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

Consider the series

    \[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n}. \]

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^2}. Then, the series \sum b_n converges. Using the limit comparison test we have

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

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

    \[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n}. \qquad \blacksquare\]

One comment

  1. Anonymous says:

    The integral goes to infinite, not 1. lim n-> infinite of sin(1/n) / (1/n) goes to infinite because 1/n goes to 0 and since it is in the denominator the whole expression goes to infinite

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