Answer
Converges to $\dfrac{1}{2}$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{n^2 \sin (1/n)}{2n-1}$
But $ \lim\limits_{n \to \infty} \dfrac{n^2 \sin (1/n)}{2n-1}=\dfrac{0}{0}$
Need to apply L-Hospital's rule.
So, $ \lim\limits_{n \to \infty} \dfrac{-\cos (1/n) \cdot (1/n^2)}{-2n/n^2+2/n^3}=\lim\limits_{n \to \infty} \dfrac{-\cos (1/n)}{-2+\dfrac{2}{n}}$
or, $=\dfrac{1}{2}$
Hence, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is Convergent and converges to $\dfrac{1}{2}$
