## Calculus (3rd Edition)

$$\sum_{i=1}^{n+1}\sin \left(\frac{\pi}{i}\right)$$
Given $$\sin (\pi)+\sin \left(\frac{\pi}{2}\right)+\sin \left(\frac{\pi}{3}\right)+\ldots+\sin \left(\frac{\pi}{n+1}\right)$$ Since the first term is $\sin (\pi)$ and the last is $\sin \left(\frac{\pi}{n+1}\right)$, then $$\sin (\pi)+\sin \left(\frac{\pi}{2}\right)+\sin \left(\frac{\pi}{3}\right)+\ldots+\sin \left(\frac{\pi}{n+1}\right)=\sum_{i=1}^{n+1}\sin \left(\frac{\pi}{i}\right)$$