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

$$\sum_{k=1}^{\infty} (-1)^k \frac{k}{1+2k^2}$$

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.

$$\lim_{x\to\infty} \frac{k}{1+2k^2} = \frac{1}{2k} \rightarrow 0$$ Therefore it converges.

But with the direct comparison test it is: $$ \mid (-1)^k \frac{k}{1+2k^2}\mid = \frac{k}{1+2k^2} =\frac{1}{\frac{1}{k}+2k}\geq \frac{1}{k+2k} = \frac{1}{3k}$$

Which is similar to the harmonic series$\sum_{k=1}^{\infty}\frac{1}{k}$ 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!

  • $\begingroup$ Yes, the series diverges absolutely. The alternating series test tells you only if it converges. You must decide if it converges conditionally or absolutely... $\endgroup$ – PhysicsMathsLove Sep 5 '18 at 10:25
  • 1
    $\begingroup$ You showed that the series converges, but it does not converge absolutely. $\endgroup$ – mechanodroid Sep 5 '18 at 10:25
  • $\begingroup$ Indeed you have already proved that the series is convergent but not absolutely convergent. $\endgroup$ – Rigel Sep 5 '18 at 10:26

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.