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 use the Leibniz rule. First, we have

since as . Furthermore, since is monotonically increasing, is monotonically decreasing. Thus, by the Leibniz rule the series converges.

This convergence is conditional since

Then, we use the limit comparison test with the series . We have

(Again, I’m pretending I can use L’Hopital’s on the integer valued functions, which I can’t. Technically we need to consider the equivalent real-valued functions, take the limit, and then return to the integer-valued functions.) Thus, diverges since diverges

I don’t understand why (π/2 – arctan (log n)) = log n in your proof of conditional convergence. I would prove conditional convergence as follows:

π/2 – arctan (log n) = arctan (1/log n)

Then I would use the limit comparison test and find that arctan (1/log n) / (1/logn) ➝ 1 as n ➝ ∞. Therefore the series arctan (1/log n) diverges if the series (1/ log n) diverges. The series (1/log n) diverges since (1/log n) > 1/n and 1/n diverges.

Thus, the series π/2 – arctan (log n) diverges.