0
$\begingroup$

Is $\sum \cos(n \pi) \frac{n}{n^2+1}$ conditionally or absolutely convergent?

My Working

Clearly, $\sum \cos(n \pi) \frac{n}{n^2+1} = \sum (-1)^n \frac{n}{n^2+1}$, and it can be shown by using the alternating series test that this series converges. So it at least converges conditionally.

Now to test for absolute convergence, I need to test whether $\sum |(-1)^n \frac{n}{n^2+1}| = \sum \frac{n}{n^2+1}$ converges or not. I've got a gut feeling that this series is divergent, but I can't seem to be able to prove it. I've tried the comparison and ratio tests but to no avail.

$\endgroup$
3
  • 2
    $\begingroup$ Note $n/(n^2+1)\ge n/(n^2+n^2)=1/(2n)$. $\endgroup$
    – David Mitra
    Apr 19, 2014 at 11:49
  • $\begingroup$ @user, your first part is correct, and the second one follows from David's comment above...and your gut feeling. $\endgroup$
    – DonAntonio
    Apr 19, 2014 at 11:51
  • $\begingroup$ Hah so simple, can't believe I didn't see it. Thanks! $\endgroup$
    – user40333
    Apr 19, 2014 at 12:02

1 Answer 1

Reset to default
1
$\begingroup$

Simply notice that $\frac{n}{n^2+1}\sim_{+\infty}\frac{1}{n}$, hence the series doesn't converge absolutely.

$\endgroup$

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.