Answer
Converges to $\dfrac{1}{2} $
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]$
Since, $ \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]=(\infty)(\infty)$
Need to apply L-Hospital's rule.
So, $\lim\limits_{n \to \infty} \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]=\lim\limits_{n \to \infty} \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}] \times \dfrac{[n+ \sqrt {n^2-n}]}{[n +\sqrt {n^2-n}]}$
or, $\lim\limits_{n \to \infty} \dfrac{n}{n +\sqrt {n^2-n}} (\dfrac{1/n}{1/n})=\lim\limits_{n \to \infty} \dfrac{1}{1+\sqrt {(1-1/n)}}=\dfrac{1}{2} $
Hence, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{2} $ and {$a_n$} is Convergent and converges to $\dfrac{1}{2} $
