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Apostol's Calculus Vol.1 10.20 #20

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    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]​

    3. 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?
     
  2. jcsd
  3. Sep 16, 2012 #2
    Keep in mind
    [itex]\arctan(1/x)=sgn(x)\pi/2 - \arctan(x) [/itex]
     
  4. Sep 16, 2012 #3
    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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