How to find the sum upto n terms?

Question

Consider the series:

$$S(n) = \sin(\pi/3) + (1/2)\sin(2\pi /3) + (1/3)\sin(3\pi/3) +\cdots+ (1/n)\sin(n \pi/3)$$

How to find the sum and generalize it? I tried taking out $√3 /2$ as a common factor, but I failed.

Answer 1

Since the sine function is periodic with period $2\pi$ the value of $\sin \frac{2\pi n}{3}$ depends only on the residue class of $n$ modulo $6$. That is,$$ \sin \frac{2\pi n}{3}= \begin{cases} \frac{\sqrt 3}{2}, &n \equiv 1,2 \pmod{6} \\ \frac{-\sqrt 3}{2}, &n \equiv 4,5 \pmod{6} \\ 0, &n \equiv 0,3 \pmod{6} \end{cases} $$ That makes the sum equal to $$\frac{\sqrt 3}{2}\left (\frac{1}{1}+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+...+a_n \right),$$

Where the form of $a_n$ depends on the congruence class of $n.$ I don't see a way to get a compact formula for this. It's possible to compute the infinite sum, but I don't suppose you're up to that yet.