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