<s>[INST] Testing for convergence of a series Test the following series for convergence
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k}$$.
I am just wondering if the following method is ok:
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k} = \sum_{k=1}^\infty \sin (\frac{1}{k} - (\frac{1}{k} -k)) \sin \frac{1}{k} \\ \leq \sum_{k=1}^\infty (\sin \frac{1}{k} \cos(\frac{1}{k} -k) - \cos \frac{1}{k}\sin(\frac{1}{k}-1)) \sin \frac{1}{k} \\ \leq 
\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k} - \cos \frac{1}{k}\sin(\frac{1}{k}-1) \sin \frac{1}{k} 
\\\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k}
\\\sum_{k=1}^\infty \sin \frac{1}{k} \sin\frac{1}{k} 
\\\sum_{k=1}^\infty \frac{1}{k^2}$$ which converges.
Is there any more elegant proof? Help appreciated thanks. 
 [/INST] Use the fact that $\sum \frac{ \sin (n)}{n} $ converges. Hence, by limit comparison test,
$$ \frac{ \sin (n) \sin ( \frac{1}{n} ) }{\frac{ \sin (n)}{n}} = \frac{ \sin( \frac{1}{n} )}{\frac{1}{n}} \to 1 $$
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