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

Consider the series

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

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

The given series converges absolutely.

Proof. First, we have

    \[ n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 \]

for all positive integers n. (Since \frac{\sin x}{x} < 1 for x > 0.) Therefore,

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

Then, we write the Taylor expansion of \sin \frac{1}{n},

    \begin{align*}  \sin \frac{1}{n} &= \frac{1}{n} - \frac{1}{6n^3} + \cdots \\[9pt]  \implies n \sin \frac{1}{n} &= 1 - \frac{1}{6n^2} + \cdots \\[9pt]  &\geq 1 - \frac{1}{6n^2} \\[9pt]  &\geq 1 - \frac{1}{n^2}. \end{align*}


    \begin{align*}  \sum_{n=2}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right) &\leq \sum_{n=2}^{\infty} \log \left( \frac{1}{1 - \frac{1}{n^2}} \right) \\[9pt]  &= \sum_{n=2}^{\infty} \log \left( 1 - \frac{1}{n^2-1} \right). \end{align*}


    \[ \log \left( 1 - \frac{1}{n^2-1} \right) < \frac{1}{n^2-1} \]

for n \geq 2 since

    \begin{align*}   e^{\frac{1}{n^2-1}} &= 1 + \left( \frac{1}{n^2-1} \right) + \cdots > 1 + \frac{1}{n^2-1}\\[9pt]  \implies \frac{1}{n^2} &> \log \left( 1 + \frac{1}{n^2-1} \right). \end{align*}

Furthermore, \frac{1}{n^2-1} > \frac{1}{n^2}. Hence,

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

therefore, by the comparison test

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

converges. So, finally, we have

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

converges. This implies

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

is absolutely convergent. \qquad \blacksquare


  1. S says:

    Very nice. I think there is a typo in the solution here, and the convergence proved in exercise 10.9.3 seems more applicable in the solution here.

  2. Mohammad Azad says:

    This is ridiculous! Did Apostol really want us to go crazy or is there a better solution? Thank you for your time RoRi

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