $0<a<|x_0|$. (Sorry not enough characters to put $0<a$ in the title)




I can easily look at the proof in my book, but I really want to learn how to think about analysis and be able to prove these things on my own.



I tried to make it easier by assuming that $x_0>0$ and $forall (n in mathbb{N}) a_n>0$. In this case, we can apply the $M$-Test with $M_n = a_n x_0^n$.



However, I'm not sure how to modify this to work in the general case.






real-analysis analysis power-series uniform-convergence absolute-convergence




real-analysis analysis power-series uniform-convergence absolute-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 15:54









José Carlos Santos

175k24134243




175k24134243










asked Dec 28 '18 at 15:31









OviOvi

12.5k1040117




12.5k1040117












  • $begingroup$
    What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 16:00






  • 1




    $begingroup$
    OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
    $endgroup$
    – D_S
    Dec 28 '18 at 16:42




















  • $begingroup$
    What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 16:00






  • 1




    $begingroup$
    OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
    $endgroup$
    – D_S
    Dec 28 '18 at 16:42


















$begingroup$
What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:00




$begingroup$
What about $sum_{n=1}^infty x^n/n$ and $x_0=-1$?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:00




1




1




$begingroup$
OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
$endgroup$
– D_S
Dec 28 '18 at 16:42






$begingroup$
OP should say if "$sum a_nx^n$ converges absolutely for some $x_0$."
$endgroup$
– D_S
Dec 28 '18 at 16:42












2 Answers
2






active

oldest

votes


















2












$begingroup$

Since the series $displaystylesum_{n=0}^infty a_n{x_0}^n$ converges, you know that $lim_{ntoinfty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $lim_{ntoinfty}lvert a_n{x_0}^nrvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $displaystylesum_{n=0}^infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<lvert x_0rvert$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
    $endgroup$
    – Ovi
    Dec 28 '18 at 16:27










  • $begingroup$
    Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
    $endgroup$
    – José Carlos Santos
    Dec 28 '18 at 16:29



















0












$begingroup$

Hint: $|a_nx^n|=left | a_nleft ( frac{x}{x_0} right )^{n}cdot x^{n}_0 right |=left | a_nx_0^{n}cdotleft ( frac{x}{x_0} right )^{n} right |, |x|le a<|x_0|$ and $sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|to 0$ as $nto infty.$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Since the series $displaystylesum_{n=0}^infty a_n{x_0}^n$ converges, you know that $lim_{ntoinfty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $lim_{ntoinfty}lvert a_n{x_0}^nrvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $displaystylesum_{n=0}^infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<lvert x_0rvert$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
      $endgroup$
      – Ovi
      Dec 28 '18 at 16:27










    • $begingroup$
      Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
      $endgroup$
      – José Carlos Santos
      Dec 28 '18 at 16:29
















    2












    $begingroup$

    Since the series $displaystylesum_{n=0}^infty a_n{x_0}^n$ converges, you know that $lim_{ntoinfty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $lim_{ntoinfty}lvert a_n{x_0}^nrvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $displaystylesum_{n=0}^infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<lvert x_0rvert$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
      $endgroup$
      – Ovi
      Dec 28 '18 at 16:27










    • $begingroup$
      Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
      $endgroup$
      – José Carlos Santos
      Dec 28 '18 at 16:29














    2












    2








    2





    $begingroup$

    Since the series $displaystylesum_{n=0}^infty a_n{x_0}^n$ converges, you know that $lim_{ntoinfty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $lim_{ntoinfty}lvert a_n{x_0}^nrvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $displaystylesum_{n=0}^infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<lvert x_0rvert$.






    share|cite|improve this answer









    $endgroup$



    Since the series $displaystylesum_{n=0}^infty a_n{x_0}^n$ converges, you know that $lim_{ntoinfty}a_n{x_0}^n=0$. This is equivalent to the ssertion that $lim_{ntoinfty}lvert a_n{x_0}^nrvert=0$. This is anough to be able to apply the Weierstrass $M$-test to the series $displaystylesum_{n=0}^infty a_nx^n$ in $[-a,a]$. Don't forget to use the fact that $a<lvert x_0rvert$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 28 '18 at 15:37









    José Carlos SantosJosé Carlos Santos

    175k24134243




    175k24134243












    • $begingroup$
      But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
      $endgroup$
      – Ovi
      Dec 28 '18 at 16:27










    • $begingroup$
      Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
      $endgroup$
      – José Carlos Santos
      Dec 28 '18 at 16:29


















    • $begingroup$
      But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
      $endgroup$
      – Ovi
      Dec 28 '18 at 16:27










    • $begingroup$
      Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
      $endgroup$
      – José Carlos Santos
      Dec 28 '18 at 16:29
















    $begingroup$
    But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
    $endgroup$
    – Ovi
    Dec 28 '18 at 16:27




    $begingroup$
    But we cannot just pick $M_n = |a_n x_0^n|$, right? Because I think the sum $sum_{n=0}^infty |a_n{x_0}^n|$ may not converge.
    $endgroup$
    – Ovi
    Dec 28 '18 at 16:27












    $begingroup$
    Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
    $endgroup$
    – José Carlos Santos
    Dec 28 '18 at 16:29




    $begingroup$
    Of course we can't! Pick $M_n=lvert a_nrvert a^n$ instead.
    $endgroup$
    – José Carlos Santos
    Dec 28 '18 at 16:29











    0












    $begingroup$

    Hint: $|a_nx^n|=left | a_nleft ( frac{x}{x_0} right )^{n}cdot x^{n}_0 right |=left | a_nx_0^{n}cdotleft ( frac{x}{x_0} right )^{n} right |, |x|le a<|x_0|$ and $sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|to 0$ as $nto infty.$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: $|a_nx^n|=left | a_nleft ( frac{x}{x_0} right )^{n}cdot x^{n}_0 right |=left | a_nx_0^{n}cdotleft ( frac{x}{x_0} right )^{n} right |, |x|le a<|x_0|$ and $sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|to 0$ as $nto infty.$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: $|a_nx^n|=left | a_nleft ( frac{x}{x_0} right )^{n}cdot x^{n}_0 right |=left | a_nx_0^{n}cdotleft ( frac{x}{x_0} right )^{n} right |, |x|le a<|x_0|$ and $sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|to 0$ as $nto infty.$






        share|cite|improve this answer









        $endgroup$



        Hint: $|a_nx^n|=left | a_nleft ( frac{x}{x_0} right )^{n}cdot x^{n}_0 right |=left | a_nx_0^{n}cdotleft ( frac{x}{x_0} right )^{n} right |, |x|le a<|x_0|$ and $sum a_nx_0^{n} $ converges so $|a_nx_0^{n}|to 0$ as $nto infty.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 16:17









        MatematletaMatematleta

        12.2k21020




        12.2k21020






























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