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I think I can prove it by Abel's test, because $\sum\frac{\sin k}k$ is convergent (can be proved by Dirichlet's test) and $k\sin(1/k)$ is monotone and bounded. But to prove that $k\sin(1/k)$ is monotone, I need to use derivative, which my analysis course hasn't touched.

Is there any way to prove it by relying on simple inequalities, such as $\sin x\le x$, etc.?

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You can use Dirichlet's test here also. Take $b_n=\sin k$ and $a_n=\sin \frac{1}{k}$.

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oh right! thanks! – han Mar 14 '12 at 9:39

If you already know that the series $\sum_{k= 1}^{+\infty}\frac{\sin k}k$ is convergent then you are almost done. You can write $\sin\frac 1k=\frac 1k-\frac 1{6k^3}+\frac{\varepsilon(k)}{k^3}$ where $\lim_{k\to \infty}\varepsilon(k)=0$. The series $\sum_{k= 1}^{+\infty}\sin k-\frac 1{6k^3}+\frac{\varepsilon(k)}{k^3}$ is absolutely convergent hence convergent, and you are done.

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