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

Consider the series

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

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

The series converges absolutely.

Proof. We know from the previous exercise (Section 10.20, Exercise #30) that the series

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

converges absolutely. Using the limit comparison test we have

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

Now we consider these as functions of a real-variable and make the substitution n = \frac{1}{x}, and use L’Hopital’s rule three times to take the limit,

    \begin{align*}  \lim_{n \to \infty} \frac{\frac{1}{n} - \sin \frac{1}{n}}{\frac{1}{n^2} \sin \frac{1}{n}} &= \lim_{x \to 0} \frac{x - \sin x}{x^2 \sin x} \\[9pt]  &= \lim_{x \to 0} \frac{1 - \cos x}{2x \sin x + x^2 \cos x} \\[9pt]  &= \lim_{x \to 0} \frac{sin x}{(2+x^2) \sin x + 4x \cos x} \\[9pt]  &= \lim_{x \to 0} \frac{\cos x}{-2x \sin x + (6+x^2)\cos x} \\[9pt]  &= \frac{1}{6}. \end{align*}

Therefore, since \sum b_n converges absolutely we have established the absolute convergence of

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


  1. Eiji says:

    You could also solve it using Taylor’s Polynomial. 1 – n(sin (1/n)) = 1 – n(1/n – ((1/n)^3)/3! + …) > ((1/n)^2)/6. Since we know that ((1/n)^2)/6 converges, 1 – n(sin (1/n)) converges.

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