Home » Blog » Determine the convergence of the series (-1)n(1 – n sin (1/n))

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

by RoRi

Consider the series

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

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

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

since

    \[ 1 - n \sin \frac{1}{n} = 1 - \frac{\sin \frac{1}{n}}{\frac{1}{n}} > 0 \qquad \text{for all } n. \]

Furthermore, using the Taylor series expansion of sine we have

    \begin{align*} 1 - n \sin \frac{1}{n} &= 1 - n \left( \frac{1}{n} - \frac{1}{6n^3} + \cdots \right) \\[9pt] &= 1 - 1 + \frac{1}{6n^2} - \cdots \\[9pt] &< \frac{1}{6n^2}. \end{align*}

Since \sum \frac{1}{6n^2} converges, by the comparison test we then have that

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

converges; hence,

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

converges absolutely. \qquad \blacksquare

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