Closed form for $\sum_{i = 1}^n i \cos(\beta i)$

Question

Is there a closed form solution to the following sum:

$$\sum_{i = 1}^n i \cos(\beta i)?$$

Thank you very much.

Answer 1

Hint: We have $$\sum_{k=1}^n k\cos(\beta k) = \Re\sum_{k=1}^nk\,(e^{i\beta})^k,$$ and $$\sum_{k=1}^nk\,z^k = z\,\frac d{dz}\sum_{k=1}^n z^k = z\,\frac d{dz}\Big(\frac{z(1-z^n)}{1-z}\Big).$$

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Using the hints above, you should arrive at $$\sum_{k=1}^nk\cos(\beta k) = \frac{(n+1)\cos(n\beta) - n\cos((n+1)\beta)-1}{4\sin^2(\beta/2)}.$$