Answer
Converges to $0$
Work Step by Step
We know that $-1\leq sinx \leq 1$
Therefore, we can write
$\frac{-1}{1+\sqrt n}\leq \frac{sin2n}{1+\sqrt n}\leq \frac{1}{1+\sqrt n}$
Also note that
$\lim\limits_{n \to \infty} \frac{-1}{1+\sqrt n}=\frac{-1}{\infty}=0$
and
$\lim\limits_{n \to \infty} \frac{1}{1+\sqrt n}=\frac{1}{\infty}=0$
Therefore, by sandwich theorem $\lim\limits_{n \to \infty} \frac{sin2n}{1+\sqrt n}=0$
Hence, the given sequence converges to $0$.
