Answer
$\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\sin (\dfrac{\pi}{2}+\dfrac{1}{n})$
or, $\lim\limits_{n \to \infty} \sin (\dfrac{\pi}{2}+\dfrac{1}{n})=\sin (\lim\limits_{n \to \infty}\dfrac{\pi}{2}+\lim\limits_{n \to \infty}\dfrac{1}{n})$
and, $\sin (\lim\limits_{n \to \infty}\dfrac{\pi}{2}+\lim\limits_{n \to \infty}\dfrac{1}{n})=\sin \dfrac{\pi}{2}==1$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent.
