Confusion if the series converges or not (alternating series test)

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I have to test for convergence and absolute convergence for the following series:



$$sum_k=1^infty (-1)^k frack1+2k^2$$



Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.



$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.



But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$



Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!






calculus convergence absolute-convergence







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  • Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
    – PhysicsMathsLove
    Sep 5 at 10:25







  • 1




    You showed that the series converges, but it does not converge absolutely.
    – mechanodroid
    Sep 5 at 10:25










  • Indeed you have already proved that the series is convergent but not absolutely convergent.
    – Rigel
    Sep 5 at 10:26














up vote
1
down vote

favorite












I have to test for convergence and absolute convergence for the following series:



$$sum_k=1^infty (-1)^k frack1+2k^2$$



Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.



$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.



But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$



Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!






calculus convergence absolute-convergence







share|cite|improve this question





















  • Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
    – PhysicsMathsLove
    Sep 5 at 10:25







  • 1




    You showed that the series converges, but it does not converge absolutely.
    – mechanodroid
    Sep 5 at 10:25










  • Indeed you have already proved that the series is convergent but not absolutely convergent.
    – Rigel
    Sep 5 at 10:26












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have to test for convergence and absolute convergence for the following series:



$$sum_k=1^infty (-1)^k frack1+2k^2$$



Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.



$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.



But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$



Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!






calculus convergence absolute-convergence







share|cite|improve this question













I have to test for convergence and absolute convergence for the following series:



$$sum_k=1^infty (-1)^k frack1+2k^2$$



Because of the alternating series test, I have to verify if the series decreases monotonically and then show that the limit goes to zero.
I don't have any problems showing that it decreases monotonically, but I have trouble showing if the limit is zero.



$$lim_xtoinfty frack1+2k^2 = frac12k rightarrow 0$$
Therefore it converges.



But with the direct comparison test it is:
$$ mid (-1)^k frack1+2k^2mid = frack1+2k^2 =frac1frac1k+2kgeq frac1k+2k = frac13k$$



Which is similar to the harmonic series$sum_k=1^inftyfrac1k$ hence the series should diverge.
So my question is, does it diverge or converge? And how do I know if it converges absolutely?
Thank you for your time and help!






calculus convergence absolute-convergence




calculus convergence absolute-convergence






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asked Sep 5 at 10:23









ina26

847




847











  • Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
    – PhysicsMathsLove
    Sep 5 at 10:25







  • 1




    You showed that the series converges, but it does not converge absolutely.
    – mechanodroid
    Sep 5 at 10:25










  • Indeed you have already proved that the series is convergent but not absolutely convergent.
    – Rigel
    Sep 5 at 10:26
















  • Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
    – PhysicsMathsLove
    Sep 5 at 10:25







  • 1




    You showed that the series converges, but it does not converge absolutely.
    – mechanodroid
    Sep 5 at 10:25










  • Indeed you have already proved that the series is convergent but not absolutely convergent.
    – Rigel
    Sep 5 at 10:26















Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25





Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely...
– PhysicsMathsLove
Sep 5 at 10:25





1




1




You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25




You showed that the series converges, but it does not converge absolutely.
– mechanodroid
Sep 5 at 10:25












Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26




Indeed you have already proved that the series is convergent but not absolutely convergent.
– Rigel
Sep 5 at 10:26










1 Answer
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You're doing nothing wrong.



Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.



Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    You're doing nothing wrong.



    Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.



    Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      You're doing nothing wrong.



      Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.



      Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        You're doing nothing wrong.



        Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.



        Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.






        share|cite|improve this answer












        You're doing nothing wrong.



        Since $|a_k|$ decreases monotonically, and $a_k to 0$, you can conclude that $sum a_k$ converges, by the alternating series test.



        Now, we say that a series $sum a_k$ coverges absolutely if $sum |a_k|$ converges. But as you show, $sum |a_k|$ doesn't converge, so we have a conditionally convergent series, one that converges, but not absolutely.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 5 at 10:36









        JuliusL33t

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