Home » Blog » Determine the convergence of the series (-1)n arctan (1 / (2n+1))

Determine the convergence of the series (-1)n arctan (1 / (2n+1))

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n \arctan \frac{1}{2n+1}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges conditionally.

Proof. We can see that the series converges by the Leibniz rule since

    \[ \lim_{n \to \infty} \arctan \frac{1}{2n+1} = 0. \]

(Since \lim_{ x \to 0} \arctan x = 0). Furthermore, \arctan \frac{1}{2n+1} is monotonically decreasing since

    \[ \left( \arctan \frac{1}{2x+1} \right)' = \frac{1}{1 + \left( \frac{1}{2x+1} \right)^2} = \frac{-1}{1+2x + 2x^2} = \frac{-1}{(1+x)^2 + x^2} \]

is always negative.

Then, we can see that the convergence is conditional since

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

We use the limit comparison test with the series \sum \frac{1}{2n}. We have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\arctan \left(\frac{1}{2n+1} \right)}{\frac{1}{2n}} \\[9pt] &= \lim_{n \to \infty} \frac{ \frac{-1}{2n^2+2n+1}}{\frac{-1}{2n^2}} &(\text{L'Hopital's}) \\[9pt] &= \lim_{n \to \infty} \frac{2n^2}{2n^2+2n+1} \\[9pt] &= 1. \end{align*}

(Technically, to use L’Hopital’s and take this limit, I should look at the real-valued functions instead of the functions only taking values on the integers, and then say that this implies the limit of the integer-valued functions goes to 1.) By the limit comparison test we then know the two series either both converge or both diverge. Since \sum \frac{1}{2n} diverges we have

    \[ \sum_{n =1}^{\infty} \arctan \left( \frac{1}{2n+1} \right) \]

diverges as well. Therefore, the convergence of the series in the question is conditional. \qquad \blacksquare

2 comments

  2. Van Gogh says:

    To prove conditional convergence, you can also state that arctan (1/(2n+1)) > 1/(2n+1) and then prove that the series of 1/(2n+1) is divergent using the limit comparison test

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):