Answer
Converges to $-2$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}$
$\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}= \lim\limits_{n \to \infty} (\dfrac{1}{\sqrt {n^2-n}-\sqrt {n^2+n}}) \dfrac{\sqrt {n^2-n}-\sqrt {n^2+n}}{\sqrt {n^2-n}-\sqrt {n^2+n}}$
or, $\lim\limits_{n \to \infty} \dfrac{\sqrt {n^2-1}+\sqrt {n^2+n}}{-1-n}[\dfrac{1/n}{1/n}]=-2$
Hence, $\lim\limits_{n \to \infty} a_n=-2$ and {$a_n$} is Convergent and converges to $-2$
