## University Calculus: Early Transcendentals (3rd Edition)

Sequence converges to $0$.
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=\sin n \pi$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\sin n \pi$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(0)$ $\lim\limits_{n \to \infty}a_n=0$ Therefore, the sequence converges to $0$.