$sum_limits{n=1}^{infty}frac{n}{x^n}$ ratio or root test? different results?












0












$begingroup$



Show that $sum_limits{n=1}^{infty}frac{n}{x^n}$ converges absolutely.




I used the ratio test:



$lim_{ntoinfty}frac{frac{n}{x^n}}{frac{n+1}{x^{n+1}}}=lim_{ntoinfty}frac{n}{n+1}x=x$, so the dominion where the function converges is $|x|<1$.



However the resolution proposed to use the root test:



$lim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}leqslantlim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}=frac{1}{|x|}$ therefore the function converges when $frac{1}{|x|}>1implies |x|>1$ which contradicts the ratio test.



Question:



Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?



Thanks in advance!






calculus real-analysis sequences-and-series







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$endgroup$








  • 7




    $begingroup$
    You applied the ratio test incorrectly.
    $endgroup$
    – Mark Viola
    Mar 3 '18 at 19:35






  • 4




    $begingroup$
    The ratio test is the n+1th term over the nth, not the other way around.
    $endgroup$
    – Andrew Li
    Mar 3 '18 at 19:37












  • $begingroup$
    @MarkViola Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47










  • $begingroup$
    @AndrewLi Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47
















0












$begingroup$



Show that $sum_limits{n=1}^{infty}frac{n}{x^n}$ converges absolutely.




I used the ratio test:



$lim_{ntoinfty}frac{frac{n}{x^n}}{frac{n+1}{x^{n+1}}}=lim_{ntoinfty}frac{n}{n+1}x=x$, so the dominion where the function converges is $|x|<1$.



However the resolution proposed to use the root test:



$lim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}leqslantlim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}=frac{1}{|x|}$ therefore the function converges when $frac{1}{|x|}>1implies |x|>1$ which contradicts the ratio test.



Question:



Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?



Thanks in advance!






calculus real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 7




    $begingroup$
    You applied the ratio test incorrectly.
    $endgroup$
    – Mark Viola
    Mar 3 '18 at 19:35






  • 4




    $begingroup$
    The ratio test is the n+1th term over the nth, not the other way around.
    $endgroup$
    – Andrew Li
    Mar 3 '18 at 19:37












  • $begingroup$
    @MarkViola Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47










  • $begingroup$
    @AndrewLi Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47














0












0








0


0



$begingroup$



Show that $sum_limits{n=1}^{infty}frac{n}{x^n}$ converges absolutely.




I used the ratio test:



$lim_{ntoinfty}frac{frac{n}{x^n}}{frac{n+1}{x^{n+1}}}=lim_{ntoinfty}frac{n}{n+1}x=x$, so the dominion where the function converges is $|x|<1$.



However the resolution proposed to use the root test:



$lim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}leqslantlim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}=frac{1}{|x|}$ therefore the function converges when $frac{1}{|x|}>1implies |x|>1$ which contradicts the ratio test.



Question:



Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?



Thanks in advance!






calculus real-analysis sequences-and-series







share|cite|improve this question









$endgroup$





Show that $sum_limits{n=1}^{infty}frac{n}{x^n}$ converges absolutely.




I used the ratio test:



$lim_{ntoinfty}frac{frac{n}{x^n}}{frac{n+1}{x^{n+1}}}=lim_{ntoinfty}frac{n}{n+1}x=x$, so the dominion where the function converges is $|x|<1$.



However the resolution proposed to use the root test:



$lim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}leqslantlim_{ntoinfty}(frac{n}{|x|^n})^{frac{1}{n}}=frac{1}{|x|}$ therefore the function converges when $frac{1}{|x|}>1implies |x|>1$ which contradicts the ratio test.



Question:



Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?



Thanks in advance!






calculus real-analysis sequences-and-series




calculus 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










asked Mar 3 '18 at 19:28









Pedro GomesPedro Gomes

1,7532721




1,7532721








  • 7




    $begingroup$
    You applied the ratio test incorrectly.
    $endgroup$
    – Mark Viola
    Mar 3 '18 at 19:35






  • 4




    $begingroup$
    The ratio test is the n+1th term over the nth, not the other way around.
    $endgroup$
    – Andrew Li
    Mar 3 '18 at 19:37












  • $begingroup$
    @MarkViola Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47










  • $begingroup$
    @AndrewLi Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47














  • 7




    $begingroup$
    You applied the ratio test incorrectly.
    $endgroup$
    – Mark Viola
    Mar 3 '18 at 19:35






  • 4




    $begingroup$
    The ratio test is the n+1th term over the nth, not the other way around.
    $endgroup$
    – Andrew Li
    Mar 3 '18 at 19:37












  • $begingroup$
    @MarkViola Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47










  • $begingroup$
    @AndrewLi Thanks a lot!
    $endgroup$
    – Pedro Gomes
    Mar 3 '18 at 19:47








7




7




$begingroup$
You applied the ratio test incorrectly.
$endgroup$
– Mark Viola
Mar 3 '18 at 19:35




$begingroup$
You applied the ratio test incorrectly.
$endgroup$
– Mark Viola
Mar 3 '18 at 19:35




4




4




$begingroup$
The ratio test is the n+1th term over the nth, not the other way around.
$endgroup$
– Andrew Li
Mar 3 '18 at 19:37






$begingroup$
The ratio test is the n+1th term over the nth, not the other way around.
$endgroup$
– Andrew Li
Mar 3 '18 at 19:37














$begingroup$
@MarkViola Thanks a lot!
$endgroup$
– Pedro Gomes
Mar 3 '18 at 19:47




$begingroup$
@MarkViola Thanks a lot!
$endgroup$
– Pedro Gomes
Mar 3 '18 at 19:47












$begingroup$
@AndrewLi Thanks a lot!
$endgroup$
– Pedro Gomes
Mar 3 '18 at 19:47




$begingroup$
@AndrewLi Thanks a lot!
$endgroup$
– Pedro Gomes
Mar 3 '18 at 19:47










1 Answer
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You have calculated $large lim {a_nover a_{n+1}}$ instead of $large lim {a_{n+1}over a_n}$ therefore the result must be $ge 1$. There is no contradiction.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You have calculated $large lim {a_nover a_{n+1}}$ instead of $large lim {a_{n+1}over a_n}$ therefore the result must be $ge 1$. There is no contradiction.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have calculated $large lim {a_nover a_{n+1}}$ instead of $large lim {a_{n+1}over a_n}$ therefore the result must be $ge 1$. There is no contradiction.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have calculated $large lim {a_nover a_{n+1}}$ instead of $large lim {a_{n+1}over a_n}$ therefore the result must be $ge 1$. There is no contradiction.






        share|cite|improve this answer









        $endgroup$



        You have calculated $large lim {a_nover a_{n+1}}$ instead of $large lim {a_{n+1}over a_n}$ therefore the result must be $ge 1$. There is no contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 14:22









        Mostafa AyazMostafa Ayaz

        15.2k3939




        15.2k3939






























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