The series converges

Prove that the series

    \[\mathcal{S}  = \sum_{n=1}^{\infty} \frac{\sin ( \sin n)}{n}\]

converges. Examine if the convergence is absolute.

Solution

Lemma: Let \alpha \in \mathbb{R} such that \alpha / \pi \notin \mathbb{Q} then the sequence

    \[\omega_n = \sin ( \sin \alpha ) + \sin ( \sin ( 2 \alpha)) + \cdots + \sin ( \sin (n \alpha))\]

is bounded.

With the above lemma in mind the series converges as a consequence of Dirichlet’s theorem. The fact that the series does not converge absolutely can be seen by applying the Jordan inequality

(1)   \begin{equation*} \left | \sin a \right | \geq \frac{2 \left | a \right |}{\pi} \quad , \quad a \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \end{equation*}

that

\begin{aligned} \left | \frac{\sin \left ( \sin n \right )}{n} \right | + \left | \frac{\sin \left ( \sin \left ( n+1 \right ) \right )}{n+1} \right | &\geq \left | \frac{\sin \left ( \sin n \right )}{n+1} \right | + \left | \frac{\sin \left ( \sin \left ( n+1 \right ) \right )}{n+1} \right | \\ &\geq \frac{2}{\pi} \left | \frac{\sin n}{n+1} \right | + \frac{2}{\pi} \left | \frac{\sin (n+1)}{n+1} \right |\\ &\geq \frac{2}{\pi \left ( n+1 \right )} \left | \sin n \cos (n+1) - \cos n \sin (n+1) \right | \\ &= \frac{2}{\pi (n+1)} \left | \sin 1 \right | \end{aligned}

hence the series diverges absolutely.

 

 

 

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