Answer
converges to $\dfrac{1}{2} $
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]$
Here, $ \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]=\infty \cdot \infty$
This shows that the limit has an Indeterminate form so, we will use L-Hospital's rule.
This implies that $ \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}]= \lim\limits_{n \to \infty} [n-\sqrt {n^2-n}] \times \dfrac{[n+ \sqrt {n^2-n}]}{[n +\sqrt {n^2-n}]}$
and $\lim\limits_{n \to \infty} \dfrac{n}{n +\sqrt {n^2-n}} (\dfrac{\dfrac{1}{n}}{\dfrac{1}{n}})=\lim\limits_{n \to \infty} \dfrac{1}{1+\sqrt {(1-\dfrac{1}{n})}}=\dfrac{1}{2} $
Hence, {$a_n$} is Convergent and converges to $\dfrac{1}{2} $
