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Apostol's Calculus Vol.1 10.20 #20

  • Thread starter Masaki
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  • #1
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Homework Statement


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

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,
[itex]f(x) = \displaystyle\left(\frac{\pi}{2}-\text{arctan}(\log{x})\right) - \frac{1}{x} > 0[/itex]​
for all [itex]x \geq 1[/itex], 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?
 

Answers and Replies

  • #2
Keep in mind
[itex]\arctan(1/x)=sgn(x)\pi/2 - \arctan(x) [/itex]
 
  • #3
8
0
Keep in mind
[itex]\arctan(1/x)=sgn(x)\pi/2 - \arctan(x) [/itex]
Thanks for the reply! I think the problem now becomes manageable.
Let [itex]a_n = \arctan(1/\log{n})[/itex] and [itex]b_n = 1/\log{n}[/itex]. Then, we have, with using l'Hôpital's rule,
[itex] \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.[/itex]​
Since [itex]\sum b_n = \sum (1/\log{n})[/itex] diverges, by the limit comparison test, [itex]\sum a_n[/itex] also diverges.
 

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