Apostol's Calculus Vol.1 10.20 #20

Masaki

Homework Statement

Determine (absolute/conditional) convergence or divergence of the series
$\displaystyle\sum_{n=1}^{\infty} (-1)^n \left(\frac{\pi}{2}-\text{arctan}({\log{n}})\right)$​

The Attempt at a Solution

It is easy to see that the series is (conditionally) convergent by Leibniz's rule, but I am not sure how to deny the absolute convergence of the series. To use the comparison test, I must show, for example,
$f(x) = \displaystyle\left(\frac{\pi}{2}-\text{arctan}(\log{x})\right) - \frac{1}{x} > 0$​
for all $x \geq 1$, which seems quite difficult (in fact, I have no idea how to prove it). Is there any other, better way to show that the series is not absolutely convergent?

$\arctan(1/x)=sgn(x)\pi/2 - \arctan(x)$
$\arctan(1/x)=sgn(x)\pi/2 - \arctan(x)$
Let $a_n = \arctan(1/\log{n})$ and $b_n = 1/\log{n}$. Then, we have, with using l'Hôpital's rule,
$\displaystyle \lim_{n\rightarrow\infty} \frac{a_n}{b_n} = \lim_{n\rightarrow\infty} \frac{a_n'}{b_n'} = \lim_{n\rightarrow\infty} \frac{ \frac{1}{1 + 1/(\log{n})^2}\cdot (1/\log{n})' }{(1/\log{n})'} = \lim_{n\rightarrow\infty} \frac{1}{1 + 1/(\log{n})^2} = 1.$​
Since $\sum b_n = \sum (1/\log{n})$ diverges, by the limit comparison test, $\sum a_n$ also diverges.