Answer: $\mathbf D$
Explanation:
$a_n = \left(1-\frac{1}{\sqrt2}\right)\ldots\left(1-\frac{1}{\sqrt{n+1}}\right)$
Now,
$\because \left(1-\frac{1}{\sqrt{n+1}}\right)\leq\left(1-\frac{1}{n+1}\right)$
$\Rightarrow \left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{n+1}\right) = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}\ldots\frac{n}{n+1}$
$\Rightarrow a_n \leq\frac{1}{n+1}$
On substituting $n \to \infty \\\Rightarrow a_n = 0$
$\therefore \mathbf D$ is the correct answer.