Answer
Divergent
Work Step by Step
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.
