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

$$\sum_{n=2}^{\infty} \frac{\sin(n)}{\log(n)}$$


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.


  • 2
    $\begingroup$ How do you know that it converges? It doesn't seem like it would converge in my mind. It definitely doesn't converge absolutely. $\endgroup$ – cheesyfluff Nov 24 '15 at 4:21
  • $\begingroup$ Hi @cheesyfluff - the problem statement asks to show that it converges. But you're right in some sense, I have worked on problems that have asked, "find all solutions to...", and the answer was indeed that there were no solutions ... $\endgroup$ – User001 Nov 24 '15 at 4:23
  • $\begingroup$ You have 3-4 positive terms followed by 3-4 negative terms and the number of positive and negative terms averages out. So you sequence is like an alternating conditionally convergent series - converges conditionallu. $\endgroup$ – A.S. Nov 24 '15 at 4:46

Note that

$$\left|\sum_{k=1}^n \sin k\right|= \frac{|\sin(n/2)\sin[(n+1)/2]|}{\sin(1/2)}\leqslant \frac{1}{\sin(1/2)}.$$


$$\begin{align} 2 \sin(1/2)\sum_{k=1}^n\sin k &= \sum_{k=1}^n2 \sin(1/2)\sin k \\ &= \sum_{k=1}^n2 \sin(1/2)\cos (k +\pi/2) \\ &= \sum_{k=1}^n[\sin(k + 1/2 + \pi/2)-\sin (k -1/2 + \pi/2)] \\ &= \sin(n + 1/2 + \pi/2) - \sin(1/2 + \pi/2) \\ &= 2 \sin(n/2)\cos[(n+1)/2 +\pi/2] \\ &= 2 \sin(n/2)\sin[(n+1)/2] \end{align} \\ \implies \sum_{k=1}^n\sin k = \frac{\sin(n/2)\sin[(n+1)/2]}{\sin(1/2)}$$

  • 6
    $\begingroup$ @ASKASK One can derive that by noting that $\sin(k)$ is the sum of two geometric series: $$\sin(k)=\frac{1}2\left[(e^i)^k-(e^{-i})^k\right]$$ and then using standard formulas for sum over geometric series. $\endgroup$ – Milo Brandt Nov 24 '15 at 4:19
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    $\begingroup$ @Lebron James: The Dirichlet test is applicable for convergence of the series You can also derive that equality without complex variables using trig identities and a telescoping sum. $\endgroup$ – RRL Nov 24 '15 at 4:37
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    $\begingroup$ @LebronJames One could prove the identity mentioned here by induction and trigonometric identities. I wouldn't recommend it, but you can. You can also prove it by considering the sum of a bunch of unit vectors on the plane separated by a constant angle (for this, consider the sum minus the sum rotated by this angle). Also, I would suspect that if you're asked to avoid complex analysis, that's more of a "Don't use contour integrals to solve this" than a "Don't think about complex numbers at all" $\endgroup$ – Milo Brandt Nov 24 '15 at 4:39
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    $\begingroup$ @LebronJames Good to see you're still here during the NBA season. To expand on the last comment by RRL, simply note $$\sin (1)\sin (n)=\frac12\left(\cos (n-1)-\cos (n+1)\right)$$Then, telescope. $\endgroup$ – Mark Viola Nov 24 '15 at 5:10
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    $\begingroup$ Thanks so much for your answer and derivation @RRL :-) $\endgroup$ – User001 Nov 24 '15 at 23:12

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