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Determine the convergence of the series (-1)n (1 – cos (1/n))

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n \left(1 - \cos \frac{1}{n} \right). \]

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


The series is absolutely convergent.

Proof. First, we have

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

Then, using the Taylor expansion of cosine we have

    \begin{align*}  1 - \cos \frac{1}{n} &= 1 - \left(1 - \frac{1}{2n^2} + o \left( \frac{1}{n^3} \right) \\[9pt]  &= \frac{1}{2n^2} - o \left( \frac{1}{n^3} \right) \\[9pt]  &\leq \frac{1}{2n^2}. \end{align*}

Since the series \sum \frac{1}{2n^2} converges, we have that

    \[ \sum_{n=1}^{\infty} \left( 1 - \cos \frac{1}{n} \right) \]

converges. Hence,

    \[ \sum_{n=1}^{\infty} (-1)^n \left( 1 - \cos \frac{1}{n} \right) \]

converges absolutely. \qquad \blacksquare

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