Riemann-Stieltjes sum

Question

Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \left[\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \right]$$

Answer 1

First, by trigonometric identities involving sums and products, we have $$\begin{align} &\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \\ &=4\cos\left(\frac{2\pi(k+1/2)}n \right)\cos\left(\frac{2\pi(k+1/2)}n\right)\cos\frac{\pi}n\sin\frac{\pi}n \\ &=2\cos^2\left(\frac{2\pi(k+1/2)}n\right)\sin\frac{2\pi}n.\\ &=\left( 2\cos^2\left(\frac{2\pi(k+1/2)}n\right)-1+1\right)\sin\frac{2\pi}n \\ &=\cos\left(\frac{4\pi(k+1/2)}n\right)\sin\frac{2\pi}n+\sin\frac{2\pi}n.\end{align} $$

By Euler's theorem, the sum over $k=1$ through $k=n$ becomes $$ \Re \left( \sum_{k=1}^n e^{ 4\pi i\frac{k+1/2}n } \right) \sin\frac{2\pi}n + n\sin\frac{2\pi }n. $$

By geometric sum formula, we have $$ \Re \left( e^{2\pi i /n} e^{4\pi i/n} \frac{ e^{4\pi i }-1}{ e^{4\pi i/n} -1} \right)\sin\frac{2\pi}n + n\sin\frac{2\pi}n. $$ We see that the quantity inside paranthesis is zero because $e^{4\pi i}=1$. Then as $n\rightarrow\infty$, we are left with $$ n\sin \frac{2\pi}n = \frac{\sin\frac{2\pi}n}{\frac1n}\rightarrow 2\pi. $$

Therefore, the limit is $2\pi$.