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Determine the convergence of the series (-1)n (π/2 – arctan (log n))

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n \left( \frac{\pi}{2} - \arctan (\log n) \right). \]

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

    \[ \lim_{n \to \infty} \left( \frac{\pi}{2} - \arctan (\log n) \right) = 0 \]

since \arctan (\log n) \to \frac{\pi}{2} as n \to \infty. Furthermore, since \arctan (\log n) is monotonically increasing, \frac{\pi}{2} - \arctan (\log n) 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*}

Then, we use the limit comparison test with the series \sum \frac{1}{\log n}. 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*}

(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, \sum \left( \frac{\pi}{2} - \arctan (\log n)\right) diverges since \sum \frac{1}{\log n} diverges. \qquad \blacksquare

One comment

  1. Eijiro Asada says:

    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.

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