Edit: I am seeking a solution that uses only calculus and real analysis methods — not complex analysis. This is an old advanced calculus exam question, and I think we are not allowed to use any complex analysis that could make the problem statement a triviality.
Show that the series
$$\sum_{n=2}^{\infty} \frac{\sin(n)}{\log(n)}$$
converges.
Any hints or suggestions are welcome.
Some thoughts:
The integral test is not applicable here, since the summands are not positive.
The Dirichlet test does seem applicable either, since if I let 1/log(n) be the decreasing sequence, then the series of sin(n) does not have bounded partial sums for every interval.
Thanks,
Best Answer
Note that
$$\left|\sum_{k=1}^n \sin k\right|= \frac{|\sin(n/2)\sin[(n+1)/2]|}{\sin(1/2)}\leqslant \frac{1}{\sin(1/2)}.$$
Derivation:
$$\begin{align} 2 \sin(1/2)\sum_{k=1}^n\sin k &= \sum_{k=1}^n2 \sin(1/2)\sin k \\ &= \sum_{k=1}^n2 \sin(1/2)\cos (k +\pi/2) \\ &= \sum_{k=1}^n[\sin(k + 1/2 + \pi/2)-\sin (k -1/2 + \pi/2)] \\ &= \sin(n + 1/2 + \pi/2) - \sin(1/2 + \pi/2) \\ &= 2 \sin(n/2)\cos[(n+1)/2 +\pi/2] \\ &= 2 \sin(n/2)\sin[(n+1)/2] \end{align} \\ \implies \sum_{k=1}^n\sin k = \frac{\sin(n/2)\sin[(n+1)/2]}{\sin(1/2)}$$