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Determine the convergence of the series (sin (1/n))3/2

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} \right)^{\frac{3}{2}}. \]

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

The series converges.

Proof. Let b_n = \frac{1}{n^{\frac{3}{2}}}. Then, the series \sum b_n converges since it is of the form \sum \frac{1}{n^s} for s > 1. Using the limit comparison test we have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\left( \sin \frac{1}{n} \right)^{\frac{3}{2}}}{\left( \frac{1}{n} \right)^{\frac{3}{2}}} \\[9pt] &= \lim_{n \to \infty} \left( \frac{\sin \frac{1}{n}}{\frac{1}{n}} \right)^{\frac{3}{2}} \\[9pt] &= 1. \end{align*}

Hence, the series \sum b_n and \sum a_n both converge or both diverge. Since \sum b_n converges we have establish the convergence of

    \[ \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} \right)^{\frac{3}{2}}. \qquad \blacksquare\]

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