What should I use to find the convergence of the series $narctanfrac1n^3$?

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I thought of using comparison test and used the other series as $V_n = frac1n^2$ . Now using the limit comparsion rule and LH rule to evaluate limit $$fracarctan(frac1n^3)frac1n^3$$
I get the answer as 1 so it should converge as $frac1n^3$ converges (p series test) but the book says it is diverging. What am I doing wrong?




sequences-and-series limits divergent-series




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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 14 at 7:41














up vote
0
down vote

favorite












I thought of using comparison test and used the other series as $V_n = frac1n^2$ . Now using the limit comparsion rule and LH rule to evaluate limit $$fracarctan(frac1n^3)frac1n^3$$
I get the answer as 1 so it should converge as $frac1n^3$ converges (p series test) but the book says it is diverging. What am I doing wrong?




sequences-and-series limits divergent-series




share|cite|improve this question


















  • 1




    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 14 at 7:41












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I thought of using comparison test and used the other series as $V_n = frac1n^2$ . Now using the limit comparsion rule and LH rule to evaluate limit $$fracarctan(frac1n^3)frac1n^3$$
I get the answer as 1 so it should converge as $frac1n^3$ converges (p series test) but the book says it is diverging. What am I doing wrong?




sequences-and-series limits divergent-series




share|cite|improve this question














I thought of using comparison test and used the other series as $V_n = frac1n^2$ . Now using the limit comparsion rule and LH rule to evaluate limit $$fracarctan(frac1n^3)frac1n^3$$
I get the answer as 1 so it should converge as $frac1n^3$ converges (p series test) but the book says it is diverging. What am I doing wrong?





sequences-and-series limits divergent-series





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edited Aug 14 at 7:53









Jack D'Aurizio♦

271k31266632




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asked Aug 14 at 7:40









ANJANI KUMAR JHA

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  • 1




    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 14 at 7:41












  • 1




    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 14 at 7:41







1




1




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 14 at 7:41




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 14 at 7:41










2 Answers
2






active

oldest

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up vote
1
down vote



accepted










By
$$arctan xleq x$$
for $x>0$, then
$$sum_n=1^infty narctandfrac1n^3<sum_n=1^infty dfrac1n^2=dfracpi^26$$






share|cite|improve this answer



























    up vote
    0
    down vote













    Your method is O.K.:



    By LH we get $fracarctan xx to 1$ as $x to 0$, hence there is $c>0$ such that



    $0 < fracarctan xx le 2$ for $x in (0,c)$. This gives



    $ arctan(1/n^3) le 2/n^3$ for $n$ large.






    share|cite|improve this answer




















    • So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
      – ANJANI KUMAR JHA
      Aug 14 at 8:19











    • Yes, the book is wrong.
      – Fred
      Aug 14 at 8:49










    Your Answer




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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    By
    $$arctan xleq x$$
    for $x>0$, then
    $$sum_n=1^infty narctandfrac1n^3<sum_n=1^infty dfrac1n^2=dfracpi^26$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      By
      $$arctan xleq x$$
      for $x>0$, then
      $$sum_n=1^infty narctandfrac1n^3<sum_n=1^infty dfrac1n^2=dfracpi^26$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        By
        $$arctan xleq x$$
        for $x>0$, then
        $$sum_n=1^infty narctandfrac1n^3<sum_n=1^infty dfrac1n^2=dfracpi^26$$






        share|cite|improve this answer












        By
        $$arctan xleq x$$
        for $x>0$, then
        $$sum_n=1^infty narctandfrac1n^3<sum_n=1^infty dfrac1n^2=dfracpi^26$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 at 7:46









        Nosrati

        20.4k41644




        20.4k41644




















            up vote
            0
            down vote













            Your method is O.K.:



            By LH we get $fracarctan xx to 1$ as $x to 0$, hence there is $c>0$ such that



            $0 < fracarctan xx le 2$ for $x in (0,c)$. This gives



            $ arctan(1/n^3) le 2/n^3$ for $n$ large.






            share|cite|improve this answer




















            • So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
              – ANJANI KUMAR JHA
              Aug 14 at 8:19











            • Yes, the book is wrong.
              – Fred
              Aug 14 at 8:49














            up vote
            0
            down vote













            Your method is O.K.:



            By LH we get $fracarctan xx to 1$ as $x to 0$, hence there is $c>0$ such that



            $0 < fracarctan xx le 2$ for $x in (0,c)$. This gives



            $ arctan(1/n^3) le 2/n^3$ for $n$ large.






            share|cite|improve this answer




















            • So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
              – ANJANI KUMAR JHA
              Aug 14 at 8:19











            • Yes, the book is wrong.
              – Fred
              Aug 14 at 8:49












            up vote
            0
            down vote










            up vote
            0
            down vote









            Your method is O.K.:



            By LH we get $fracarctan xx to 1$ as $x to 0$, hence there is $c>0$ such that



            $0 < fracarctan xx le 2$ for $x in (0,c)$. This gives



            $ arctan(1/n^3) le 2/n^3$ for $n$ large.






            share|cite|improve this answer












            Your method is O.K.:



            By LH we get $fracarctan xx to 1$ as $x to 0$, hence there is $c>0$ such that



            $0 < fracarctan xx le 2$ for $x in (0,c)$. This gives



            $ arctan(1/n^3) le 2/n^3$ for $n$ large.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 14 at 7:54









            Fred

            37.9k1238




            37.9k1238











            • So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
              – ANJANI KUMAR JHA
              Aug 14 at 8:19











            • Yes, the book is wrong.
              – Fred
              Aug 14 at 8:49
















            • So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
              – ANJANI KUMAR JHA
              Aug 14 at 8:19











            • Yes, the book is wrong.
              – Fred
              Aug 14 at 8:49















            So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
            – ANJANI KUMAR JHA
            Aug 14 at 8:19





            So that means it is converging and the book is wrong about it being divergent, no? Thank you so much. Also, I used the expansion for arc tan function and it came out to be convergent that way as well.
            – ANJANI KUMAR JHA
            Aug 14 at 8:19













            Yes, the book is wrong.
            – Fred
            Aug 14 at 8:49




            Yes, the book is wrong.
            – Fred
            Aug 14 at 8:49












             

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