<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] By Taylor's theorem:
$$\begin{align}
\sin k\sin \frac 1k&=\frac 12\left[\cos\left(k-\frac1k\right)-\cos\left(k+\frac1k\right)\right]=\\
&=\frac 12\left[\cos k-\frac1k\sin\xi_k-\cos k+\frac1k\sin\xi'_k\right]=\\
&=\frac1{k}\sin\left(\frac{\xi_k-\xi'_k}2\right)\cos\left(\frac{\xi_k+\xi'_k}2\right)
\end{align}
$$
where $\xi_k,\xi'_k\in[k-\frac1k,k+\frac1k]$. Note that $|\xi_k-\xi'_k|\leq 2/k$
Since $|\sin x|\leq|x|$ and $|\cos x|\leq 1$, the absolute value of the latter expresion is bounded by $1/k^2$.
Nice problem :)
</s>