University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac {\sin n\pi}{2}$ The given sequence diverges because: $\dfrac {\sin n\pi}{2}=0,1,0,-1,0,1,0,-1,0,1,0-1.....$so on This means the the given sequence does not have a finite value. Hence, the sequence is divergent.