Answer: For \(\alpha\) not an integer multiple of \(\pi\), we have \(\sin\alpha \neq 0\), so we may divide by \(2\sin\alpha\).

Using the identity

\(\cos[(2n-1)\alpha]-\cos[(2n+1)\alpha] \equiv 2\sin\alpha\sin(2n\alpha)\),

sum both sides from \(n=1\) to \(N\):

\(2\sin\alpha \sum_{n=1}^{N} \sin(2n\alpha) = \sum_{n=1}^{N} \big(\cos[(2n-1)\alpha]-\cos[(2n+1)\alpha]\big).\)

The right-hand side telescopes:

\(\big(\cos\alpha-\cos3\alpha\big)+\big(\cos3\alpha-\cos5\alpha\big)+\cdots+\big(\cos[(2N-1)\alpha]-\cos[(2N+1)\alpha]\big)\)

so

\(2\sin\alpha \sum_{n=1}^{N} \sin(2n\alpha)=\cos\alpha-\cos[(2N+1)\alpha].\)

Now divide by \(2\sin\alpha\):

\(\sum_{n=1}^{N} \sin(2n\alpha)=\frac{\cos\alpha-\cos[(2N+1)\alpha]}{2\sin\alpha}.\)

Using \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\) and \(\csc\alpha=\frac{1}{\sin\alpha}\), this becomes

\(\sum_{n=1}^{N} \sin(2n\alpha)=\frac12\cot\alpha-\frac12\csc\alpha\cos[(2N+1)\alpha].\)

To consider \(\sum_{n=1}^{\infty} \sin\left(\frac{2}{3}n\pi\right)\), take \(\alpha=\frac{\pi}{3}\). Then

\(\sin\left(\frac{2}{3}n\pi\right)=\sin(2n\alpha)\)

and the partial sum formula gives

\(\sum_{n=1}^{N} \sin\left(\frac{2}{3}n\pi\right)=\frac12\cot\frac{\pi}{3}-\frac12\csc\frac{\pi}{3}\cos\left(\frac{(2N+1)\pi}{3}\right).\)

Here \(\cos\left(\frac{(2N+1)\pi}{3}\right)\) takes the repeating values \(\frac12,-1,\frac12,\dots\), so the partial sums do not approach a single limit. Therefore the infinite series does not converge.

Consider the identity

\(\cos[(2n-1)\alpha]-\cos[(2n+1)\alpha] = 2\sin\alpha\sin(2n\alpha).\)

Sum from \(n=1\) to \(N\):

\(2\sin\alpha\sum_{n=1}^{N}\sin(2n\alpha)=\sum_{n=1}^{N}\left(\cos[(2n-1)\alpha]-\cos[(2n+1)\alpha]\right).\)

The sum on the right telescopes:

\(\left(\cos\alpha-\cos3\alpha\right)+\left(\cos3\alpha-\cos5\alpha\right)+\cdots+\left(\cos[(2N-1)\alpha]-\cos[(2N+1)\alpha]\right)\)

so all the intermediate terms cancel, leaving

\(2\sin\alpha\sum_{n=1}^{N}\sin(2n\alpha)=\cos\alpha-\cos[(2N+1)\alpha].\)

If \(\alpha\) is not an integer multiple of \(\pi\), then \(\sin\alpha\neq 0\), so dividing by \(2\sin\alpha\) gives

\(\sum_{n=1}^{N}\sin(2n\alpha)=\frac{\cos\alpha-\cos[(2N+1)\alpha]}{2\sin\alpha}.\)

Now write this as

\(\sum_{n=1}^{N}\sin(2n\alpha)=\frac12\frac{\cos\alpha}{\sin\alpha}-\frac12\frac{\cos[(2N+1)\alpha]}{\sin\alpha}\)

and hence

\(\sum_{n=1}^{N}\sin(2n\alpha)=\frac12\cot\alpha-\frac12\csc\alpha\cos[(2N+1)\alpha].\)

For the series \(\sum_{n=1}^{\infty}\sin\left(\frac{2}{3}n\pi\right)\), take \(\alpha=\frac{\pi}{3}\). Then the partial sums are

\(S_N=\frac12\cot\frac{\pi}{3}-\frac12\csc\frac{\pi}{3}\cos\left(\frac{(2N+1)\pi}{3}\right).\)

But \(\cos\left(\frac{(2N+1)\pi}{3}\right)\) is periodic and does not tend to a limit as \(N\to\infty\). Therefore the partial sums \(S_N\) do not tend to a limit either.

So the infinite series \(\sum_{n=1}^{\infty}\sin\left(\frac{2}{3}n\pi\right)\) does not converge.