Answer
Converges to $1$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})^n$
But $ \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{\ln (1-\dfrac{1}{n^2})}{1/n}=\dfrac{0}{0}$
Need to apply L-Hospital's rule.
So, $\lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{-2n}{n^2-1}=0$
Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$
Thus, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} e^{n \ln n (1-\dfrac{1}{n^2})}=e^0=1$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent and converges to $1$
