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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.

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    $\begingroup$ Note $n/(n^2+1)\ge n/(n^2+n^2)=1/(2n)$. $\endgroup$ 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

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

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