Consider the series
![\[ \sum_{n=1}^{\infty} (-1)^n \left( \frac{\pi}{2} - \arctan (\log n) \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-74f50f388013cad97c9cb3d211920a37_l3.png "Rendered by QuickLaTeX.com")
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
![\[ \lim_{n \to \infty} \left( \frac{\pi}{2} - \arctan (\log n) \right) = 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9b32f4c42f452c0945b1d2fe26494553_l3.png "Rendered by QuickLaTeX.com")
since 
as 
. Furthermore, since 
is monotonically increasing, 
is monotonically decreasing. Thus, by the Leibniz rule the series converges.
This convergence is conditional since
![\begin{align*} \sum_{n=1}^{\infty} \left| (-1)^n \left( \frac{\pi}{2} - \arctan (\log n) \right) \right| &= \sum_{n=1}^{\infty} \left( \frac{\pi}{2} - \arctan (\log n) \right) \\[9pt] &= \sum_{n=1}^{\infty} \arccot (\log n). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b666ea644adc5025deb9454da5c4555d_l3.png "Rendered by QuickLaTeX.com")
Then, we use the limit comparison test with the series 
. We have
![\begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\arccot(\log n)}{\frac{1}{\log n}} \\[9pt] &= \lim_{n \to \infty} \frac{\frac{-1}{n(1+\log^2n)}}{\frac{-1}{n \log^2 (n)}} \\[9pt] &= \lim_{n \to \infty} \frac{\log^2 n}{1 + \log^2 n} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0d1bfb19b5ac404fd71ababd3f04f456_l3.png "Rendered by QuickLaTeX.com")
(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.
I think rori wanted to write
pi/2 – arctan(log(x)) = arccot(log(x)).