Answer
$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{n^2-2n+1}{n-1}$
and $ \lim\limits_{n \to \infty} \dfrac{n^2-2n+1}{n-1}= \lim\limits_{n \to \infty} \dfrac{(n-1)^2}{(n-1)}=\lim\limits_{n \to \infty} (n-1)=\infty$
Thus, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent