## Calculus (3rd Edition)

The sequence $a_n$ converges to $\pi$.
We have $$\lim_{n\to \infty}a_n=\lim_{n\to \infty}n \sin \frac{\pi}{n} \\ = \pi\lim_{n\to \infty}\frac{ \sin \frac{\pi}{n} }{\pi/n}= \pi\lim_{\pi/n\to 0}\frac{ \sin \frac{\pi}{n} }{\pi/n}= \pi.$$ Here, we used the fact that $\lim_{x\to 0}\frac{ \sin x}{x}=1$. Hence, by Theorem 1, the sequence $a_n$ converges to $\pi$.