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?

 

[susskind_leon]

Keep in mind
\arctan(1/x)=sgn(x)\pi/2 - \arctan(x)

 

[Masaki]

Keep in mind
\arctan(1/x)=sgn(x)\pi/2 - \arctan(x)

Thanks for the reply! I think the problem now becomes manageable.
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}<br /> = \lim_{n\rightarrow\infty} \frac{a_n&#039;}{b_n&#039;}<br /> = \lim_{n\rightarrow\infty} \frac{ \frac{1}{1 + 1/(\log{n})^2}\cdot (1/\log{n})&#039; }{(1/\log{n})&#039;}<br /> = \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.