Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges conditionally.

*Proof.* We can see that the series converges by the Leibniz rule since

(Since ). Furthermore, is monotonically decreasing since

is always negative.

Then, we can see that the convergence is conditional since

We use the limit comparison test with the series . We have

(Technically, to use L’Hopital’s and take this limit, I should look at the real-valued functions instead of the functions only taking values on the integers, and then say that this implies the limit of the integer-valued functions goes to 1.) By the limit comparison test we then know the two series either both converge or both diverge. Since diverges we have

diverges as well. Therefore, the convergence of the series in the question is conditional

You may enjoy the curious fact that the sum equals ArcTan(Exp(\pi/2)) – \pi/4

To prove conditional convergence, you can also state that arctan (1/(2n+1)) > 1/(2n+1) and then prove that the series of 1/(2n+1) is divergent using the limit comparison test